Download Chapter 4 Real-valued functions

Chapter 4
Real-valued functions
The subject as well as the methods of study of a class of mappings depend crucially on
structures of the sets which the denition domains and values are chosen from. The mathematical analysis is interested especially in the mappings from Rm to Rk for m, k ∈ N. These
mappings appear frequently in many elds of applications of mathematics. Namely such
the mappings are used in physics, technical sciences, natural sciences, economy and social
sciences (like psychology and sociology). This is the reason why we will be interested in
various kinds of this class of mappings and in their properties.
4.1 Functions and operations with them
Mostly, we will be interested in the following class of mappings:
Denition 4.1.1. Let A, B be sets. The mapping f : A → R is called the real-valued
function (or simply function ). The mapping g : B → Rk , k ∈ N is called the vector
function and the mapping h : (A ⊂ Rm ) → B, m ∈ N is called the abstract function.
Especially, the mapping f : (A ⊂ Rm ) → R, m ∈ N is called the real-valued function of
m real variables. (For m = 1 this simplies to f : (A ⊂ R) → R).
The examples of such the mappings can be found on various gures in previous chapters.
An abstract function g : (T ⊂ R) → R2 was used when dening the curve in Denition
2.4.1 and in Example 2.4.1.
The fact that we have dened operations of addition, subtraction, multiplication and
division allows us to introduce all these operations also for the functions and build up with
help of these operations new kinds of mappings.
Denition 4.1.2. Let A be an arbitrary nonempty set and let f, g : A → R. We dene:
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4.1 Functions and operations with them
26
1. Addition of the functions f and g by
f + g : A → R;
(f + g)(x) = f (x) + g(x), x ∈ A.
2. Subtraction of the functions f and g by
f − g : A → R;
(f − g)(x) = f (x) − g(x), x ∈ A.
3. Product of the functions f and g by
f · g : A → R;
(f · g)(x) = f (x) · g(x), x ∈ A.
4. If there is such a subset M in A that g(x) 6= 0 for x ∈ M then we dene fraction of
the functions f and g by
f /g : M → R;
(f /g)(x) = f (x)/g(x), x ∈ M.
As a special case we obtain multiplication of a real-valued function by a real number.
We see, by Denition 4.1.2, that we can dene, over any set of mappings f : A → B , the
same operations those are dened on the set B . The following theorem holds true:
THEOREM 4.1.1. Let A be an arbitrary set and let F be the set of all real-valued functions
f : A → R. Then the structure:
a) (F ; +, ·, =) is the commutative and unitary ring;
b) (F, R; +, ·, =) is the vector space over the eld of real numbers R.
Proof. It is sucient to verify the properties of the ring from Denition 1.6.4 and properties
of the vector space from Denition 1.6.5, respectively.
Let us remind that the neutral element with respect to addition is given by the function:
e1 : x 7→ 0, x ∈ A. The inverse element (with respect to addition) - also called additive
inverse to an element f is given by the function: f¯ : x 7→ −f (x), x ∈ A. The neutral
element with respect to multiplication is given by the function: e2 : x 7→ 1, x ∈ A.
The proof can be concluded by the fact that the set of all real numbers (that is the set
of values of considered mappings) is both the commutative and unitary ring as well as the
vector space (over the eld R).
We see that the structure (F ; +, ·, =) is not a domain or a eld since there is no an
multiplicative inverse to any f dierent from e1 .
There is the following generalization of the statement b) of Theorem 4.1.1:
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4.2 Monotonic functions
27
THEOREM 4.1.2. Let F be a set of all mappings f : A → Rk , k ∈ N. The the
mathematical structure (F, R, +, ·, =) (with the operations dened on the set of ordered
k -tuples from Theorem 1.6.4) is the vector space over the eld of real numbers R.
Proof. The proof of the theorem consists of simple verication of the axioms of a vector
space.
Problems
1. Prove in details Theorem 4.1.2.
2. Let the mappings: f, g : A → R be bounded in R. Find out whether the real-valued
functions f ± g, f · g and if exists then f ◦ g (all of them acting from A to R) are
bounded in R.
3. Is the following function:
f : x 7→
x+1
, x ∈ R \ {−1, 1},
x2 − 1
bounded in R?
4. Find the denition domains of the following functions
f : x 7→
h : x 7→
√
g : x 7→ √
3x − x2 ,
x2
1
1
+ 2
,
+ 5x + 4 x − 2
1
1
+
.
1 + |1 − x| 1 + x2
Answers
2 The functions are bounded in R.
3 The function is not bounded in R.
4
√
√
√
D(f ) = (−∞, − 3) ∪ [0, 3]; D(g) = (−∞, −4) ∪ ( 2, +∞); D(h) = R.
4.2 Monotonic functions
We will discuss some special kinds of real-valued functions of one real variable in this section.
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4.2 Monotonic functions
28
4.2.1 Even and odd functions, periodic functions
Denition 4.2.1. Let f : (A ⊂ R) → R. We say that the function f is
a) even i the following holds true
x ∈ A ⇒ [−x ∈ A & f (x) = f (−x)] ,
b) odd i the following holds true
x ∈ A ⇒ [−x ∈ A & f (x) = −f (−x)] ,
c) p-periodic (0 6= p ∈ R) i the following holds true
x ∈ A ⇒ [x ± p ∈ A & f (x ± p) = f (x)] .
Such a number p is then called the period of the function f .
4.2.2 Monotonic functions
Starting with the fact that the set of all real numbers R is ordered by the relation of natural
ordering, we can compare by the same ordering also the values of real-valued functions,
especially we can say that a function f : (A ⊂ R) → R is non-negative (f ≥ 0) i for all x
in A we have f (x) ≥ 0. And by this we can dene new and important class of monotonic
functions
Denition 4.2.2. Let f : (A ⊂ R) → R and let M ⊂ A. We say that the function f is
a) increasing on M i
x1 , x2 ∈ M, x1 6= x2 ⇒ [f (x1 ) − f (x2 )](x1 − x2 ) > 0
b) decreasing on M i
x1 , x2 ∈ M, x1 6= x2 ⇒ [f (x1 ) − f (x2 )](x1 − x2 ) < 0
c) nondecreasing on M i
x1 , x2 ∈ M, x1 6= x2 ⇒ [f (x1 ) − f (x2 )](x1 − x2 ) ≥ 0
d) noninreasing on M i
x1 , x2 ∈ M, x1 6= x2 ⇒ [f (x1 ) − f (x2 )](x1 − x2 ) ≤ 0
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4.2 Monotonic functions
29
e) If a function f is of one of above kinds, then we call it monotonic on M .
f) Is a function f is either increasing or decreasing on M we say that it is strictly monotonic on M .
