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analysis_1_chunk_35 | other hand, min S ∈S, and inf S ≤s, for all s ∈S. In particular, inf S ≤min S. Thus,
inf S ≤min S and inf S ≥min S, which implies that inf S = min S.
9We could have choosen to translate by −l + c, for any c > 0. Hence the choice of c = 1 was arbitrary, but I
needed to choose something explicit, so I went for 1.
17 | analysis_1 | pdf |
analysis_1_chunk_36 | 2.3.2
Archimedean property of R
As we have already mentioned, given any two real numbers x, y we can always compare them,
that is, we can decide whether either x = y, or x < y or x > y. On the other hand, whenever
it makes sense, for example, if x, y are both non-negative real numbers with x < y, we may
ask a more gene... | analysis_1 | pdf |
analysis_1_chunk_37 | holds, in full generality, for any pair of positive numbers x, y.
Proposition 2.30 (Archimedeand property of R). Let x, y be real numbers, with x > 0, y ≥0.
Then there exists n ∈N∗such that nx > y.
Proof. If y = 0, then take n = 1. Then nx = 1 · x = x > 0 and we are done.
Let us now assume that y > 0. We make a proof b... | analysis_1 | pdf |
analysis_1_chunk_38 | largest (resp. the smallest) lower bound (resp. upper bound) of S, then whenever we take
a number c larger than inf S (resp. smaller than sup S), we must be able to find an element
s ∈S contained between inf S and c (resp. between c and sup S), that is, inf S ≤s < c (resp.
18 | analysis_1 | pdf |
analysis_1_chunk_39 | c < d ≤sup S).
Using this reasoning, we can characterize the infimum (resp. supremum) of S in the following
alternative way.
Proposition 2.32. Let S ⊂R be a non-empty set.
(1) A real number u is the supremum of S if and only if
(i) u is an upper bound for S, and
(ii) for all ε > 0, there is sε ∈S, such that sε > u −ε.
(... | analysis_1 | pdf |
analysis_1_chunk_40 | positive real number, but we are thinking that we have fixed one specific value for ϵ). Again,
we have to find an element sϵ ∈S (this element that we construct will depend on the initial
choice of ϵ, that is why we denote it as sϵ, to remind ourselves about the dependence from ϵ)
such that 1 −ϵ < sϵ.
If ϵ > 1 then 1 −ϵ < ... | analysis_1 | pdf |
analysis_1_chunk_41 | bound, by itw definition, then if we take any ϵ > 0, l + ϵ cannot be a lower bound for S.
This means that there exists an element of S (which will depend on the choice of ϵ in general,
cf. Example 2.33), call it sϵ, such that sϵ < l + ϵ, which shows that l satisfies also property (ii)
of Proposition 2.32.
We then prove t... | analysis_1 | pdf |
analysis_1_chunk_42 | then it must always coincide with inf S. Hence, the important bit of information contained
in Proposition 2.34 is that the minimum of any set S ⊆N indeed exists, a property that we
had already mentioned in Section 2.2.
Example 2.35. Let
S := {x ∈R | x ∈N∗and x is divisible by at least 5 distinct prime numbers} .
Then, ... | analysis_1 | pdf |
analysis_1_chunk_43 | Apply (2.35.d) with ε′ := 1
2. This yields an element sε′ of S such that
d < sε′ < d + ε′ = d + 1
2
Apply then again the above property of S, but now for ε′′ := sε′ −d > 0. Then, we can find
sε′′ ∈S such that
d < sε′′ < d + ε′′ = sε′ < d + ε′ = d + 1
2.
In particular, 0 < sε′ −sε′′ < d + 1
2 −d = 1
2. This gives a contr... | analysis_1 | pdf |
analysis_1_chunk_44 | 3.
To this end, let us consider S := {x ∈R | x2 ≤3}. First of all, S is a non-empty subset of R,
since 1 ∈S. Moreover, S is bounded: in fact, 3 is an upper bound and −3 is a lower bound for
S. [Prove it! Remember that for real numbers x > y > 0, then x2 > y2 > 0.] As S is bounded
then by Corollary 2.26 both the infimum ... | analysis_1 | pdf |
analysis_1_chunk_45 | for S. This immediately yields the desired contradiction, since sup S −1
n < sup S and sup S is
by definition the least of all upper bounds.
As sup S > 1, then sup S −1
n > 0 for all n ∈N∗. Hence to show that for some n ∈N∗, sup S −1
n
is an upper bound for S, it suffices to show that (sup S −1
n)2 > 3, since for x > 0, x... | analysis_1 | pdf |
analysis_1_chunk_46 | n2 + 2 sup S
n
< (sup S)2 + 1
n + 2 sup S
n
,
it suffices to show that there exists n ∈N∗such that (sup S)2 + 1
n + 2 sup S
n
< 3. But this is
equivalent to finding n ∈N∗such that n >
d
1+2 sup S . The existence of one such n ∈N∗is again
guaranteed by the archimidean property of R, cf. Proposition 2.30.
2.4.2
Integral par... | analysis_1 | pdf |
analysis_1_chunk_47 | (3) [x] = −[−x];
(4) {x} ∈] −1, 1[ and {x} = −{−x}
(5) x = [x] + {x};
(6) x ∈Z if and only if x = ⌊x⌋= ⌈x⌉= [x].
Example 2.42.
(1) [−4] = −4 and hence {−4} = 0. In general, if z ∈Z, then [z] = z,
{z} = 0.
