Applications of differentiability.

Extrema of functions
Lecture 10

Mathematics - 1st year, English

Faculty of Computer Science, UAIC

e-mail: corina.forascu@gmail.com

Corina Forăscu

December 6, 2022

Corina Forăscu (FII) Lecture 10 December 6, 2022

Outline of the lecture

1 Optimization problems in Euclidean spaces

2 Unrestrained extrema

3 Restricted extrema

Corina Forăscu (FII) Lecture 10 December 6, 2022

Optimization problems in Euclidean spaces

One of the possible applications of differentiability: establishing extremal

points for functions involved in optimization problems.
More precisely, these problems are targeting the minimization (or the
maximization) of a so-called cost functional (profit function), in presence or
absence of some restrictions.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Example 1. Least squares method

Suppose that we are experimenting on a certain physical quantity:

“input” data: a1 , a2 , . . . , ap ;
we obtain the “output” values b1 , b2 , . . . , bp ;
we represent the points (ak , bk ) (k = 1, p) in the Euclidean plane;
we can (visually) estimate the nature of the unknown function ϕ satisfying
ϕ(ak ) = bk .
More precisely, we estimate that the function ϕ has a certain type
(polynomial, exponential, trigonometric, etc.), whose characteristics (real
parameters) c1 , c2 , . . . , cn have to be identified.

Corina Forăscu (FII) Lecture 10 December 6, 2022

For this purpose we use the method of least squares, by considering the
problem of the minimization of the expression
p
∑ ( ϕ(ak ; c1 , c2 , . . . , cn ) − bk )2 ,
k =1
with respect to (c1 , c2 , . . . , cn ) ∈ ’n .
Solving this problem (of unrestrained extrema) means finding
(c10 , c20 , . . . , cn0 ) ∈ ’n such that
( )
p p  2
min ∑ ( ϕ(ak ; c1 , c2 , . . . , cn ) − bk ) 2
= ∑ ϕ(ak ; c10 , c20 , . . . , cn0 ) − bk .
k =1 k =1
We can conclude that the physical quantity under study has the law
y = ϕ(x; c10 , c20 , . . . , cn0 ).
We remark that if the graphic of {(ak , bk )|k = 1, p } suggests us that ϕ is
linear, then we can take n = 2 and ϕ(x ) := c1 x + c2 .
In this case, the method of least squares leads us to the problem of
minimization of the expression
p
∑ (c1 ak + c2 − bk )2 .
k =1
Corina Forăscu (FII) Lecture 10 December 6, 2022
Example 2. Realization of a maximum profit/minimum cost

In an economic theory, the space ’n is interpreted as the space of consumer

goods output:
each consumer good is characterized by a certain index i ∈ {1, 2, . . . , n };
the consumer goods output is a vector x = (x1 , x2 , . . . , xn ) ∈ ’n , where the
component xi is the quantity in which the consumer good i is produced.
In such a context, a system of prices is characterized by a function which
associates to each consumer goods output a certain value.
It is naturally considered that such function is a linear function, hence
characterized by a vector p = (p1 , p2 , . . . , pn ) ∈ ’n , where pi is the
“unitary” price of the consumer good i; the system of prices is therefore
n
given by ∑ pi xi .
i =1

Corina Forăscu (FII) Lecture 10 December 6, 2022

When a production company is producing certain consumer goods, its
interest is to realize the corresponding production such that the production
costs to be minimal and/or the production profit to be maximal.
A cost function and/or a profit function convenably chosen, named objective
functions, are used by economical theorists.
In the situation that the set of consumer goods output is ’n , we deal with an
unrestained extrema problem.
If the set of consumer goods output is K ’n then we deal with a restricted
extrema problem.
In both cases, it is a linear programming problem if the objective function is
linear and K is a linear subspace of ’n .
If the objective function is quadratic or convex, then the respective problem is
called quadratic, respectively convex optimization problem.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Example 3. Maximal informational entropy

The entropy was introduced as a mathematical quantity by Claude E.

