MT2502 Analysis
Mike Todd
19th November 2015
2
0.1 0.1
What What is Anal Analys ysis is? ?
Many mathematical objects, such as derivative and integral, and even the real numbers themselves, are best (most rigorously) defined as the limit of an infinite infinite process. process. The problem problem is to understa understand nd what it mea means ns to find the limit of an infinite process, since we can never really go to the ‘end’ of an infinite process. For example, the sine function is a nice continuo continuous us (a notion we’ll explore more later) function, so you might expect the function S (x) =
sin x 1
sin 2x sin3x sin3x + − sin − . . . 2 3
to be continuous, but in fact it’s not:
−2π
−π
0
π
2π
This particular example, now part of Fourier analysis (see MT2507) arose from the study of heat conduction (18th/19th century): at the time it wasn’t even even considered considered to be a genuine genuine function. function. In fact many of the advances advances in what now would be thought of as applied maths, from calculus onwards, were originally made without the logical underpinning of analysis, which led to confusion and paradox in the 18th/19th century. Analysis now provides a framework for these problems and has since blossomed into a beautiful and useful theory.
2
0.1 0.1
What What is Anal Analys ysis is? ?
Many mathematical objects, such as derivative and integral, and even the real numbers themselves, are best (most rigorously) defined as the limit of an infinite infinite process. process. The problem problem is to understa understand nd what it mea means ns to find the limit of an infinite process, since we can never really go to the ‘end’ of an infinite process. For example, the sine function is a nice continuo continuous us (a notion we’ll explore more later) function, so you might expect the function S (x) =
sin x 1
sin 2x sin3x sin3x + − sin − . . . 2 3
to be continuous, but in fact it’s not:
−2π
−π
0
π
2π
This particular example, now part of Fourier analysis (see MT2507) arose from the study of heat conduction (18th/19th century): at the time it wasn’t even even considered considered to be a genuine genuine function. function. In fact many of the advances advances in what now would be thought of as applied maths, from calculus onwards, were originally made without the logical underpinning of analysis, which led to confusion and paradox in the 18th/19th century. Analysis now provides a framework for these problems and has since blossomed into a beautiful and useful theory.
Chapter 1
The rationals and the reals The real number system is a very intuitive idea and has been used since at least the ancient ancient Greek times. Howeve However, r, while ostensibly ostensibly higher-pow higher-powered ered notions like derivative and integral (1820s) use this intuitive notion, a rigorous formal definition of the real numbers came much later (1860s-early 20th century). In this section, we’ll discuss the construction of the reals using precise definitions, logic, set theory and the rational number system (although we’ll have to leave out some of the construction in the interests of time). First recall that the set of rational numbers Q is defined as the collection of all fractions of the form pq where p Z and q N.
∈
∈ ∈
The first result is that adding/subtracting/multiplying two rational numbers yields a rational number: Q is closed under the usual algebraic operations. Proposition 1.1. If a, b Q , then a + b, + b, a b,ab a then b Q.
∈
∈
−
=0 ∈ Q . Moreo Moreover, ver, if b
The proof is an exercise. While the rationals are a nice set from many points of view, they are insufficient ficient for even even the most basic mathematics. mathematics. Eg a right-angle right-angled d triangle with sides of length 1 and 1, by Pythagoras’ Theorem has hypotenuse of length 2. But But 2, which by definition is the positive solution x to x2 = 2, isn’t rational:
√
√
3
4
CHAPTER 1. THE RATIONALS AND THE REALS
Proposition 1.2. There is no rational number x such that x2 = 2.
Proof. Suppose that, on the contrary, there is some x = pq where p, q N such that x2 = 2. First notice that if p and q have a common divisor (an integer which divides both) then we can factorise this out. Hence w.m.a. p and q have no common divisors.
∈
Since x 2 = 2, p2 =2 q 2
⇒ p2 = 2q 2,
(1.0.1)
we know that p must be even (note that the only way a square of a number can be even is if the number itself is even). So p is even, so there must exist n N such that p = 2n. Hence p2 = (2n)2 = 2q 2 2n2 = q 2 , so similarly q is even. So p and q have a common divisor of 2, contradicting our assumption. Hence the proposition must be true.