Example 4.2.1. Are the following functions monotonic or periodic?
a)
f : x 7→
b)
sgn : x 7→
c)
¨
χ : x 7→
8
>
<
>
:
1
, x ∈ R \ {0};
x
1 for x > 0,
0 for x = 0,
−1 for x < 0,
x ∈ R;
1 for x ∈ Q,
, x ∈ R,
0 for x ∈ R \ Q
(the so-called Dirichlet function );
d)
g : x 7→ [x], x ∈ R;
(the so-called oor function, or, when dened over the set of positive numbers, also
called integral part of a number ).
These functions are plotted on Fig. 4.1.
Solution: We will discuss all the functions, using also Fig. 4.1
a) Let x1 6= x2 be from D(f ), then
1
1
(x1 − x2 )2
−
(x1 − x2 ) = −
.
x1 x2
x1 x2
This means that the function f is decreasing on (−∞, 0) and also on (0, ∞), however,
it is not decreasing nor increasing on D(f ). The function f is obviously not periodic.
b,d) The functions signum and oor are nondecreasing on R. In fact, for any x1 , x2 ≥ 0 or
x1 , x2 ≤ 0 that are dierent (x1 6= x2 ) we have
(sgn(x1 ) − sgn(x2 ))(x1 − x2 ) ≥ 0.
(?)
For mutually dierent x1 , x2 such that x1 ≥ 0 and x2 ≤ 0 or x1 ≤ 0 and x2 ≥ 0 we
have that the left-hand side of (?) is given by
2(x1 − x2 ) ≥ 0,
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4.2 Monotonic functions
30
or by
−2(x1 − x2 ) ≥ 0.
Similarly, we can proof the same about the oor function.
c) The Dirichlet function is not monotonic on R. It is, at the same time, nondecreasing
and nonincreasing on (separately) Q and on R \ Q. Moreover, it is p-periodic with
arbitrary rational period p ∈ Q. In fact, for any rational numbers x, p, the numbers
x ± p are also rational, and for any irrational x and rational p the numbers x ± p are
also irrational. Therefore, for any real x and for any rational p the following equality
holds true:
χ(x ± p) = χ(x).
y
y
f
sgn
1
1
−1
1
x
x
−1
−1
y
y
2
floor
1
1
−2
− 32
−1
− 12
0
1
2
1
−1
0
3
2
1
x
2
3
x
−1
−2
Figure 4.1:
4.2.3 Properties of (strictly) monotonic functions
In this subsection we will be interested in relation between monotonic functions and injective
functions. The following theorem holds true
THEOREM 4.2.1. If a function f : (A ⊂ R) → R is strictly monotonic on A then it is
injective.
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4.2 Monotonic functions
31
Proof. Let f be increasing on A. Then for all x1 , x2 ∈ A such that x1 6= x2 we have
[f (x1 ) − f (x2 )](x1 − x2 ) > 0.
Thus, f (x1 ) 6= f (x2 ). For a decreasing function, the proof can be done in the same way.
We can easily see that the implication in Theorem 4.2.1 cannot be inverted, since there
are functions that are injective but are not monotonic. As an example of such a function we
have:
1
f : x 7→ , x ∈ R \ {0}.
x
THEOREM 4.2.2. Let a function f : (A ⊂ R) → R is increasing (decreasing) on A. Then
the inverse function f −1 : f (A) → A is increasing (decreasing) on f (A).
Proof. We will prove the theorem for a decreasing function. Theorem 4.2.1 ensures the
existence of the inverse function f −1 (see also Theorem 2.2.1). Now, let y1 , y2 ∈ f (A) be
such that y1 6= y2 . The there are x1 , x2 ∈ A such that
y1 = f (x1 ), y2 = f (x2 ),
and x1 6= x2 . (Otherwise, we would have y1 = y2 ). Thus,
[f −1 (y1 ) − f −1 (y2 )](y1 − y2 ) = (x1 − x2 )[f (x1 ) − f (x2 )] < 0,
and this inequality veries that f −1 is decreasing on f (A).
All the notions discussed above for real-valued functions of one real variable can be
applied also for the sequences of real numbers.
Especially, the following theorem holds true for the sequences of real numbers:
THEOREM 4.2.3. A sequence of real numbers (an )n∈N is increasing (decreasing) i
∀n ∈ N : an < an+1
(an > an+1 ).
Example 4.2.2. Let the function g : (A ⊂ R) → (B ⊂ R) be increasing on A and let the
function f : (E ⊃ B) → R be increasing on E . Then the composite function f ◦ g : A → R
is increasing on A. Prove this!
Solution: For any two x1 , x2 ∈ A such that x1 < x2 we have g(x1 ) < g(x2 ) and also
f [g(x1 )] < f [g(x2 )]. Therefore, we have
{f [g(x1 )] − f [g(x2 )]}(x1 − x2 ) > 0.
The same inequality holds true also for x1 > x2 . So, we can conclude that f ◦ g is increasing
on A.
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4.2 Monotonic functions
32
4.2.4 Bounded functions
The following simple statement holds true for the bounded functions (see denition 2.1.4).
THEOREM 4.2.4. A real-valued function f : A → R is bounded (in R) on a set M ⊂ A
if and only if
∃k ∈ R ∀x ∈ M : |f (x)| ≤ k.
Proof. We split the proof, as usually, into two implications:
1. By using denition 2.1.4 we have that the values H(f |M ) of the restriction f |M is a
bounded set in R. Thus, we know that there are real numbers k1 ≤ k2 such that
∀y ∈ H(f |M ) : k1 ≤ y ≤ k2 .
For any y ∈ H(f |M ) we have an x ∈ M such that y = f (x). Then we can rewrite the last
statement into the form
∀x ∈ M : −k ≤ f (x) ≤ k ⇔ |f (x)| ≤ k,
where k = max{|k1 |, |k2 |}. This concludes the proof of the necessary condition.
2. The inequality |f (x)| ≤ k, x ∈ M means that for x ∈ M we have −k ≤ f (x) ≤ k . So, we
have that the lower (upper) bound of the set H(f |M ) is −k (k ).