Shannon (1947);
it is a function corresponding to the quantity of information offered by a
certain source, by the means of a certain language, electrical signal or data
file.
This function, denoted by H is defined on the set of random variables
 
1 2 ... n
X =
p1 p2 . . . pn

and has the expression

n
H (X ) = − ∑ pk · log2 pk ,
k =1

where pk is the probability (pk ∈ (0, 1), k = 1, n and ∑nk =1 pk = 1) that a

particular message k is actually transmitted by the above source.
Corina Forăscu (FII) Lecture 10 December 6, 2022
A particular problem is to find an optimal distribution of the random variable
X such that the value H (X ) is maximal.
In other words, we study the problem of maximizing
n
− ∑ pk log2 pk
k =1

with respect to (p1 , p2 , . . . , pn ) subject to the restriction ∑nk =1 pk = 1,

pk ∈ (0, 1), k = 1, n.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Definition
Let ∅ 6= A ⊆ ’n , f : A → ’ a function and x0 ∈ A. We say that x0 is:
• a minimum point of f if f (x) ≥ f (x0 ), ∀x ∈ A;
• a maximum point of f if f (x) ≤ f (x0 ), ∀x ∈ A;
• a strict minimum point of f if f (x) > f (x0 ), ∀x ∈ A r {x0 };
• a strict maximum point of f if f (x) < f (x0 ), ∀x ∈ A r {x0 };
• a local minimum point of f if there exists V ∈ V (x0 ) such that
f (x) ≥ f (x0 ), ∀x ∈ A ∩ V ;
• a local maximum point of f if there exists V ∈ V (x0 ) such that
f (x) ≤ f (x0 ), ∀x ∈ A ∩ V ;
• a local strict minimum point of f if there exists V ∈ V (x0 ) such that
f (x) > f (x0 ), ∀x ∈ A ∩ (V r {x0 });
• a local strict maximum point of f if there exists V ∈ V (x0 ) such that
f (x) < f (x0 ), ∀x ∈ A ∩ (V r {x0 });
• an extremum (strict extremum, local extremum, local strict extremum) point
of f if x0 is a minimum (strict minimum, local minimum, respectively a local
strict minimum) point or a maximum (strict maximum, local maximum,
respectively a local strict maximum) point of f .
Corina Forăscu (FII) Lecture 10 December 6, 2022
Remarks. Any global extremum point for a function f is always a local extremum
point for f . The converse is not true.
1. If D ⊆ ’n is an open set and f : D → ’ is a function, the problem of
determining the global and/or local extremum points (and the associated extreme
values) for f is called an unrestricted extremum problem.
2. Let D ⊆ ’n be an open set, f : D → ’ a function and A ⊆ ’n a set of
constraints. The problem of determining the global and/or local extremum points
(and the associated extreme values) for f |A∩D is called a restricted extremum
problem.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Fermat theorem
Theorem (Fermat)
Let A ⊆ ’n be a non-empty set, f : A → ’ a function and x0 ∈ Å. If x0 is a
local extremum point of f and f has partial derivatives (of first order) in x0 , then

∂f ∂f
(x0 ) = · · · = (x0 ) = 0.
∂x1 ∂xn

Remark. The conclusion of the theorem can be written as (∇f ) (x0 ) = 0’n .
If, moreover f is Fréchet differentiable in x0 , this is also equivalent to

df (x0 ) = 0L(’n ;’) .

Definition
Let A ⊆ ’n , x0 ∈ Å and f : A → ’ a function, Fréchet differentiable in x0 . The
element x0 is called a critical point (or stationary point) of f if df (x0 ) = 0L(’n ;’)
(i.e. (∇f ) (x0 ) = 0’n ).
Corina Forăscu (FII) Lecture 10 December 6, 2022
Remark. Fermat theorem states that if x0 ∈ Å is a local extremum point of a
Fréchet differentiable function f in x0 , then x0 is critical point of f . The converse
is not true, as we can see from the following example.
Example. Let f : ’2 → ’,

f (x, y ) = x 2 − y 2 , x, y ∈ ’.

∂f
Then (0, 0) is a critical point of f , since (0, 0) = 2x |x =0,y =0 = 0 and
∂x
∂f
(0, 0) = −2y |x =0,y =0 = 0.
∂y
On the other hand, f (x, 0) = x 2 > f (0, 0) = 0 > −y 2 = f (0, y ) for each
x 6= 0 and each y 6= 0. This implies that (0, 0) is not a local minimum point,
nor a local maximum point for f .