(2) Considering the number x = π2 + π,
π2 + π =13.0111970546791518572971343831556540195108688066158964473882939
685... | analysis_1 | pdf |
analysis_1_chunk_48 | find rational numbers between two arbitrary real numbers.
Example 2.43. For example, is there a rational number c, such that 0 < c < π? The left
inequality, that is, 0 < c, is an easy one to guarantee. It suffices to choose c to be a positive
rational number. But, how do we guarantee that the inequality on right holds as ... | analysis_1 | pdf |
analysis_1_chunk_49 | We know that
√
2 is not a rational number. Then, how can we show that rational numbers are
arbitrarily close to
√
2? We could try to approximate
√
2 by means of rational numbers.
So, for example, what is a rational number that is close within
1
10 of
√
2?
Proposition 2.44
tells us that such approximation certainly exis... | analysis_1 | pdf |
analysis_1_chunk_50 | Easy case; we assume a = 0:
We have
1
b
+ 1 > 1
b and
1
b
+ 1 is a positive integer. We conclude that
0 <
1
1
b
+ 1 < b ,
so we can take c =
1
⌊1
b⌋+1.
General case:
Let us define the number n :=
j
1
b−a
k
+ 1. Then,
n =
1
b −a
+ 1 ⇒n >
1
b −a ⇒1
n < b −a
a = an
n < ⌊an⌋+ 1
n
≤an + 1
n
= a + 1
n < a + b ... | analysis_1 | pdf |
analysis_1_chunk_51 | 2.4.4
Irrational numbers are dense in R
The same property of density in R that we showed holds for Q, in the previous section, holds
also for the complement R \ Q of Q in R. The set R \ Q is called the set of irrational numbers.
Proposition 2.47. If a < b are real numbers, then there is c ∈R \ Q, such that a < c < b.
T... | analysis_1 | pdf |
analysis_1_chunk_52 | ordering on R?
Example 2.51. −
−
√
3
≤−
√
3 and | −
√
3| ≥−
√
3.
The inequalities in the above example hold for any real number: that is, for x ∈R
−|x| ≤x ≤|x|.
(2.51.a)
The absolute value behaves well with respect to the multiplication.
25 | analysis_1 | pdf |
analysis_1_chunk_53 | Example 2.52. |5 · (−3)| = | −15| = 15 = 5 · 3 = |5| · | −3|. Similarly,
(−
√
2) · (−4)
=
4
√
2
= 4
√
2 =
√
2 · 4 =
−
√
2
· | −4|.
We can generalize Example 2.52: indeed, for all x, y ∈R
|x| · |y| = |x · y|.
To prove this, you can just list all possible combinations for the signs of x, y (that is,
... | analysis_1 | pdf |
analysis_1_chunk_54 | still be able to say something. What the second part of the example suggests is that Is this a
general property of the absolute value over the real numbers?
Indeed, it is. A deep property of the absolute value is the so-called triangle inequality,
whose name is rooted in geometric considerations that we already clear a... | analysis_1 | pdf |
analysis_1_chunk_55 | Proposition 2.57 (Triangle inequality). For all x, y ∈R
|x + y| ≤|x| + |y|.
Proof. Recall that x ≤|x| and y ≤|y|. So, if x + y ≥0, then |x + y| = x + y ≤|x| + |y|.
Similarly, x ≥−|x| and y ≥−|y|. So, if x + y ≤0, then |x + y| = −x −y ≤|x| + |y|.
Exercise 2.58. Prove that for any x, y ∈R
|x −y| ≥||x| −|y||
27 | analysis_1 | pdf |
analysis_1_chunk_56 | 3
COMPLEX NUMBERS
When we work over the real numbers, we will be often working with functions of the form
f : R →R. We will be intersted in understanding and studying the properties (e.g., deriva-
tives, integrals, monotonicity) of certain classes of functions (e.g., continuous, differnetiable,
integrable functions). Of... | analysis_1 | pdf |
analysis_1_chunk_57 | where the second equality follows from the first by flipping the signs and multiplying the first
equation by 2. We rewrite the above equation in the form
aT 2 −2vT + 2c = 0,
(3.1.a)
where a, v, c are fixed real numbers, while the unknown that we are trying to compute is given
by T. As you probably already know, the above e... | analysis_1 | pdf |
analysis_1_chunk_58 | 3.1
Definition
As discussed in the previous section, one obstruction to finding real solutions already for
quadratic equations is the lack, within the real numbers, of the square root for negative real
numbers.
To define the complex numbers, we introduce a new number i called the imaginary unit.
The number i is the square... | analysis_1 | pdf |
analysis_1_chunk_59 | x, y ∈R, or in the form x+iy,both representations stand for the same complex number, as the
imaginary unit i commutes with all real numbers; that is,
s · i = i · s,
∀s ∈R.
Considering the notation for complex numbers introduced in Definition 3.2, in the form
x + yi, taking y = 0 and letting x vary in R, we immediately o... | analysis_1 | pdf |
analysis_1_chunk_60 | 3.2
Operations between complex numbers
We can add and multiply complex numbers using the standard formal properties of addition
and multiplication, always remembering that i2 = −1.
Example 3.6.
(1) (5 + 3i) + (2 −i) = (2 + 5) + (3 −1)i = 7 + 2i. In general:
(x1 + y1i) + (x2 + y2i) = (x1 + x2) + (y1 + y2)i.
(2) (1 −2i)(... | analysis_1 | pdf |
analysis_1_chunk_61 | complex plane identifies instead with the set of complex numbers whose real part is 0. Numbers
of this form are called purely imaginary numbers. For this reason, the line {x = 0} is called
the imaginary axis.