4.2.5 Supremum of a function
By using the previous knowledge we can formulate and prove the following theorem on
supremum of a function
THEOREM 4.2.5. Let a real-valued function f : A → R be bounded from above in R.
Then the following statements are mutually equivalent:
•
sup f = b ∈ R,
(1)
∀x ∈ A : f (x) ≤ b,
(2)
•
• let y ∈ R be such that
(∀x ∈ A : f (x) ≤ y) ⇒ b ≤ y
∀x ∈ A : f (x) ≤ b,
(3)
•
(∀y ∈ R : y < b) ∃x0 ∈ A : y < f (x0 )
∀x ∈ A : f (x) ≤ b,
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(4)
4.2 Monotonic functions
33
•
∀² > 0∃x0 ∈ A : b − ² < f (x0 ).
Proof. 1. (1) ⇔ (2) follows directly from Denition 1.5.4 of the supremum of a ordered set
B = H(f ) and from Denition 2.1.4 of the supremum of a mapping f .
2. (2) ⇔ (3) follows from Theorem 1.5.3 for B = h(f ).
3. (3) ⇔ (4) can be obtained from Theorem 3.1.1.
4. (4) ⇔ (1) follows directly from Theorem 3.1.1.
An analogical theorem holds true also in R̄.
Example 4.2.3. Let the functions f, g : A → R be bounded from above in R. Then
sup(f + g) ≤ sup f + sup g,
(A)
sup(λf ) = λ sup f, λ ∈ R+ ,
(B)
| sup f − sup g| ≤ sup |f − g|.
(C)
Prove these relations!
Solution: 1. Since the function f + g is bounded from above, all the suprema in Eq. (A)
do exist. It follows from the rst property of the supremum that for all x ∈ A we have
f (x) ≤ sup f, g(x) ≤ sup g,
therefore
f (x) + g(x) ≤ sup f + sup g.
The real number sup f + sup g is an above bound of the function f + g , and from the second
property of the supremum we have
sup(f + g) ≤ sup f + sup g.
2. For all x ∈ A and λ > 0 we have f (x) ≤ sup f ⇒ λf (x) ≤ λ sup f ⇒ sup λf ≤ λ sup f .
From the other side for x ∈ A we have
1
1
λf (x) ≤ sup λf ⇒
sup λf ⇒ sup f ≤ sup λf ⇒ λ sup f ≤ sup λf.
λ
λ
So, the formula (B) holds true.
3. Property (A) implies
sup f = sup(f − g + g) ≤ sup(f − g) + sup g ≤ sup |f − g| + sup g.
The inequality sup(f −g) ≤ sup |f −g| follows from that: ∀x ∈ A : f (x)−g(x) ≤ |f (x)−g(x)|.
Furthermore, we have
sup f − sup g ≤ sup |f − g|.
Similarly,
sup g − sup f ≤ sup |g − f | = sup |f − g|.
The last two inequalities imply the required formula (C).
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4.2 Monotonic functions
34
Problems
1. Show that for n ∈ N the function c 7→ xn is even for even n and is odd for odd n.
2. Find a function that is at the same time even and odd.
3. Can a periodic function be monotone? What is the period of such a function equal to?
4. Can a function be, at the same time, increasing and even (decreasing and odd)?
5. Explain what are the diculties when trying to dene monotonicity of a function from
Rm with m ≥ 2 (of a vector function from Rm to Rk ). What are the diculties when
trying to dene the boundedness of a vector function of a vector argument?
6. Prove Theorem 4.2.2 for an increasing function.
7. Prove Theorem 4.2.3 and formulate an analogical theorems for nondecreasing function
and nonincreasing function, respectively.
8. Formulate and prove statements analogical to those from Theorem 4.2.5 concerning
the inmum of a function.
9. Prove the following. Let the functions f, g : A → R be bounded from below in R.
Then:
inf(f + g) ≥ inf f + inf g,
sup(λf ) = λ inf f, λ < 0.
10. Examine the function: f : x 7→ x − [x] := {x} (called also fractional part of x) for
periodicity, parity and monotonicity.
11. Examine the composite function f ◦ g (when we know that (1) both functions f and
g are decreasing or (2) one of them is increasing and the second one decreasing) for
monotonicity.
Answers
2 f : x 7→ 0, x ∈ (−a, a), a ∈ R̄.
3 Yes. Any real number can be the period of such a function.
4 No (no).
10 The function f is 1-periodic and is not even nor odd.
34
4.3 Polynomials and rational functions
35
4.3 Polynomials and rational functions
In this section we will dene and work with very often used kinds of functions - polynomials
and their ratios, i.e. rational functions. The values of these functions are dened very simply
by operations of addition, subtraction multiplication and constant, especially, only integernumber exponents of variable appear in polynomials. These mappings are often used in the
theory of approximation of functions and we will meet them again in chapter 10 devoted to
integration.
4.3.1 Polynomials
For x ∈ C and n ∈ N we have dened the n-th power of x by
xn = x
| · x ·{z· · · · x} .
n−times
Denition 4.3.1. The mapping Pn,m : Cm → C (alternatively, Pn,m : Rm → R) dened by
X
x = (x1 , x2 , . . . , xm ) 7→
ai1 ...im xi11 · · · · · ximm , x ∈ Cm
0≤i1 +···+im ≤n
is called the polynomial of n-th degree (order) (of many complex (real) variables), where the
coecients ai1 ...im ∈ C and at least on of the coecients such that i1 + . . . im = n is dierent
from zero.
The degree of a polynomial P is denoted by deg P . For example, the real-valued function
P1,m : x = (x1 , . . . , xm ) 7→ a1 x1 + · · · + am xm + b, x ∈ Rm
is the polynomial of the rst degree for ai ∈ R, i = 1, 2, . . . , m, b ∈ R if at least of of the
coecients a is dierent from zero. (This polynomial, in case b = 0, is a linear function
(mapping)).
The mapping
Pn,1 =: P : x 7→ an xn + an−1 xn−a + · · · + a1 x + a0 , x ∈ C
(1)
for ai ∈ C, i = 0, 1, 2, . . . , n ∈ N and an 6= 0 is the polynomial of n-th degree (of one complex
variable).
The function
P2,2 : (x, y) 7→ xy + x + y + 3, (x, y) ∈ R2
is the polynomial of the degree two.
The constant nonzero function
x 7→ a ∈ R \ {0}, x ∈ R,
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4.3 Polynomials and rational functions
36
is the polynomial of degree zero.