Corina Forăscu (FII) Lecture 10 December 6, 2022

Definition
Let A ⊆ ’n , x0 ∈ Å and f : A → ’ a function, Fréchet differentiable in x0 . If x0
is a critical point of f , but is not a local extremum for f , we say that x0 is a
saddle point of f .

Corina Forăscu (FII) Lecture 10 December 6, 2022

Sufficient conditions of extremum
Case n=1

Theorem
Let A ⊆ ’ be an interval, x0 ∈ Å and f : A → ’ a function n-times (n ≥ 2)
derivable in x0 . Suppose that f 0 (x0 ) = f 00 (x0 ) = . . . = f (n−1) (x0 ) = 0 and
f (n) (x0 ) 6= 0.
i) If n is even, then x0 is a local extremum point of f . More precisely, if
f (n) (x0 ) > 0 then x0 is a local minimum point of f , while if f (n) (x0 ) < 0
then x0 is a local maximum point of f .
ii) If n is odd, then x0 is not a local extremum point of f .

Corina Forăscu (FII) Lecture 10 December 6, 2022

Case n>1
We will characterize extremum points of a function f in x0 according to the
Fréchet derivative of order 2 in x0 . Recall that
n
  ∂2 f
d2 f (x0 ) (v) = ∑ ∂x i ∂x j
(x0 )vi vj , ∀v = (v1 , v2 , . . . , vn ) ∈ ’n .
i ,j = 1

Theorem
Let A ⊆ ’n a non-empty set and f : A → ’ a function which is C 2 on a
neighbourhood of a critical point x0 ∈ Å.
i) If d2 f (x0 ) (v) > 0, ∀v ∈ ’n r {0’n }, then x0 is a local minimum point

for f .
ii) If d2 f (x0 ) (v) < 0, ∀v ∈ ’n r {0’n }, then x0 is a local maximum point

for f .
iii) If d2 f (x0
) is an undefined quadratic

form (i.e. ∃v0 , v00 ∈ ’n such that
2 0 2 00
d f (x0 ) (v ) < 0 and d f (x0 ) (v ) > 0), then x0 is a saddle point for f .

Corina Forăscu (FII) Lecture 10 December 6, 2022

Remarks.
1. If d2 f (
x0 ) is a positive or a negative semidefined quadratic form (i.e.,
d2 f (x0 ) (v) ≥ 0, ∀v ∈ ’n or, respectively, d2 f (x0 ) (v) ≤ 0, ∀v ∈ ’n ), then

we cannot determine the nature of the critical point.

∂2 f ∂2 f
2. By the hypotheses of the above result, we have that (x0 ) = (x0 ),
∂xi ∂xj ∂xj ∂xi
∀i 6= j, by Schwarz theorem, so d2 f (x0 ) is indeed a quadratic form (i.e., coming
from a symmetric bilinear form). We call the matrix associated to this quadratic
form (relative to the canonical basis) the Hessian of f :

∂2 f
 
Hf ( x 0 ) : = (x0 ) .
∂xi ∂xj 1 ≤ i ,j ≤ n

Corina Forăscu (FII) Lecture 10 December 6, 2022

By Lecture 7, we have several methods for determining if Hf (x0 ) is positive
definite, negative definite or neither.
Proposition
Let ∅ 6= A ⊆ ’n and f : A → ’ a function which is C 2 on a neighbourhood of a
critical point x0 ∈ Å.
i) If all the eigenvalues of Hf (x0 ) are positive, then x0 is a local minimum point
for f .
ii) If all the eigenvalues of Hf (x0 ) are negative, then x0 is a local maximum
point for f .
iii) If Hf (x0 ) has at least one positive and one negative eigenvalue, then x0 is a
saddle point for f .

Of course, we cannot say anything if all the eigenvalues are either all non-negative
or all non-positive.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Proposition
Let ∅ 6= A ⊆ ’n and f : A → ’ a function having partial derivatives of order 2
which are continuous in a neighbourhood of a critical point x0 of f . Let
2f
∆k = det ∂x∂ ∂x (x0 ) , k = 1, n, the principal minors of the Hessian
i j 1 ≤ i ,j ≤ k
Hf ( x 0 ) .
i) If ∆k > 0, ∀k = 1, n, then x0 is a local minimum point for f .
ii) If (−1)k +1 ∆k < 0, ∀k = 1, n, then x0 is a local maximum point for f .
iii) If there exist j, k = 1, n such that ∆j < 0 and (−1)k +1 ∆k > 0, then x0 is a
saddle point for f .