∈
⇒
This proposition implies that there are ‘holes’ in the rational numbers. In the same vein, the following example shows that while you can take a sequence of rational numbers which intuitively converge, they need not converge to a rational number. Example 1.1. Let x1 = 2 and for n 1, let xn+1 =
(i.e., x1 = 2, x2 = 22 +
1 2
= 32 , x3 =
xn 1 + 2 xn
( 32 ) 2
+
1
( 32 )
Claim. The sequence is decreasing, i.e., x1
Proof. We’ll show that x n
3 4
+
2 3
x2
x3
=
− x +1 0 for any n. n
=
17 12 .)
···.
5
1.1. ORDERING AND BOUNDING
First, since x 1 = 2 xn
− x +1 = x n
n
3
2
= x 2 , the claim is proved for n = 1. For n 2:
− − − − − − − − − − −−
1 = 2xn 1 = 2xn
xn 1 xn + = 2 xn 2 xn 2
x4n
1
+
1 + 4 4x2
2
1
xn
1
4x2n
n 1
−
1
1 1 = (x2n xn 2xn
1 2 = 2xn
1 = 2xn
2)
x2n 4
1
+1+
(x2n 1 2)2 4x2n−1
1 x2n−1
− 2
0.
The sequence is also bounded: 0 xn
∀n ∈ N.
x1 = 2
Suppose now that, contrary to the statement of the proposition, (xn )n does converge to a number x Q. But since xn+1 = x2n + x1n , any such limit must satisfy x 1 x 1 x = + = x 2 = 2, 2 x 2 x which by Proposition 1.2 is impossible if x Q.
∈
⇒
⇒ ∈
It is strange that a bounded decreasing sequence in Q doesn’t converge in Q. For such reasons we need to extend our set of numbers.
1.1
Ordering and bounding
Before defining the reals, we’ll need some abstract, but widely applicable, definitions. Definition 1.1. An ordered set (X, <) consists of a set X and a relation < on X such that
1. The trichotomy law holds: exactly one of the following is true: x < y,
x = y,
y < x.
6
CHAPTER 1. THE RATIONALS AND THE REALS
2. The transitivity law holds: if x, y, z x < z.
∈ X and x < y and y < z then
We’ll also use the notation x y which means that x < y or x = y; x > y, which means y < x; and x y which means y x. Ordered sets can be very abstract, but a fairly concrete example is (Q, <), i.e., the rational numbers with the usual ordering <. Definition 1.2. Given an ordered set (X, <) and A
⊆ X ,
1. u X is called an upper bound for A if
∈
∀a ∈ A,
a u.
If there is an upper bound, we say that A is bounded above
∈
2. X is called an lower bound for A if
∀a ∈ A,
a.
If there is a lower bound, we say that A is bounded below. If A is bounded both above and below, we say that A is bounded. N.B. There may be lots of upper/lower bounds for given sets A, which contrasts with the following notions. Definition 1.3. Let (X, <) be an ordered set and A
⊆ X .
1. An element M A is called a maximum for A if M is an upper bound for A, i.e., we require
∈
∀a ∈ A,
a M and M A.
∈
∈
2. An element m A is called a minimum for A if m is an lower bound for A, i.e., we require
∀a ∈ A,
m a and m A.
∈
7
1.1. ORDERING AND BOUNDING
Note that there can be at most one maximum for A: suppose we had two, say M 1 and M 2 . Then by definition, M 2
M 1 and M 1
M 2 , so M 1 = M 2 .
Similarly, there’s at most one minimum. Assuming it exists, we denote the unique maximum by max A,
{ ∈ }
max x, or max x A ,
∈
x A
and similarly the unique minimum by min A,
min x, or min x A .
{ ∈ }
∈
x A
Problem: Maxima/minima may not exist. Eg, set
{ ∈ Q : 0 < q 1}.
I := q
This set is bounded above (eg by 70) and below (eg by 0), but while it has a maximum (i.e., 1), it has no minimum. To cope with this kind of issue, define: Definition 1.4. Let (X, <) be an ordered set and A subset of X .
⊆ X a non-empty
1. Let U (A) denote the set of upper bounds for A. Then an element u U (A) is called a least upper bound/supremum for A if
∈
∀v ∈ U (A),
u v.