The function
x 7→ 0, x ∈ C
is called the zero polynomial.
The set of all polynomials of form (1) forms the innitely dimensional vector space over R
which basis is {1, x, x2 , x3 , . . . } (see section 1.6).
Roots of the polynomials
In the next text we will be interested in polynomials of the form (1).
Denition 4.3.2. The complex number z is called the root of the polynomial (1) i the
equality P (z) = 0 holds true.
It is obvious that the polynomial of degree zero has no root. The base of study the roots
of the roots of the polynomials is the so-called fundamental theorem of algebra :
THEOREM 4.3.1. A polynomial of the form (1) has at least one root (from C).
We have not enough of knowledge at this time to prove this theorem. We refer the reader
to the literature, namely the book [11] (pages 221-223) or [8] (pages 147-155).
THEOREM 4.3.2. Let z ∈ C be a root of the polynomial P (1) of degree n equal or
greater than 1. Then there is a polynomial Q of degree n − 1 such that
P (x) = Q(x)(x − z),
∀x ∈ C.
Proof. Since P (z) = 0 we can write
€
Š
P (x) = P (x) − P (z) = an xn + an−1 xn−1 + · · · + a1 x + a0 − an z n + an−1 z n−1 + · · · + a1 z + a0 =
= an (xn − z n ) + an−1 (xn−1 − z n−1 ) + · · · + a1 (x − z) =
”
—
= (x − z) an (xn−1 + zxn−1 + · · · + z n−2 x + z n−1 ) + · · · + a1 =
= (x − z)Q(x),
where Q : x 7→ an (xn−1 + zxn−1 + · · · + z n−2 x + z n−1 ) + · · · + a1 is really the polynomial of
degree n − 1 since we have supposed an 6= 0.
Denition 4.3.3. We say that the complex number z is the root of multiplicity k of the
polynomial P (1) i the following equality holds true:
P (x) = (x − z)k Q(x), ∀x ∈ C and Q(z) 6= 0.
THEOREM 4.3.3. Let P be a polynomial of the form (1) of degree n ≥ 1. Then P has
just n roots when counting any root with its multiplicity.
36
4.3 Polynomials and rational functions
37
Proof. By induction. If the degree of P is 1 then the single root can be easily computet.
So, the statement holds true for n = 1. Let the theorem holds true for degree n. Let P has
degree n + 1. Then Theorem 4.3.1 implies existence of a root z ∈ C of P such that with
respect to Theorem 4.3.2 we can write
P (x) = (x − z)Q(x), x ∈ C,
where Q is a polynomial of degree n. By assumption, Q satises theorem, thus P satises
it, too.
Corollary 4.3.1. Any polynomial P : x 7→ an xn + · · · + a1 x + a0 , x ∈ C of degree n ≥ 1
can be written as
P (x) = an (x − z1 )k1 (x − z2 )k2 · · · · · (x − zi )ki ,
x ∈ C,
where z1 , . . . , zi are the roots of P and k1 , . . . , ki are their multiplicities. Moreover, we have
k1 + · · · + ki = n.
THEOREM 4.3.4. Let P, Q be polynomials of one variable. The equality P = Q holds
true i all the corresponding coecients of both polynomials (i.e. coecients standing by
equal powers of the variable) are equal.
Proof. 1. Let P = Q and
P : x 7→ an xn + · · · + a1 x + a0 , x ∈ C,
Q : x 7→ bm xm + · · · + b1 x + b0 , x ∈ C.
We can suppose without lost of generality that n ≥ m. If the corresponding coecients are
not mutually equal then we can denote by i the maximal index with the property ai 6= bi .
Then
P − Q : x 7→ (ai − bi )xi + · · · + (a1 − b1 )x + (a0 − b0 ), x ∈ C.
If i = 0 then for all complex x we have P (x) 6= Q(x). If i > 0 then the polynomial P − Q has
positive degree and with respect to Theorem 4.3.3 has only nite number of roots - namely
i. Thus, there exists x0 ∈ C such that P (x0 ) 6= Q(x0 ). In both cases, we are in contradiction
with the assumption.
2. If all the corresponding coecients of the polynomials P and Q are mutually equal then
for all the complex numbers x we have P (x) = Q(x) and this means (by Theorem 2.1.1)
that P = Q.
THEOREM 4.3.5. Let P be a polynomial of the form (1) with real coecients. If a
complex number z is a root of P then also the complex conjugate z̄ is the root of P . The
multiplicities of the roots z and z̄ are equal.
37
4.3 Polynomials and rational functions
38
Proof. If the root z of P is real then the theorem holds true trivially. We will suppose
z ∈ C \ R and split the proof into two steps.
1. Let P (z) = 0. Since the coecients are real, we can write
P (z̄) = an (z̄)n + · · · + a1 z̄ + a0 = an (z n ) + · · · + a1 z̄ + a0 =
= an z n + · · · + a1 z + a0 = 0̄ = 0.
2. Now, we will prove that if the root z of P has the multiplicity k then z̄ has also the
multiplicity k . Let us suppose that the multiplicity of z is k and the multiplicity of z̄ is l.
Furthermore, we will suppose that k > l. Then we can write (by Denition 4.3.3):
∀x ∈ C : P (x) = (x − z)l (x − z̄)l Q(x)
where Q(z̄) 6= 0 and Q(z) = 0. However, this is in contradiction with the result of the rst
part of this proof. The same can be derived starting with k < l. Thus, it must be k = l.
Corollary 4.3.2. The value P (x) of any polynomial P :
P : x 7→ an xn + · · · + a1 x + a0 , x ∈ C,
with real coecients, can be written in the following form:
P (x) = an (x − x1 )k1 · · · · · (x − xi )ki (x2 + p1 x + q1 )l1 · · · · · (x2 + pj x + qj )lj ,
(2)
where x1 , . . . , xi are all the real roots of P (with corresponding multiplicities k1 , . . . , ki ).
The numbers p1 , q1 , . . . , pj , qj are real and the polynomials x 7→ x2 + pk x + qk , x ∈ C, k =
1, 2, . . . , j have no real roots and k + 1 + · · · + k + i + 2(l1 + · · · + lj ) = n.
Proof. The proof follows Corollary 4.3.1 and from that for any z ∈ C we have
(x − z)(x − z̄) = x2 + px + q,
where the numbers p and q are real.