Again, we cannot conclude anything by just knowing that ∆k ≥ 0, ∀k = 1, n or

(−1)k +1 ∆k ≤ 0, ∀k = 1, n.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Case n=2

Remark. In the particular case n = 2, the above result tells us that if

f : A ⊆ ’2 → ’ is C 2 on a neighbourhood of a critical point x0 ∈ Å and we
∂2 f ∂2 f ∂2 f
denote p := 2 (x0 ), q := (x0 ), r := 2 (x0 ), then:
∂x ∂x ∂y ∂y
i) when p > 0 and pr − q 2 > 0, x0 is a local minimum point for f ;
ii) when p < 0 and pr − q 2 > 0, x0 is a local maximum point for f ;
iii) when pr − q 2 < 0, x0 is a saddle point for f ;
iv) when pr − q 2 = 0, we cannot establish the nature of x0 by this method.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Method of least squares revisited
Coming back to the method of least squares, with ϕ being linear, we see that the
function f : ’2 → ’, defined by
`
f ( c1 , c2 ) = ∑ (c1 ak + c2 − bk )2
k =1

C2 ’2 and has as critical points the elements c10 , c20 ∈ ’2

is of class on
satisfying
`

∂f
c10 , c20 = 2 ∑ (c10 ak + c20 − bk )ak = 0;




∂c1 k =1
∂f `
c , c = 2 ∑ (c10 ak + c20 − bk ) = 0.
0 0



∂c2 1 2

k =1
In order to solve this equation, equivalent to the linear system
` ` `

 c10 ∑ ak2 + c20 ∑ ak = ∑ bk ak ;


k =1 k =1 k =1
` `
 c10 ∑ ak2 + `c20 = ∑ bk ,


k =1 k =1

Corina Forăscu (FII) Lecture 10 December 6, 2022

we compute its determinant
` `
∑ ak2 ∑ ak
!2
` `
k =1
`
k =1 =` ∑ ak2 − ∑ ak ,
∑ ak ` k =1 k =1
k =1

which is non-zero when not all a1 , a2 , . . . , a` are equal, by

Cauchy-Buniakowski-Schwarz inequality.
In this case, we have
  
` ∑k` =1 ak bk − ∑k` =1 ak ∑k` =1 bk
c10 =  2
` ∑k` =1 ak2 − ∑k` =1 ak
and      
∑k` =1 bk ∑k` =1 ak2 − ∑k` =1 ak ∑k` =1 ak bk
c20 =  2 .
` ∑k` =1 ak2 − ∑k` =1 ak

Corina Forăscu (FII) Lecture 10 December 6, 2022

We then have
∂2 f `
p = 2 c10 , c20 = 2 ∑ ak2 ,


∂c1 k =1
∂2 f `
c10 , c20 = 2 ∑ ak and


q=
∂c1 ∂c2 k =1
2
∂ f
r = 2 c10 , c20 = 2`.


∂c2
Again by Cauchy-Buniakowski-Schwarz inequality, we have p > 0 and
pr − q 2 > 0, i.e. c10 , c20 is a global minimum point for f (not only a local

minimum point, because c10 , c20 is the unique critical point of f ).

Corina Forăscu (FII) Lecture 10 December 6, 2022

Restricted extrema

Besides the situations we have to solve problems of extremum without constraints,

there are also problems where we have to find extrema of functions constrained to
some conditions. In this case, the respective extrema are called restricted extrema.
Setting. Let D be an open set (with respect to the Euclidean metric) in
’n × ’m (n, m ∈ Ž∗ ) and f : D → ’, g : D → ’m functions of class C 1 on D.
We look for the extremum points of f restricted to the supplimentary
condition g (x, y) = 0, where (x, y) ∈ D.
by denoting A := {(x, y) ∈ D |g (x, y) = 0’m } and by g1 , g2 , . . . , gm the
components of g , we see that the above problem demands the finding of
extremum points of f |A .
Terminology: we say that a point (x0 , y0 ) ∈ A is a local extremum
(minimum, maximum) or saddle point of f , conditioned to the restrictions
g1 (x, y) = 0, g2 (x, y) = 0, . . ., gm (x, y) = 0, if it is a local extremum
(minimum, maximum) or saddle point of f |A .