Using this representation complex numbers become vectors, and the sum of complex num-
bers is equal to the sum o... | analysis_1 | pdf |
analysis_1_chunk_62 | Images/sum_complex.png
Figure 3: Sum of complex numbers as vectors
Example 3.8.
(1) |3 −2i| =
√
32 + 22 =
√
25 = 5,
(2) | −3i| =
√
32 + 02 = 3
Using the representation of a complex number z ∈C as z = x + yi, then the formula for
the modulus |z| of z can be written as
|z| = |x + yi| =
p
x2 + y2.
As we can represent the ... | analysis_1 | pdf |
analysis_1_chunk_63 | Images/tr_ineq.png
Figure 4: Triangle inequality.
Conjugation is compatible with all operations by explicit computation: namely,for z1, z2 ∈
C, z3 ∈C∗,
z1 + z2 = z1 + z2,
z1 · z2 = z1 · z2,
z1
z3
= z1
z3
.
To verify the formulas above, it suffices to write all numbers involved as x + iy and expand all
the expressions ... | analysis_1 | pdf |
analysis_1_chunk_64 | ../Images/Complex_conjugate_picture_svg.png
Figure 5: Conjugate of a complex number.
We can use the above formula to better understand division between two complex numbers.
Given two complex numbers z and w, with w ̸= 0, we would like explicitly write z
w in the form
x + yi.
Example 3.11. We can try to turn the denomin... | analysis_1 | pdf |
analysis_1_chunk_65 | 4 .
Take now a non-zero complex number z, we have seen that its distance from the origin is
|z|. Let φ be its argument. The number
z
|z| has distance 1 from the origin, so it lies on the
trigonometric (or, unit) circle
S1 := {z ∈C | |z| = 1} = {x + yi ∈C | x, y ∈R, and x2 + y2 = 1}.
33 | analysis_1 | pdf |
analysis_1_chunk_66 | Hence, the real part of
z
|z| (resp. the imaginary part of z) is just cos(φ) (resp. sin(φ)), where
φ is the angle (measured in radiants) formed by the positive part of the real axis and the half
line passing through the origin and the point z on the complex plane, cf. Figure 6.
../Images/complex_angle.png
Figure 6
Thus... | analysis_1 | pdf |
analysis_1_chunk_67 | (3.15.b)
sin(φ + ψ) = cos(φ) sin(ψ) + sin(φ) cos(ψ).
Thus, the example above can be immediately generalized to show that for two non-zero
complex numbers z1, z2 ∈C, then arg z1 · z2 = arg z1 + arg z2, while we already saw that
|z1 · z2| = |z1| · |z2|,
z1 · z2 = |z1| · |z2| · (cos(arg z1 + arg z2) + sin(arg z1 + arg z2)... | analysis_1 | pdf |
analysis_1_chunk_68 | The above example shows that the polar form is really useful to compute, for example,
powers of complex numbers.
We can also use the polar form to divide complex numbers. As with multiplication when
the moduli (plural of the modulus) multiplied and the arguments added up, with division, we
have to do the inverse. That ... | analysis_1 | pdf |
analysis_1_chunk_69 | Proposition 3.19. Let φ, ψ ∈R and k ∈Z. Then,
(1) eiφ · eiψ = ei(φ+ψ);
(2) eiφ+2kπ = eiφ.
Proof.
(1) Use the trigonometric formulas in (3.15.b).
(2) As we measure angles in radiants, this is a simple consequence of the 2π-periodicity of
the sine and cosine functions.
We have mentioned above that we can use Euler’s form... | analysis_1 | pdf |
analysis_1_chunk_70 | t1 = b +
√
b2 −4ac
2a
,
t2 = b −
√
b2 −4ac
2a
.
In the above expression, what do we mean when we write
√
b2 −4ac, with a, b, c ∈C? By
definition of square root as the inverse operation to taking the second power of a given number,
√
b2 −4ac denotes those complex numbers t ∈C satisfying t2 = b2 −4ac. There are indeed
two... | analysis_1 | pdf |
analysis_1_chunk_71 | number t is a root of the equation W 2 −(b2 −4ac) = 0 (in the indeterminate W), also −t will
be a root of that same equation. Hence we can rewrite the formulas for the two solutions as
t1 = b + t
2a ,
t2 = b −t
2a ,
where t2 = b2 −4ac.
As we work over the complex numbers, it is not hard to see – and we shall see it bel... | analysis_1 | pdf |
analysis_1_chunk_72 | Moreover
p(T) = (T −t1) · (T −t2) · r(T).
Iterating, we obtain that we can factor any degree d polynomial p(T) into d linear factors, that
is, there d complex numbers t1, . . . , td (some of them may happen to coincide) such that
p(T) = (T −t1)(T −t2)(T −t3) . . . (T −td−1)(T −td).
(3.22.a)
The multiplicity of a root t... | analysis_1 | pdf |
analysis_1_chunk_73 | 3.5.1
Solving complex equations
We are going to learn how to solve some slightly more general equations, thanks to the polar
form.
Problem 3.25. For a ∈C∗and n ∈N∗, how many solutions does the equation T n = a in the
indeterminate T has? Can we compute them all?
Using what we discussed in the first part of Section 3.5, ... | analysis_1 | pdf |
analysis_1_chunk_74 | coefficients and we have to find real solutions for those (which, a priori, we do not know if it is
always possible).