The equality (2) is called the factorization of a polynomial.
Example 4.3.1. Perform the factorization (2) of the polynomial
P : x 7→ x6 + x4 − x2 − 1,
x ∈ R.
Solution: One can easily see that 1 is a root of this polynomial. Subsequently, we obtain:
P (x) = (x − 1)(x5 + x4 + 2x3 + x + 1) = (x − 1)(x + 1)(x4 + 2x2 + 1) =
= (x − 1)(x + 1)(x2 + 1)2 .
38
4.3 Polynomials and rational functions
39
4.3.2 Rational functions
In the rest of this section, we will be interested in basic properties of the rational functions
(ratios of the polynomials), and, especially, their factorization - this procedure will play an
important role in integration of the rational functions.
Denition 4.3.4. Let P, Q : Cm → C be two polynomials of degree n with real coecients
(see Denition 4.3.1). Let Q(x) 6= 0 for x ∈ M ⊂ C. Then the fraction:
P/Q : M → C
is called the rational function (of m complex variables).
For example,
S : (x, y, z) 7→
xyz + x + y − 1
, (x, y, z) ∈ M = {(x, y, z) ∈ R3 ⊂ C3 ; x2 + y 2 6= 1},
x2 + y 2 − 1
is the rational function of three variables.
Lemma 4.3.1. Let P, Q be polynomials of the form (1) (of one variable and with real
coecients) and let deg P < deg Q. Let Q have a real root a of multiplicity k (k ≥ 1) and
such that
Q(x) = (x − a)k Q1 (x), x ∈ C, where Q1 (a) 6= 0.
Then there is a real number A and a polynomial P1 such that
P (x)
A
P1 (x)
=
+
, x ∈ {y ∈ C : Q(y) 6= 0} =: M,
k
Q(x)
(x − a)
(x − a)k−1 Q1 (x)
and deg P1 < deg Q − 1.
Proof. For A ∈ R and x ∈ M we have
P (x)
A
P (x) − AQ1 (x)
=
+
.
k
Q(x)
(x − a)
(x − a)k Q1 (x)
Since Q1 (a) 6= 0, we have that there exists A ∈ R such that P (a) − AQ1 (a) = 0. If the
polynomial x 7→ P (x) − AQ1 (x), x ∈ C is not zero then P (x) = (x − a)P1 (x), x ∈ C and
deg P1 = deg(P − AQ1 ) − 1 ≤ max[deg P, deg Q1 ] − 1 < deg Q − 1.
Lemma 4.3.2. Let P, Q be polynomials of the form (1) (of one variable and with real
coecients) and let deg P < deg Q. Let Q have a root z ∈ C \ R of multiplicity k (k ≥ 1)
and such that
Q : x 7→ (x − z)k (x − z̄)k Q1 (x), x ∈ C,
39
where
Q1 (z) 6= 0 6= Q1 (z̄).
4.3 Polynomials and rational functions
40
Then there exists real numbers B, C and a polynomial P1 such that
P (x)
Bx + C
P1 (x)
=
+
, x ∈ {y ∈ C : Q(y) 6= 0} =: M,
k
k
k−1
Q(x)
(x − z) (x − z̄)
(x − z) (x − z̄)k−1 Q1 (x)
and
deg P1 < deg Q − 2.
Proof. By using Theorem 4.3.5, the proof is very similar to that of Lemma 4.3.1.
Denition 4.3.5. Let A, B, C be real numbers and let a ∈ R and z ∈ C and k ∈ N. The
rational functions
x 7→
A
, x ∈ C \ {a}
(x − a)k
x 7→
Cx + C
, x ∈ C \ {z, z̄}
(x − z)k (x − z̄)k
are called the partial fractions.
Lemmas 4.3.1 and 4.3.2 allows, if we have the factorization of the denominator, to expand
any rational function into the partial fractions - these are considered in many applications
as simple(r) functions.
THEOREM 4.3.6. Let P, Q be polynomials of the form (1) (of one variable and with
real coecients) and let deg P < deg Q. Let Q has the factorization (2). Then there exist
real numbers A11 , . . . , A1k1 ; . . . ; Ai1 , . . . , Aiki ; P11 , Q11 , . . . , Pl11 , Q1l1 ; . . . ; P1j , Qj1 , . . . , Pljj , Qjlj (their
number is equal to the degree of the polynomial Q) such that for any x ∈ C \ {y ∈ C :
Q(y) = 0} we have
"
#
#
"
Aiki
A1k1
A11
Ai1
P (x)
=
+
·
·
·
+
+
·
·
·
+
+
·
·
·
+
+
Q(x)
(x − x1 )k1
x − x1
(x − xi )ki
x − xi
"
#
2
3
Pljj x + Qjlj
Pl11 x + Q1l1
P11 x + Q11
P1j x + Qj1 5
4
+
·
·
·
+
+
·
·
·
+
+
·
·
·
+
.
(x2 + p1 x + q1 )l1
x2 + p 1 x + q 1
(x2 + plj x + qlj )lj
x2 + plj x + qlj
Proof. The proof can be done by induction with respect to the degree of the polynomial Q,
however it is technically quite complicated and we will omit it. We refer the reader to the
literature [6], chapter IV, section 2, where you can nd a detailed proof.
We will demonstrate the usage of the expansion from Theorem 4.3.6 in the following
example in which we will see how to nd the coecients of the expansion into the partial
fractions.
Example 4.3.2. Expand into the partial fractions the following rational function:
2x6
,
f : x 7→ 6
x + x4 − x2 − 1
40
x ∈ C \ {1, −1, i, −i}.
4.3 Polynomials and rational functions
41
Solution: If we denote the nominator by P and the denominator by Q then we know, from
Example 4.3.1, that
Q(x) = (x − 1)(x + 1)(x2 + 1)2 ,
and
P (x) = 2Q(x) − 2x4 + 2x2 + 2.
So, by division of polynomials, we get
P (x)
−2x4 + 2x2 + 2
=2+
,
Q(x)
Q(x)
and the expansion of the right-hand side of the previous equation into the partial fractions
has the form
−2x4 + 2x2 + 2
A
B
Cx + D
Ex + F
=
+
+ 2
+
.
Q(x)
x − 1 x + 1 (x + 1)2
x2 + 1
This exactly means that
−2x4 + 2x2 + 2 = A(x5 + x4 + 2x3 + x + 1) + B(x5 − x4 + 2x3 − 2x2 + x − 1) +
+(Cx + D)(x2 − 1) + (Ex + F )(x4 − 1).