Corina Forăscu (FII) Lecture 10 December 6, 2022

The above problem may look as an unrestrained extrema one, but the
extremum points (x0 , y0 ) of f |A will be elements of Å if and only if
g1 = g2 = · · · = gm = 0 on a neighbourhood of (x0 , y0 ), which in fact
means that we would not effectively deal with a restricted extrema problem.
Hence in general we cannot apply the results from the previous section.
If D is bounded, then A is bounded, too; thus, since g is continuous, A is a
closed, bounded set, hence compact.
By the continuity of f , f |A will surely have a minimum and a maximum point
in A. Therefore, the problem remains to determine these extremum points.
In order to approach this problem, we could try to solve first the system
gk (x, y) = 0, ∀ k = 1, m, with respect to y. Finding a solution y = ϕ(x),
x ∈ B, would lead to studying the unrestrained extrema of the function
x 7→ f (x, ϕ(x)), with x ∈ B. The procedure is, however, difficult to apply in
practice, because the equation g (x, y) = 0’n is not easily solved.
This is why we will follow the method of Lagrange multipliers, described in
the sequel.

Corina Forăscu (FII) Lecture 10 December 6, 2022

Method of Lagrange multipliers
Theorem (existence of Lagrange multipliers)
Let ∅ 6= D ⊆ ’n × ’m an open set, f : D → ’ and g : D → ’m C 1 -functions
on D. Let g1 , g2 , . . . , gm be the components of g and (x0 , y0 ) ∈ D a local
extremum point of f , conditioned to the restrictions g1 (x, y) = 0, g2 (x, y) = 0,
D (g ,g ,...,g )
. . ., gm (x, y) = 0. If D (y1 ,y2 ,...,y m) (x0 , y0 ) 6= 0, then there exist
1 2 m
λ1 , λ2 , . . . , λm ∈ ’ such that (x0 , y0 ) is a critical point for the C 1 -function
L : D → ’ defined by

L(x, y) = f (x, y) + λ1 g1 (x, y) + λ2 g2 (x, y) + · · · + λm gm (x, y), (x, y) ∈ D,

i.e. 
∂L
(x0 , y0 ) = 0, ∀k = 1, n;





 ∂x k

(?) ∂L
(x0 , y0 ) = 0, ∀j = 1, m;



 ∂y j


 g (x , y ) = 0, ∀j = 1, m.
j 0 0
Corina Forăscu (FII) Lecture 10 December 6, 2022
Remarks.
1. The numbers λ1 , λ2 , . . . , λm are called Lagrange multipliers.
2. The conclusion of the above result can be rephrased as follows: the
conditioned local extremum points of f are critical points of the associated
function L = f + ∑m k = 1 λ k gk .
3. The function L : ’m × D → ’ defined by

L (λ, x, y) = f (x, y) + λ1 g1 (x, y) + λ2 g2 (x, y) + · · · + λm gm (x, y),

for λ = (λ1 , . . . , λm ) ∈ ’m , is called the Lagrangian associated to f and the

restriction g .
Then the triple (λ0 , x0 , y0 ) satisfies the system (?) (i.e., (x0 , y0 ) is a critical
point for L, satisfying the restriction g (x0 , y0 ) = 0’m ) if and only if (λ0 , x0 , y0 ) is
a critical point for L , since

∂L
(λ0 , x0 , y0 ) = gj (x0 , y0 ), j = 1, m.
∂λj

Corina Forăscu (FII) Lecture 10 December 6, 2022

4. Suppose now that (λ0 , x0 , y0 ) satisfies the system (?) and f , g are functions
of class C 2 on D (or at least on a neighbourhood of (λ0 , x0 , y0 )). Since
L(x, y) − L(x0 , y0 ) = f (x, y) − f (x0 , y0 ), ∀(x, y) ∈ A,
we see that (x0 , y0 ) is a conditioned local extremum point of f if and only if
(x0 , y0 ) is a local extremum point for L|A .
A practical way to determine sufficient conditions of conditioned local extremality
is the following:
Let us denote x = (x̃1 , . . . , x̃n ) ∈ ’n and y = (x̃n+1 , . . . , x̃n+m ) ∈ ’m ; the
second order Fréchet differential of L in (x0 , y0 ) is
n +m
∂2 L
(??) ∑ ∂x̃ i ∂x̃ j
(x0 , y0 )dx̃i dx̃j .
i ,j = 1