Instead, we are going to take a slightly different approach, which is more natural given that
we are working in the field of complex numbers: the main idea is to write a in exponential form
and do some reaso... | analysis_1 | pdf |
analysis_1_chunk_75 | where ψ is an angle such that einψ = eiφ. Since R is a positive real number, then
n√
R is well
defined and it is in turn positive and real, which means that
n√
Reiψ is the polar form of a
complex number (uniquely determined by R and ψ).
How do we compute all possible choices that we have for φ? Since einψ = eiφ, we must... | analysis_1 | pdf |
analysis_1_chunk_76 | 4
SEQUENCES
Definition 4.1. A sequence is a function x : N →R.
Traditionally, we denote the value of the function x at n ∈N by xn, that is, xn := x(n).
We denote instead by (xn) the whole sequence.
Let us start by looking at a few simple examples of sequences.
Example 4.2.
(1) Let us fix a real number C ∈R. Then the cons... | analysis_1 | pdf |
analysis_1_chunk_77 | all natural number values of the index n. For example, the sequence (xn) defined as
xn := 1
n
is only well-defined when n ̸= 0.
In discussing sequences (and their limits, or lack thereof), we will mostly be concerned with
properties of a sequence which are eventually true. That means that we will look for properties
of a... | analysis_1 | pdf |
analysis_1_chunk_78 | Similarly to what we did for the case of subset of R, we would like to define the concept
of boundedness, boundedness from above/below also in the case of sequences. To this end, it
suffices to notice that given a sequence (xn)n≥l, then it uniquely defines a subset S ⊂R given
by all the values that the sequence takes,
S :=... | analysis_1 | pdf |
analysis_1_chunk_79 | S is a finite subset of R, it follows that it is bounded and possesses both maximum and
minimum, 1 and −1, respectively.
(3) Let (xn)n∈N be an arithmetic progression with a = 0, b = 2. Then
{xn | n ∈N} = {2n | n ∈N}
where the latter is the set of even numbers. In particular, (xn) is not bounded.
We have also the followi... | analysis_1 | pdf |
analysis_1_chunk_80 | Example 4.8.
(1) Let C ∈R and let (xn)n∈N be constant sequence of value C.
Then
{xn | n ∈N} = {C}. Hence (xn)n∈N is bounded.
(2) Let (xn)n∈N be an arithmetic progression of the form xn := a + nb, a, b ∈R. Then,
(i) the sequence is constant sequence of value a if and only if b = 0;
(ii) the sequence is increasing if and... | analysis_1 | pdf |
analysis_1_chunk_81 | q ̸= 1.
(iii) q = −1, then xn = (−1)na. This sequence is bounded but not monotone;
(iv) if q = 1
2, then xn =
a
2n . This sequence is strictly decreasing and bounded;
(v) if q = −1
2, then xn = (−1)na
(2)n . This sequence is bounded but not monotone;
(vi) if q = 2, then xn = 2n. This sequence is strictly increasing and... | analysis_1 | pdf |
analysis_1_chunk_82 | Notation 4.9. We will denote a recursive sequence (xn) defined by xn := f(xn−1, . . . , xn−j)
and with assighned initial values c0, c1, c2, . . . , cj−1 with the following notation
xn = f(xn−1, . . . , xn−j)
x0 = c0
x1 = c1
x2 = c2
...
xj−1 = cj−1
Example 4.10. Let us recall the... | analysis_1 | pdf |
analysis_1_chunk_83 | Induction
Induction is a method of proving a property P(k) which depends on a parameter k which
varies among the natural numbers that are greater or equal than a fixed natural number C ∈N.
More precisely, we want to be able to prove infinitely many different statements – all the versions
of property P(k), when k ≥C in N; ... | analysis_1 | pdf |
analysis_1_chunk_84 | (2) we then proceed to show that P(k) holds for a given value k = n ∈N (where n here is to
be treated as an unspecified number), under the assumption that we already know that
P(k) holds all choices of k starting from C and up to n −1. This second step is called
the inductive step of a proof by induction. The assumption... | analysis_1 | pdf |
analysis_1_chunk_85 | will show that P(k) holds for k = n, i.e., we will show that x2n = (2n + 1)n. Thus,
x2n =
x2n−1 + (2n)2
|
{z
}
recursive formula applied to x2n
=
x2(n−1) −(2n −1)2
|
{z
}
recursive formula applied to x2n−1
+(2n)2
= x2(n−1) −((2n)2 −4m + 1)
|
{z
}
=(2n−1)2
+(2n)2 = x2(n−1) + 4n −1
=
(2n −1)(n −1)
|
{z
}
inductive hypoth... | analysis_1 | pdf |
analysis_1_chunk_86 | and the Archimedean property Corollary 2.31 shows that the second inequality in (4.13.a) is
indeed satisfied for some kb ∈N – it suffices to take kb = [
p
|b|] + 1. Thus, any given b ∈R
cannot be an upper bound for the set of values of the sequence, since for m > kb, x2m > x2kb > b.
We can also prove that the sequence is ... | analysis_1 | pdf |
analysis_1_chunk_87 | What can we say in regards to the boundedness of a geometric progression xn = aqn, when
a ̸= 0 and |q| > 1? In this section we will show that, when a ̸= 0 and |q| > 1, then the sequence
is unbounded. In order to do that, we need to show that
∀C ∈R, ∃nC ∈R, such that |xnC| ≥C,
which is to say that there are no upper or ... | analysis_1 | pdf |
analysis_1_chunk_88 | Proposition 4.16 (Bernoulli’s inequality). Let q be a positive real number satisfying q > 1.
Then qn ≥1 + n(q −1).
To prove Proposition 4.16, we first need to introduce a few new mathematical tools and
results. The first is the concept of binomial coefficient.