With respect to Theorem 4.3.4 the coecients A, B, . . . , F must obey the system of (linear)
equations that arises from comparison of terms by equal powers of the independent variable
x. The resulting system reads:
0 = A+B+E
−2 = A − B + F
0 = 2A + 2B + C
2 = 2A − 2B + D
0 = A+B−C −E
2 = A − B − D − F.
The (only) solution to this system is given by
1
5
1
A = , B = − , C = 0, D = 1, E = 0, F = − .
4
4
2
Thus, for any x ∈ C \ {1, −1, i, −i} we have
2x6
1
1
1
5
=2+
−
+ 2
−
.
6
4
2
2
2
x +x −x −1
4(x − 1) 4(x + 1) (x + 1)
2(x + 1)
41
4.4 Elementary functions
42
Problems
1. Find the factorization of the polynomials:
a) Q : x 7→ x9 + 2x6 + x3 , x ∈ C;
b) S : x 7→ x4 + 1, x ∈ C.
2. Find the expansion of the polynomials following rations functions into the partial
fractions:
a)
x 7→
x7 + 7x − 1
,
Q(x)
b)
x 7→
1
,
S(x)
x ∈ C \ {y ∈ C : Q(y) 6= 0},
x ∈ C \ {y ∈ C : S(y) 6= 0},
where Q and S are the polynomials from Problem 1.
Answers
1
a) Q(x) = x3 (x + 1)2 (x2 − x + 1)2 , x ∈ C
√
√
b) S(x) = (x2 + 2x + 1)(x2 − 2x + 1), x ∈ C
2
a)
−
for
1
7
1
31
x+7
31x + 1
+
+
+
−
−
,
x3 x2 (x + 1)2 9(x + 1) 3(x2 − x + 1)2 9(x2 − x + 1)
√
1±i 3
x ∈ C : x 6= 0, x 6= −1, x 6=
.
2
b)
1
√
2 2
for
√
√
!
x+ 2
x− 2
√
√
,
−
x2 + 2x + 1 x2 − 2x + 1
1
1
x ∈ C : x 6= √ (1 ± i), x 6= √ (−1 ± i).
2
2
4.4 Elementary functions
In this section, we will be interested in a very special and important class of the so-called
basic elementary functions. We can construct, using the functions of this class and certain
operations, the wide class of elementary functions. In fact, in theory as well as in practice,
we can often meet also non-elementary functions represented by functional series. However,
the study of non-elementary functions exceeds this book.
42
4.4 Elementary functions
43
4.4.1 Basic elementary functions
It is supposed that the reader has some skill in using the elementary functions from previous courses in mathematics. The following denition contains the list of basic elementary
functions we will be interested in.
Denition 4.4.1. The following mappings are called the basic elementary functions :
• x 7→ ax , x ∈ R, a ∈ R+ \ {1} - the exponential function
• x 7→ loga (x), x ∈ R+ , a ∈ R+ \ {1} - the logarithmic function
• x 7→ xα , x ∈ R+ , α ∈ R - the power function
• x 7→ sin(x), x ∈ R - the sine function
• x 7→ cos(x), x ∈ R - the cosine function
• x 7→ tan(x), x ∈ {v ∈ R : cos(v) 6= 0} - the tangent function
• x 7→ cot(x), x ∈ {v ∈ R : sin(v) 6= 0} - the cotangent function
• x 7→ arcsin(x), x ∈ [−1, 1] - the arcsine function
• x 7→ arccos(x), x ∈ [−1, 1] - the arccosine function
• x 7→ arctan(x), x ∈ R - the arctangent function
• x 7→ arccot(x), x ∈ R - the arccotangent function
The functions sin, cos, tan, cot are called the trigonometric functions and the functions
arcsin, arccos, arctan, arccot are called the inverse trigonometric functions.
Some of the basic elementary functions can be dened very easily on the set of rational
numbers, e.g. the exponential function. The goniometric functions can be dened in the
interval [0, π/2] by an elementary geometry: by ratios of the lengths of sides in rectangled
triangle. The values of the functions sine and cosine can be dened in the interval [0, 2π] by
help of unit circumference (see Fig. 4.2) as coordinates of the point C . The values of the
functions tangent and cotangent are also dened in the unit circumference (see again Fig.
4.2), namely
sin(x)
cos(x)
tan(x) = |BD| =
,
cot(x) = |EF | =
.
cos(x)
sin(x)
These equalities follows from the homothety of the triangles M OAC and M OBD, and
M OBD and M F EO, respectively. Similarly, the values of the functions secant and cosecant
are dened as lengths of the sides |OD| = 1/ cos(x), and |OF | = 1/ sin(x), respectively.
43
4.4 Elementary functions
44
o2
D
cot(x)
E
F
C
G
tan(x)
x
sin(x)
B
cos(x)
O
−1
A
1
o1
Figure 4.2:
Our task will be more general - we need to dene all the basic elementary functions
exactly and for all possible values of their arguments. And this cannot be done by simple
extension of a known function. For example, it is not evident how to dene a real power of
a real number despite the rational powers of a number are dened easily. The geometrical
denition of the functions sine and cosine is not self-evident, since it assumes the existence
of a bijective mapping of the interval [0, 2π) onto the unit circumference.
4.4.2 Powers with a rational exponent
We will start with a list of properties of the powers of the positive real numbers with a rational
exponent. We will need these properties in the sequel, when we will study properties of the
exponential function and the power function.
Let us consider a ∈ R+ . Then the following properties hold:
1 a0 = 1
2 For all n ∈ N we have an = an−1 · a
3 For all n ∈ N (−n ∈ Z) we have a−n =
1
an
4 For all n ∈ N we have granted the existence of the n-th root of a: a1/n =
Theorem 3.1.6, see also Denition 3.1.2)
√
5 Since for all m ∈ Z: am > 0 we have for all n ∈ N: am/n = (am )1/n = n am
44
√
n
a (by
4.4 Elementary functions
45
Bellow we continue with the properties obeyed by the powers with rational exponents (these
properties follows from the well-known rules of computation with real numbers):
6 ar > 0 for all r ∈ Q
7 If (a, b) ∈ R+ × R+ then (ab)r = ar br , and ( ab )r =
8 If (r1 , r2 ) ∈ Q2 then ar1 ar2 = ar1 +r2 , and
ar1
ar2
ar
,
br
for all r ∈ Q
= ar1 −r2 , and (ar1 )r2 = ar1 r2
9 If 0 < a < b and r ∈ Q then ar < br
10 If a > 1 and (r1 , r2 ) ∈ Q are such that r1 < r2 then ar1 < ar2 . For 0 < a < 1 one
obtains ar1 > ar2 .