2. Let L(A) denote the set of lower bounds for A. Then an element L(A) is called a greatest lower bound/infimum for A if
∈
∀m ∈ L(A),
m .
There can be at most one supremum (Exercise), so we denote this, if it exists, by sup A, sup x, sup x A .
∈
x A
Similarly for infimum, inf A,
inf x,
∈
x A
{ ∈ }
{ ∈ }
inf x A .
8
CHAPTER 1. THE RATIONALS AND THE REALS
Lemma 1.3. For A a non-empty subset of Q:
1. if A has a maximum, then max A = sup A; 2. if A has a minimum, then min A = inf A. Proof. See tutorial sheet. Example 1.2. Recall the set I := x Q : 0 < x 1 . Since I has a maximum, 1, then it shares the same supremum. However, while it doesn’t have a minimum, inf I = 0 (first 0 is clearly a lower bound; suppose that > 0 is an infimum: then for large enough n, n1 < , so since n1 I , isn’t a lower bound for I ).
{ ∈
}
∈
Example 1.3. Given the ordered set (Q, <), define
{ ∈ Q : 0 < q, q 2 < 2}.
K := q
Clearly this set is bounded. However, Proposition 1.4. K has no supremum.
Proof. Suppose that the proposition is false, so there exists a least upper bound x Q for K . We’ll show that this is impossible.
∈
Claim. 2 < x2 .
Proof of claim. Suppose that the claim is false, so x2 2. Since x Q, x2 Proposition 1.2 implies that x 2 = 2, so in fact x 2 < 2. Therefore, 22− > 0. x+1 2 1 x So choosing n N large enough, we can ensure that 22− x+1 > n (*). Then
∈
2
1 x + n
∈
2x 1 1 1 = x 2 + + 2 = x 2 + 2x + n n n n 1 2 2 2 x + (2x + 1) < x + (2 x ) (by (*)) n =2
−
So x + n1 is a rational number whose square is < 2 and hence in K , so x can’t be an upper bound on K , a contradiction.
9
1.2. ABSOLUTE VALUE 2
2 The claim implies that x2 > 2, so x 2− > 0 and so we can choose m x 2 2 1 that x 2− x > m , which rearranges to
x2 Set y := x
−1
m
∈ N so
− 2x > 2 (**). m
< x. Then y2 = x 2
1 2x − 2x + 2 > x2 − > 2 m m m
by (**). Hence y < x is an upper bound for K , contradicting x being a least upper bound. Adding this all together, the proposition is true. We want to work in a set of numbers which has no such ‘holes’, i.e., all bounded subsets have supremum/infimum. Definition 1.5. An ordered set (X, <) is called complete if every non-empty set which is bounded above has a supremum. Note that (Q, <) is not complete by Proposition 1.4. In the following result we’ll use the notion of ‘field’ see eg Section 1.4 of Howie. Theorem 1.5. There exists an ordered field denoted (R, <) such that
i) Q
⊆R
ii) (R, <) is complete. We omit the proof of this theorem, as well as the theorem which states that (R, <) is essentially unique. The set above is called the real numbers R. Note that elements of R Q are called irrational numbers .
\
1.2
Absolute Value
(Note that this short section doesn’t particularly fit into this chapter, but it’ll be useful later.)
10
CHAPTER 1. THE RATIONALS AND THE REALS
Definition 1.6. Given x as
∈ R, the absolute value of x, denoted |x|, is defined |x| := −x if x < 0,
x
if x 0.
Theorem 1.6. Given x, y
∈ R and a 0,
i) x
| | 0; ii) | − x| = |x|; iii) |x| a iff −a x a; iv) |xy| = |x||y |; v) (Triangle inequality) |x + y | |x| + |y|; vi) (Reverse triangle inequality) |x| − |y| |x − y |. Proof. i)-iv): Exercise.
−|x| x |x| and −|y| y |y|, summing: −(|x| + |y|) x + y |x| + |y|,
v): Since
so applying iii), we obtain v). vi) Exercise. N.B. Absolute value is often used as a way of finding the distance between two real numbers x, y R: i.e., this is x y .
∈
| − |
Chapter 2
Sequences Now that we’ve laid a solid foundation for the real number system we can begin to address more ‘analytic’ issues; the first of these being sequences.