Definition 4.17. If 0 ≤k ≤n are natural numbers, then
n
k
is ... | analysis_1 | pdf |
analysis_1_chunk_89 | n
X
i=0
n
i
xiyn−i.
Proof. This is an exercise in the exercise sheet for Week 4.
We are now ready to fully prove Bernoulli’s inequality.
Proof of Proposition 4.16. The inequality is an actual equality when n = 0, 1. Then, we can
assume that n ≥2. Let us apply the binomial formula, Proposition 4.19; then,
qn = (1 + (... | analysis_1 | pdf |
analysis_1_chunk_90 | As q > 1, then (q −1)i > 0, for all i ∈N∗. Thus, Pn
i=2
n
i
(q −1)i · 1n−i > 0 and
qn = (1 + (q −1))n =
n
X
i=0
n
i
(q −1)i · 1n−i
=1 + n(q −1) +
n
X
i=2
n
i
(q −1)i · 1n−i > 1 + n(q −1).
Example 4.20. Let (xn)n∈N be a geometric progression for some real numbers a and q, xn :=
aqn. Assume that |q| > 1 and a ̸=... | analysis_1 | pdf |
analysis_1_chunk_91 | 0 < q < 1 and a < 0 .
(4) (xn)n∈N is decreasing if and only if
0 ≤q ≤1 and a ≥0, or
q ≥1 and a ≤0.
(5) (xn)n∈N is strictly decreasing if and only if
0 < q < 1 and a > 0, or
q > 1 and a < 0 .
(6) (xn)n∈N is bounded from above if and only if |q| ≤1 or q > 1 and a ≤0;
(7) (xn)n∈N is bounded from below if and only if |... | analysis_1 | pdf |
analysis_1_chunk_92 | If the number y ∈R defined above exists then y is called the limit of the sequence (xn).
Once again, as we are talking about the limit of a sequence, this is only possible if the limit is
unique, when it exists. That is indeed the case.
Proposition 4.22. If a sequence (xn)n≥l converges, then its limit is unique.
Proof. ... | analysis_1 | pdf |
analysis_1_chunk_93 | 1
√n < ε.
The latter inequality is equivalent to the inequality √n > 1
ε, which in turn is equivalent to the
inequality n > 1
ε2 . Hence, for any fixed ε ∈R∗
+, we have to find an index nϵ ∈N such that
∀n ≥nε,
n > 1
ε2 .
Thus, for a fixed ε > 0, it suffices to take nε :=
1
ε2
+ 1.
Example 4.25. Let us introduce an examp... | analysis_1 | pdf |
analysis_1_chunk_94 | Remark 4.26. In fact, the argument used in Example 4.25 shows that if (xn)n≥l is a convergent
sequence, then for all 0 < ε ∈R there is an nε ∈N such that for all n, n′ ≥nε, |xn −xn′| < 2ε.
[Verify this fact using again the triangle inequality!] We will see that this observation can be
formalized into the notion of Cauc... | analysis_1 | pdf |
analysis_1_chunk_95 | bound) for the other elements of the sequence, because these elements are lying in the interval
I = [y −1, y + 1], R (resp. −R) is an upper bound (resp. a lower bound) for I, again, by
definition of R.
Example 4.28. The sequence (xn)n∈N defined by xn :=
√
n3 cannot be convergent as it is not
bounded.
Remark 4.29. The vic... | analysis_1 | pdf |
analysis_1_chunk_96 | Proof. We prove only (1). We refer to 2.3.3 and 2.3.6 in the book for the proofs of the others.
Fix 0 < ε ∈R. Let us try to explain how the proof should intuitively go. We need to show
that for big enough an index n ∈N, |(xn + yn) −(x + y)| is smaller than ε. However, as (xn),
(yn) are both convergent with limit x, y, ... | analysis_1 | pdf |
analysis_1_chunk_97 | 2 + ε
2 = ε.
This shows that (xn + yn) satisfies Definition 4.21 for convergence with respect to the finite
limit x + y.
Property (4) in Proposition 4.30 implies the following immediate corollary.
Corollary 4.32. Let (xn)n≥l be a converging sequence and let x := lim
n→∞xn be its limit. If
there exists n0 ∈N such that xn ≥... | analysis_1 | pdf |
analysis_1_chunk_98 | (i) dividing the numerator and the denominator by n is an operation that one cannot
perform for n = 0. So, after the second equality sign the expression that we wrote
does not make sense for n = 0. But this is not an issues, for when we study a
sequence for the purpose of understanding its convergence, we are only inte... | analysis_1 | pdf |
analysis_1_chunk_99 | 4n+5
. For n ≥1, we have 0 ≤3
n and 1 ≥5
n. Hence, for n ≥1:
xn = n2 + 2n + 3
4n + 5
= n + 2 + 3
n
4 + 5
n
≥n + 2
5
This shows that (xn) is not bounded and hence cannot be convergent by Proposition 4.27.
Using the method of the above exercise one can show the following result on limits of
sequences defined by means of r... | analysis_1 | pdf |
analysis_1_chunk_100 | 4.5
Squeeze theorem
Theorem 4.36 (Squeeze Theorem). Let (xn), (yn), and (zn) be three sequences. Assume
that:
(1) the sequences (xn), (zn) are convergent, and limn→∞xn = a = limn→∞zn; and
(2) there exists n0 ∈N such that for all integers n ≥n0,
xn ≤yn ≤zn.
Then the sequence (yn) is convergent, and
lim
n→∞yn = a.