Example 4.4.1. Let 1 < a ∈ R. Show that
∀² > 0 ∃n0 ∈ N : (n ≥ n0 ⇒ a1/n − 1 < ²).
Solution: Let us introduce the notation
δn := a1/n − 1.
Property 10 listed above tells us that a1/n > 1, thus δn > is positive. The binomial theorem
implies that
a = (a1/n )n = (1 + δn )n = 1 + nδn + ∆n > 1 + nδn ,
since the remainder denoted by ∆n is obviously positive. So, we have:
δn <
a−1
.
n
The Archimedean property (Theorem 3.1.4) ensures the existence of a natural number n0
such that
a−1
a−1
< n0 ,
or
< ².
²
n0
For such an n0 and for all n ≥ n0 one obviously obtains
a−1
a−1
≤
< ²,
n
n0
thus, for all n ≥ n0 we have 0 < a1/n − 1 < ², as required.
45
4.4 Elementary functions
46
4.4.3 Powers with a real exponent
The way we will dene in the real power of a real number is motivated by Example 3.1.3.
Denition 4.4.2. Let x ∈ R.
1. For any a ∈ [1, +∞) we can dene
E := {p ∈ Q; p < x} and F := {p ∈ Q; x < p}.
Then the real number y that obeys:
∀r ∈ E ∀s ∈ F : ar ≤ y ≤ as
(1)
is called the power of the number a with real exponent x. We denote the number y by ax .
2. If a ∈ (0, 1), then we dene:
x
−x
a :=
1
a
,
(1/a > 1).
Having formulated this denition of the real power of a real number, it is not obvious
to be sure that there is just one number y with given properties. We have to prove the
correctness of Denition 4.4.2.
THEOREM 4.4.1. Denition 4.4.2 is correct, i.e. for all x ∈ R and for all a > 1 there
exists just one real number y with property (1).
Proof. 1. First, we will prove the existence of the number y . Let us introduce the notation:
A := {ar ∈ R; r ∈ E}.
For all r ∈ E and for any xed s ∈ F we have: r < x < s, (r ∈ Q, s ∈ Q), therefore
(see property 10 of the powers with rational exponent) we get that ar < as for all r ∈ E .
Thus, the set A ⊂ R is bounded from above and thus (see Theorem 3.1.2) there exists its
supremum:
sup A =: y ∈ R.
The rst property of the supremum implies that ∀r ∈ E : ar ≤ y . Since y is the least upper
bound of the set A, we have that y ≤ as for any s ∈ F . We can conclude that the number
y obeys (1) and this proves its existence.
2. The uniqueness will be proven indirectly. Let us suppose there are two distinct real
numbers y1 , y2 obeying (1), this means the following holds true:
∀r ∈ E ∀s ∈ F : ar ≤ y1 ≤ as ,
ar ≤ y2 ≤ as .
As a consequence we have:
−(as − ar ) ≤ y1 − y2 ≤ as − ar .
46
4.4 Elementary functions
47
For a xed s0 ∈ F and for all r ∈ E, s ∈ F we have
|y1 − y2 | ≤ as − ar = ar (as−r − 1) < as0 (as−r − 1).
By Example 3.1.3 (see also Example 3.1.4) we have
x = sup E = inf E.
Theorem 3.1.3 implies
∀n ∈ N ∃r1 ∈ E ∃s1 ∈ F : x −
1
< r1
2n
i.e.
s1 − r1 <
& x+
1
> s1 ,
2n
1
.
n
Thus,
∀n ∈ N : |y1 − y2 | < as0 (as1 −r1 − 1) < as0 (a1/n − 1).
Now, let ² > 0. Then Example 4.4.1 implies that
∃n0 ∈ N : a1/n0 − 1 < ²a−s0 .
So, we have obtained the following statement:
∀² > 0 ∃n0 ∈ N : |y1 − y2 | < as0 (a1/n0 − 1) < ²,
or simply
∀² > 0 : |y1 − y2 | < ².
However, this contradicts our assumption: y1 6= y2 . Therefore, y1 = y2 .
The correctness of the second part of Denition 4.4.2, where the case a ∈ (0, 1) is considered, follows from the correctness of the rst part of this Denition.
The powers with a real exponent obeys the following properties (analogical with the
properties of the powers with a rational exponent):
THEOREM 4.4.2. Let (x1 , x2 ) ∈ R2 and let (a, b) ∈ R+ × R+ . Then
ax1
= ax1 −x2
ax2
 ‹x1
a
ax1
4.
= x1
b
b
1
1. ax ax2 = ax1 +x2
2.
3. (ab)x1 = ax1 bx1
5. (ax1 )x2 = ax1 x2
47
4.4 Elementary functions
48
Proof. We will prove the rst of the listed properties. The proofs of the properties 2.-5. can
be done in a similar way.
Let a ∈ [1, +∞). (For a ∈ (0, 1] is the statement evident from the second part of Denition
4.4.2). If x1 , x2 ∈ R and
E := {(r1 , s1 ) ∈ Q2 ; r1 < x1 < s1 },
F := {(r2 , s2 ) ∈ Q2 ; r2 < x2 < s2 },
then, for any (r1 , s1 ) ∈ E and for any (r2 , s2 ) ∈ F , we have:
r1 + r2 < x1 + x2 < s1 + s2 .
Thus, one obtains from Denition 4.4.2 that
ar1 ≤ ax1 ≤ as1 ,
ar2 ≤ ax2 ≤ as2 ,
ar1 +r2 ≤ ax1 +x2 ≤ as1 +s2
holds true for all (r1 , s1 ) ∈ E, (r2 , s2 ) ∈ F . After some algebra one can deduce from the
previous inequalities that
ar1 +r2 ≤ ax1 ax2 ≤ as1 +s2
∀(r1 , s1 ) ∈ E, (r2 , s2 ) ∈ F.
Finally, we can use the same as in the proof of Theorem 4.4.1, namely:
∀² > 0 : ax1 +x2 − ax1 ax2 < ².