By considering the system obtained by differentiating the relations

gk (x, y) = 0, k = 1, m, i.e.
n +m
∂gk
∑ ∂x̃ j
(x0 , y0 )dx̃j = 0
j =1

we can retrieve dx̃n+1 , . . . , dx̃n+m as a function of dx̃1 , . . . , dx̃n and replace them
in (??).
Corina Forăscu (FII) Lecture 10 December 6, 2022
This gives the quadratic form we look for as
n
d2 L(x0 ) = ∑ aij dx̃i dx̃j .
i ,j = 1

Now we can decide if x0 is a local extremum point or a saddle point for L:

if d2 L(x0 ) is a positive or negative definite form, then (x0 , y0 ) is a
conditioned local minimum, respectively maximum point of f ;
if d2 L(x0 ) is undefined, then (x0 , y0 ) is a conditioned saddle point for f ;
if d2 L(x0 ) is positive or negative semi-defined, we cannot decide the nature
of (x0 , y0 ).

Corina Forăscu (FII) Lecture 10 December 6, 2022

Shannon’s entropy revisited
Let now apply this method for finding the extremum points for
n
H̃ (p1 , p2 , . . . , pn ) = − ∑ pk log2 pk , pk ∈ (0, 1), ∀k = 1, n,
k =1

subject to ∑nk =1 pk = 1. Since we have only one restriction, we have m = 1,

g (p1 , p2 , . . . , pn ) := ∑nk =1 pk − 1 and the Lagrangian is
!
n n
L (λ1 , p1 , p2 , . . . , pn ) = − ∑ pk log2 pk + λ1 ∑ pk − 1 .
k =1 k =1

Then (λ01 , p10 , p20 , . . . , pn0 ) is a critical point for L if


∂L 0 0 0  
(λ1 , p1 , p2 , . . . , pn0 ) = − log2 pk0 + p0 1ln 2 + λ01 = 0, ∀k = 1, n



∂pk k
∂L 0 0 0 n
 0
(λ , p , p , . . . , pn ) = ∑ pk − 1 = 0.0
∂λ1 1 1 2


k =1

From the first n equations we yield p10 = p20 = . . . = pn0 = ψ(λ01 ), where ψ(λ) is
1 = λ,
the unique solution of the equation in p ∈ (0, 1): log2 p + p ln 2
Corina Forăscu (FII) Lecture 10 December 6, 2022
while from the last one we retrieve
1 1 n
p10 = p20 = · · · = pn0 = , λ01 = log2 + .
n n ln 2
From the restraint g (p1 , p2 , . . . , pn ) = 0 we get dp1 + dp2 + · · · + dpn = 0;
therefore !
n
1 1
d L(p1 , p2 , . . . , pn ) = ∑ 0
2 0 0 0
0
− 1 (dpk )2 ,
p
k =1 k ln 2 p k

which is a negative definite quadratic form in dp  1 , . . . , dpn −1 .

0 0 0 1 1 1
This means that the point (p1 , p2 , . . . , pn ) = n , n , . . . , n is a maximum point
for H̃, restricted to p1 + p2 + · · · + pn = 0.
In consequence, the maximal informational entropy is obtained for the random
variable  
1 2 ... n
X = 1 1 ... 1 .
n n n

Corina Forăscu (FII) Lecture 10 December 6, 2022

C. Canuto, A. Tabacco, Mathematical Analysis II, Springer, 2015.
B. Edwards, R. Larson, Calculus (9th Edition), Brooks/Cole-Cengage
Learning, 2014.
K. P. Evans, N. Jacob, A Course in Analysis, World Scientific, 2016.
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Corina Forăscu (FII) Lecture 10 December 6, 2022