Proof.... | analysis_1 | pdf |
analysis_1_chunk_101 | (1) When a = 0 or q = 0, 1, then the sequence is constant. If q = −1, then xn = (−1)na and
the sequence does not converge.
(2) If |q| > 1, we have already shown in Example 4.14 that the sequence is not bounded.
Hence, Proposition 4.27 implies that the sequence is also non-convergent.
(3) If |q| < 1, we show that the se... | analysis_1 | pdf |
analysis_1_chunk_102 | n→∞0 = 0 and lim
n→∞zn = lim
n→∞
9
2 ·
2
3
n = 9
2 · lim
n→∞
2
3
n = 9
2 · 0 = 0
by Example 4.38. So, the Squeeze Theorem 4.36 concludes our claim.
Example 4.40. Let (yn)n∈N be the sequence defined by yn :=
n√n. we show that lim
n→∞xn = 1.
To prove the above claim, we show that we can squeeze the sequence (yn) as ... | analysis_1 | pdf |
analysis_1_chunk_103 | Note that the sum on the right hand side for i = 4 is
n(n −1)(n −2)(n −3)
24
1
(√n)4 = n(n −1)(n −2)(n −3)
24n2
.
So, we know the desired inequality (i.e., that
n√n ≤1 +
1
√n) as soon as n ≥4 and n ≤
n(n−1)(n−2)(n−3)
24n2
. The latter is equivalent to
24n2
(n −1)(n −2)(n −3) ≤1.
However, we have just learned that
lim
n... | analysis_1 | pdf |
analysis_1_chunk_104 | squeeze |xnyn|:
0 ≤|xnyn| ≤M · |xn|,
Since lim
n→∞M · |xn| = M · lim
n→∞|xn| = 0, then also lim
n→∞|xnyn| = 0.
Example 4.43. Let (xn)n≥1 be the sequence defined by xn :=
1
n2 sin(n).
We show that
lim
n→∞
1
n2 sin(n) = 0.
Let us note that we do not know whether sin(n) does or does not converge in itself – it is possible
... | analysis_1 | pdf |
analysis_1_chunk_105 | where the last inequality follows from the fact that | sin(x)|
|x|
≤1 for all x ∈R. To show this,
just notice that |x| measures the length of the circle segment of angle (measured in radiants)
x, where we count multiple revolutions too, and | sin(x)| gives the absolute value of the y-
coordinate of the endpoint of the ... | analysis_1 | pdf |
analysis_1_chunk_106 | ◦Inductive step: we can assume that we know that statement for n and then we prove it
for n + 1 below:
yn+1 = 1 + 1
yn
≥1 + 1
2 ≥1,
and
yn+1 = 1 + 1
yn
≤1 + 1
1 = 2,
Example 4.47. Let us continue with Example 4.45. As we know that the sequence of Fibonacci
quotients is bounded, we can ask whether it converges or not.
I... | analysis_1 | pdf |
analysis_1_chunk_107 | 2
=
1
yn
−
1
1+
√
5
2
|
{z
}
we apply the definition of the sequence to yn+1, and
then as we found 1+
√
5
2
as the solution of y = 1 + 1
y
we may replace 1+
√
5
2
by 1 +
1
1+
√
5
2
=
yn −1+
√
5
2
yn 1+
√
5
2
≤|zn|
1+
√
5
2
Iterating this reasoning, we get that
zn+1 ≤|zn|
1+
√
5
2
≤
|zn−1|
(1+
√
... | analysis_1 | pdf |
analysis_1_chunk_108 | So, the Squeeze Theorem (Theorem 4.36) shows that lim
n→∞zn = 0. This in turn implies, by the
definition of zn that lim
n→∞yn = 1+
√
5
2
. Summarizing, we showed that for the Fibonacci sequence
(xn)
lim
n→∞
xn+1
xn
= 1 +
√
5
2
The number 1+
√
5
2
is also known as the Golden ratio.
The general approach to finding the limi... | analysis_1 | pdf |
analysis_1_chunk_109 | lim
n→∞xn = ¯x by showing that the ¯x satisfies the definition of limit for (xn).
Example 4.48. This method of finding the limit does not always work.
For example, consider the recursive sequence (xn)n∈N defined by
(
xn+1 = 1
2(xn + xn−1)
x0 = C.
Then applying Step 1 in the above recipe gives the equation x = 1
2(x + x). O... | analysis_1 | pdf |
analysis_1_chunk_110 | Example 4.49. Here is an example of a recursive sequence where our recipe does not work.
Let a, b ∈(0, +∞) and let (xn)n∈N be the recursive sequence defined by the recurrence relation
(
xn+1 = ax2
n
x0 = b.
Claim 1. For all n ∈N, xn = a2n−1b2n.
Proof. We prove the claim by induction on n ∈N.
◦Starting step: For n = 0, x... | analysis_1 | pdf |
analysis_1_chunk_111 | 4.5.2
Unbounded sets and infinite limits
Definition 4.50. Let (xn) be a sequence.
(1) We say that (xn) approaches +∞if for all C ∈R there is an index nC ∈N such that for
all integers n ≥nC, xn ≥C.
(2) We say that (xn) approaches −∞if for all C ∈R there is an index nC ∈N such that for
all integers n ≥nC, xn ≤C.
Notation 4... | analysis_1 | pdf |
analysis_1_chunk_112 | An example is the sequence defined by xn := −3 · 2n.
(3) On the other hand, if a ̸= 0 and q < 0, then (xn) is unbounded but it neither approaches
+∞not it approeaches −∞. For example xn = (−2)n is non-convergent but it also does
not admit limit equal to +∞or −∞.