Thus, the case ax1 +x2 6= ax1 ax2 is ruled out.
4.4.4 Exponential function
We can formulate the following denition.
Denition 4.4.3. An exponential function expa is dened as the real-valued function
expa : x 7→ ax ,
x ∈ R,
where a ∈ R+ \ {1} and the values ax are specied in Denition 4.4.2.
THEOREM 4.4.3. The exponential function expa has the following properties:
1. It is increasing on R for a ∈ (1, +∞)
2. It is decreasing on R for a ∈ (0, 1)
3. For all x ∈ R and for all a ∈ R+ \ {1} the value ax is positive
4. It is, for all a ∈ R+ \ {1}, a bijective mapping from R onto R+
48
4.4 Elementary functions
49
5. For all a ∈ R+ \ {1} we have:
inf ax = 0,
x∈R
sup ax = +∞ in the extended real line.
x∈R
Proof. 1. Let x1 , x2 ∈ R be two distinct numbers. Let us suppose x1 < x2 . Then we can use
Theorem 3.1.5 (the rational numbers are dense in the real numbers) to state that the exist
two numbers c, d in Q such that x1 < c < d < x2 . Denition of the exponential function and
the increase of the exponential function on Q imply that ax1 ≤ ac < ad ≤ ax2 , or
(ax1 − ax2 )(x2 − x1 ) > (ad − ac )(d − c) > 0.
Thus, the exponential function is, in fact, increasing for a > 1.
2. Analogically, for 0 < a < 1 we have (using the fact that the exponential function expa
with a ∈ (0, 1) is decreasing on Q):
x1 < c < d < x2 , (c, d) ∈ Q2 ⇒ ax1 =
−x1
1
a
−c
≥
1
a
−d
≥
1
a
−x2
≥
1
a
= ax2 .
3. Let x ∈ R and c ∈ Q be such that c < x. Then 0 < ac < ax for all a > 1. If 0 < a < 1
then there exists d in Q such that x < d and ax > ad > 0.
4. Let us consider the case when a > 1. Following the properties 1. and 3. we see that the
exponential function f : x 7→ ax , x ∈ R is an injective mapping from R into R+ . To show
that expa is also a surjective mapping f : R 7→ R+ it suces to show that
∀y ∈ R+ ∃x ∈ R : ax = y.
Let us suppose the opposite is true, i.e.
∃y0 ∈ R+ ∀x ∈ R : ax 6= y0
(ax < y0 or ax > y0 ).
It follows from the binomial theorem (n ∈ N) that
an = [1 + (a − 1)]n = 1 + n(a − 1) + ∆ > n(a − 1),
∆ > 0.
So, there exists n in N such that an > y0 . (It is sucient to choose n as: n(a − 1) > y0 ,
i.e. n > y0 /(a − 1)). To have a−m ≤ y0 it suces to choose m as: 1/(m(a − 1)) ≤ y0 or
N 3 m ≥ 1/(y0 (a − 1)). If it would hold true: ax < y0 for any real x then also ax < an would
be true. However, the exponential function is increasing and therefore we must have x < n
for all x ∈ R - this, however, is in contradiction with the fact that the set of all natural
numbers N is unbounded.
Similarly, for y0 < ax one must have a−m < ax and therefore for any x ∈ R this would require
−m < x that is again impossible because of the fact that the set of all real numbers R is
also unbounded.
49
4.4 Elementary functions
50
Finally, the mapping f is bijective.
The fact that f is bijective in the case a ∈ (0, 1) follows from Theorem 2.2.3 (on composite
mappings). In fact,
−x
1
x
a =
, x ∈ R,
a
can be understood as a value of the composite mapping g ◦ h : R → R+ , where g : v 7→
(1/a)v , v ∈ R and 1/a > 1 is bijective mapping from R onto R+ and h : x 7→ −x, x ∈ R is
bijective mapping from R onto R.
5. It holds true:
inf ax := inf(0, +∞) = 0 (in R),
x∈R
and
sup ax := sup(0, +∞) = 0 (in R̄).
x∈R
The property 5 from the previous Theorem tells us that the exponential function is
bounded from bellow by zero and it is unbounded from above. The graph of the exponential
function is shown in Figure 4.3.
y=ax , a>1
y=ax , 0<a<1
1
1
x
x
Figure 4.3:
4.4.5 Logarithmic function
The exponential function is the bijective map from R onto R+ . Therefore its inverse mapping
exists.
Denition 4.4.4. The inverse function with respect to the exponential function (expa :
x 7→ ax , x ∈ R, a ∈ R+ \ {1}) is called the logarithmic function. It is denoted by loga .
As a consequence of Denition 4.4.4 and Theorem 4.4.3 we have the following Theorem
THEOREM 4.4.4. The logarithmic function loga : R+ → R has the following properties:
50
4.4 Elementary functions
51
1. For any a ∈ R+ \ {1} the logarithmic function is the bijective mapping.
2. For any a > 1 the logarithmic function is increasing on R.
3. For any 0 < a < 1 the logarithmic function is decreasing on R.
4. For any a ∈ R+ \ {1} we have:
inf loga (x) = −∞ (in R̄),
x>0
sup loga (x) = +∞ (in R̄).
x>0
Proof. The property 1 follows the part 4 of Theorem 4.4.3. The properties 2 and 3 follow
from Theorem 4.2.2. The property 4 is a direct consequence of Denition 2.1.4.
The value of the logarithmic function loga at the point x ∈ R+ is called the logarithm to
the base a of the number x. If we denote this logarithm by y then it follows from Denition
4.4.4 that ay = x, or aloga (x) = x.
One uses frequently the following notation: log(x) := log10 (x), for x > 0.
The next theorem summarizes the rules of computation with logarithms.
THEOREM 4.4.5. Let a ∈ R+ \ {1}, (x, y) ∈ R+ × R+ and z ∈ R, then:
1. loga (xy) = loga (x) + loga (y).
2. loga (x/y) = loga (x) − loga (y).
3. loga (xz ) = z loga (x).
4. If furthermore b ∈ R+ \ {1}, then
logb (x) =
loga (x)
.
loga (b)
Proof. The proof of the theorem is a simple consequence of denition of the logarithm and
of Theorem 4.4.2.
The graph of the logarithmic function is shown in Figure 4.4.
51
4.4 Elementary functions
52
y=loga HxL, a>1
y=loga HxL, 0<a<1
x
1
1
Figure 4.4:
52
x