The infinite limits satisfy some algebraic rules, and do no... | analysis_1 | pdf |
analysis_1_chunk_113 | Example 4.54. Let (xn)n∈N be the sequence defined by xn := 2n + sin(n). Then
lim
n→∞xn = lim
n→∞(2n + sin(n)) = +∞,
because lim
n→∞2n = +∞and sin(n) ≥−1.
Remark 4.55. Part (3) for both of the above propositions claims that if lim
n→∞|xn| = +∞and (yn)
is bounded then lim
n→∞
yn
xn = 0. It is important to remark that one ... | analysis_1 | pdf |
analysis_1_chunk_114 | (2) xn := 2n, yn := −n lim
n→∞(xn + yn) = +∞;
(3) xn := n, yn := −2n lim
n→∞(xn + yn) = −∞;
(4) xn := 2n, yn := (−1)nn, then
xn + yn =
(
n
for n odd,
3n
for n even.
Hence, xn + yn is unbounded, thus, non-converging, and its limit cannot be ±∞.
It is a homework to cook up similar examples for multiplication and division... | analysis_1 | pdf |
analysis_1_chunk_115 | (i’) If lim
n→∞xn = +∞and q ∈R∗
+, then lim
n→∞yn = +∞.
(ii’) If lim
n→∞yn = −∞and q ∈R∗
+ ∪{+∞}, then lim
n→∞xn = −∞.
Proof.
(1) Let us prove (i). The other case is proven analogously.
Fix C ∈R. As lim
n→∞xn = +∞, there exists nC ∈N such that ∀n ≥nC, xn ≥C. Taking
n′
C := max n0, nC, then ∀n ≥n′
C, yn ≥xn ≥C. Hence, l... | analysis_1 | pdf |
analysis_1_chunk_116 | 2n
= 1
2
3
2
n−2, and lim
n→∞
1
2
3
2
n−2 = +∞according to Example 4.52.
Hence, Theorem 4.57 part (1.i) yields lim
n→∞
n!
2n = +∞.
(2) Similarly, but using part (1.ii) of Theorem 4.57, then lim
n→∞−n!
2n = −∞.
4.6
More convergence criteria
We can apply the Squeeze Theorem 4.36 to obtain more convergence criteria.... | analysis_1 | pdf |
analysis_1_chunk_117 | Proof. We show here only the 0 ≤q < 1 case; the other case is similar, and is left as a
homework.
Fix ε := 1−q
2 . In particular ε > 0 and q + ε < 1. There is an index nε ∈N, such that
∀n ≥nε,
|xn+1|
|xn|
−q
< ϵ,
or equivalently,
∀n ≥nε,
q −ε < |xn+1|
|xn|
< q + ε,
thus, |xn+1| < (q + ε)|xn|. Denoting q := q ... | analysis_1 | pdf |
analysis_1_chunk_118 | n , then (xn) is convergent to 1, while
lim
n→∞
|xn+1|
|xn|
= lim
n→∞
n+2
n+1
n+1
n
= lim
n→∞
(n + 2)n
(n + 1)2 = 1.
(4) If xn := (−1)n, then (xn) is bounded but it does not admit a finite limit, while
lim
n→∞
|xn+1|
|xn|
= lim
n→∞1 = 1.
Hence, all possible behaviors of a sequence, in terms of its convergence or lack th... | analysis_1 | pdf |
analysis_1_chunk_119 | (1) If q < 1, then both (xn) and (|xn|) converge and their limit is 0.
(2) If q > 1 or q = +∞, then (xn) and (|xn|) both are non-converging sequences. Moreover,
lim
n→∞|xn| = +∞.
Proof. We prove part (2). The proof of the case is analogous.
We start assuming that q ∈(1, +∞).
If q = +∞, instead, then there exists n2 ∈N ... | analysis_1 | pdf |
analysis_1_chunk_120 | S −ε < xnε. In particular, for any integer n ≥nε:
S −ε
< xnε
| {z }
definition of nε
≤xn
| {z }
(xn) is monotone
≤S
| {z }
S is the supremum
< S + ε.
Example 4.64 (Nepero’s number e). Let us consider the sequence (xn)n≥1 defined by
x :=
1 + 1
n
n
, n ∈N∗.
Claim. The sequence (xn)n≥1 is strictly increasing.
Proof. We n... | analysis_1 | pdf |
analysis_1_chunk_121 | Similarly,
1 +
1
n + 1
n+1
=
n+1
X
i=0
1
i!
1 −
1
n + 1
. . .
1 −i −1
n + 1
(4.64.b)
= 1 + 1 +
n
X
i=2
1
i!
1 −
1
n + 1
|
{z
}
> (1−1
n)
1 −
2
n + 1
|
{z
}
> (1−2
n)
. . .
1 −i −1
n + 1
|
{z
}
> (1−i−1
n )
+
1
n + 1
n+1
> 2 +
n
X
i=2
1
i!
1 −
1
n + 1
. . .
1 −i −1... | analysis_1 | pdf |
analysis_1_chunk_122 | assume it for n −1, then the recursive formula gives it to us also for n. Hence, by induction,
∀n ∈N, xn > 0. In particular, the division in the definition does make sense.
Next, we claim that xn ≥1 for all integers n ≥1. Indeed, a similar induction shows that this
claim: indeed, for n = 0, we have x0 = 2 ≥1. Furthermor... | analysis_1 | pdf |
analysis_1_chunk_123 | where we obtained the last inequality using that xn ≥1 ≥
1
xn .
So, (xn) is decreasing (hence bounded from above) and also bounded from below by 1. In
particular, xn is convergent, and lim
n→∞xn ≥1. Hence to find the actual limit we may just apply
limit to the recursive equation to obtain that if y is the limit, then
y ... | analysis_1 | pdf |
analysis_1_chunk_124 | Then,
lim
n→∞sup{xk|n ≤k ∈N} = lim
n→∞sup{−1, 1} = lim
n→∞1 = 1,
and
lim
n→∞inf{xk|n ≤k ∈N} = lim
n→∞sup{−1, 1} = lim
n→∞−1 = −1.
Hence,
lim
n→∞sup xn = lim
n→∞yn = 1,
and
lim
n→∞inf xn = lim
n→∞zn = −1.
In general, it is not always easy to compute the liminf and limsup of a sequence. Nonetheless,
when a sequence is co... | analysis_1 | pdf |
analysis_1_chunk_125 | 4.9
Subsequences
Definition 4.70. Let (xn)n≥l be a sequence. A subsequence (yk)k∈N of (xn)n≥l is a sequence
sequence defined by yk := xnk where nk ∈N is defined by a function
f : N →{n ∈N | n ≥l}
k 7→f(k) =: nk
which is a strictly increasing function of k.
To say that f is strictly increasing simply means that ∀k ∈N, f(k)... | analysis_1 | pdf |
analysis_1_chunk_126 | k→∞
1 + 1
k
k!2
= e2.
We can ask whether (xn)n≥1 converges and, if so, what its limit is? Is lim
n→∞xn = e2?
The next proposition illustrates the (simple) connection between the convergence of a se-
quence and that of a subsequence.
Proposition 4.72. Let (xn) be a sequence.
If lim
n→∞xn = a ∈R, then for any subseque... | analysis_1 | pdf |
analysis_1_chunk_127 | Remark 4.74. It actually follows from the definition, that if a sequence (xn) is unbounded then
it admits a subsequence (yk), yk := xnk such that either lim
k→∞yk = +∞or lim
k→∞yk = −∞. [Try
to prove this claim!]
Hence, in view of the claim, we can ask whether for a bounded sequence (xn), there always
exists a convergen... | analysis_1 | pdf |
analysis_1_chunk_128 | convergent subsequences.
For example, defining xn := sin
nπ
4
1 + 1
n
n, then setting
(1) nk := 8k + 1, lim
k→∞yk = lim
k→∞xnk =
1
√
2e;
(2) nk := 8k + 2, then lim
k→∞yk = lim
k→∞xnk = e,
(3) nk = 8k + 5, then lim
k→∞yk = lim
k→∞xnk = −1
√
2e.
Example 4.77. Other times, given a sequence (xn), it is not quite possi... | analysis_1 | pdf |
analysis_1_chunk_129 | It is not hard to show that xn is increasing and bounded for a > 0 – the proof is similar
to the case where a = 1, using binomial expansion. In particular, xn is convergent, as it is
bounded – again the proof of this is similar to the case a = 1. However, if (xn) is convergent
we may compute the limit lim
n→∞xn by comp... | analysis_1 | pdf |
analysis_1_chunk_130 | In fact, for any
n, m ∈N, n < m, then
|xm −xn| < 10−n.
Thus, for a given ϵ > 0, it suffices to take nϵ ∈N such that 10−nϵ < ϵ – this is always possible
since lim
n→∞10−n = 0 – and thus
∀n, m ≥nϵ,
|xn −xm| < 10−nϵ < ϵ.
The important fact about Cauchy sequences is that they are always convergent.
Theorem 4.82. Let (xn) be ... | analysis_1 | pdf |
analysis_1_chunk_131 | Proof. (1) =⇒(2). First we assume that (xn) is convergent, and then we show that it is Cauchy
convergent. Let x := lim
n→∞xn and 0 < ε ∈R arbitrary. Then there is an n ε
2 ∈N such that for
all integers n ≥n ε
2 , we have |xn −x| ≤ε
2. Then, for any integers n, m ≥n ε
2 we have
|xn −xm| = |(xn −x) + (x −xm)| ≤ε
2 + ε
2 ... | analysis_1 | pdf |
analysis_1_chunk_132 | 5
SERIES
Let us start this section with the following motivating example.
Example 5.1 (Zeno’s paradox). Achilles races a tortoise. Achilles runs at 10 m/s, while the
tortoise moves at 0.1 m/s. Achilles gives the tortoise a head start of 100m.
Step 1 Achilles runs to the tortoise’s starting point, in 10s, while, at the ... | analysis_1 | pdf |
analysis_1_chunk_133 | To understand how to solve the problem above, we now introduce the concept of series.
Definition 5.2. Let (xn)n≥l be a sequence. The series associated to (xn)n≥l is the sequence
(sn)n≥l defined by the formula
sn :=
n
X
i=l
xi.
Given a sequence (xn) and the associated series (sn) defined above, we will refer to the
sequenc... | analysis_1 | pdf |
analysis_1_chunk_134 | (1) (Geometric series) Taking xk :=
1
2k , then sn =
n
X
k=0
1
2k = 1 −
1
2n+1
1 −1
2
= 2
1 −
1
2n+1
; in
general, for q ∈R, we can define xn = qn then sn =
n
X
k=0
qk.
(2) (Harmonic series) Taking xk := 1
k, then sn =
n
X
k=1
1
k;
(3) Taking xk := (−1)k 1
k, then sn =
n
X
k=1
(−1)k 1
k;
(4) Taking xk :=
1
k2 , then... | analysis_1 | pdf |
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