2.1
Sequences and convergence
Definition 2.1. A sequence of real numbers is a function f : N
→R
going from the rationals to the reals. Usually we’ll denote f (n) by xn and write the sequence as (xn )n ,
(xn )n∈N ,
(xn )∞ n=1 ,
{ }
(x1 , x2 , . . .).
Note that in contrast to set notation , the order is important here, eg ( 1, 1, 1, 1, . . .) means something different to 1, 1, 1, 1, . . . = 1, 1 . Moreover, (1, 2, 3, 4, 5, . . .) = (2, 1, 3, 4, 5, . . .).
−
−
{−
−
} {− }
Sometimes we have a nice formula for the n-th term of a sequence, eg (2, 4, 6, 8, . . .) = (2n)n i.e., x n = 2n;
1 1 1 1, , , , . . . = 2 3 4
1 n
n
i.e., y n =
1 . n
If b R then the sequence (xn ) = (b , b , b , . . .) is called Example 2.1. the constant sequence b. Eg the constant sequence 300 is (300, 300, 300, 300, . . .).
•
∈
11
12
CHAPTER 2. SEQUENCES
• Given the expression x
n =
( 1)n for n
−
∈ N, we obtain
(xn )n = ( 1, 1, 1, 1, . . .).
−
• For a
n =
( 1)n 2n
−
−
∈ N, we obtain (a ) =
for n
n
1 1 1 2 , 4 , 8 , . . .
−
−
Definition 2.2. Let (xn )n be a sequence of real numbers and x
.
∈ R.
1. We say that (xn )n converges to x (as n tends to infinity) if
∀ε > 0 ∃N ∈ N s.t. ∀ n ∈ N, n N ⇒ |x − x| ε. n
In this case we write
→ x as n → ∞ is called convergent if there exists x ∈ R such that
lim xn = x, lim xn = x, xn
→∞
n
n
A sequence (xn )n limx→∞ xn = x ; otherwise the sequence is called divergent.
2. We say that (xn )n tends to infinity (as n tends to infinity) if
∀K ∈ R ∃N ∈ N s.t. ∀ n ∈ N, n N ⇒ x
n
K.
In this case we write lim xn = n
∞,
lim xn =
→∞
n
∞, x → ∞ as n → ∞ n
3. We say that (xn )n tends to minus infinity (as n tends to infinity) if
∀K ∈ R ∃N ∈ N s.t. ∀ n ∈ N, n N ⇒ x K. lim x = −∞, lim x = −∞, x → −∞ as n → ∞ →∞ n
n
n
n
n
n
In other words, if xn x then no matter how small ε > 0 is chosen, there will be some stage in the sequence (stage N ) beyond which all elements xn will lie in the interval [x ε, x + ε].
→
Example 2.2.
1
n2 n
−
is convergent:
Proof. Let ε > 0. Given N
So choosing N so that
∈ N, if n N then 1 1 1 − 0 = . n2 n2 N 2
1
N 2
< ε, i.e., N >
√ 1ε completes the proof.
13
2.1. SEQUENCES AND CONVERGENCE
Example 2.3. Let x be some real number and let (xn )n be the constant sequence (x , x , x , . . .). Then xn
→ x as n → ∞.
Proof. Let ε > 0. Then given N
∈ N, n N implies |x − x| = |x − x| = 0 ε, n
as required. Example 2.4. The sequence
1
n n
converges to 0 as n
→ ∞.
∈ N such that N 1 . Then |x − 0| = n1 − 0 = n1 N 1 ε,
Proof. Let ε > 0. Fix N
ε
n
as required. Example 2.5. The sequence
4n+300 5n+2 n
4n + 300 5n + 2
Proof. Let ε > 0. For N
4n + 300 5n + 2
−
∈
is convergent: in fact
→ 54 as n → ∞.
N, n N implies
−
4 5(4n + 300) 4(5n + 2) 1492 1492 = = < . 5 5(5n + 2) 25n + 10 25N
So setting N 1492 25ε , we complete the proof.
Example 2.6. For xn = ( 1)n , the sequence (xn )n is divergent.
−
(Idea: suppose there is a limit x and then show that there is some ε > 0 for which xn x > ε even for some very large n.)
| − |
Proof. Suppose that there is a limit, call it x R. Then set ε = 12 . Since x is a limit, there exists N such that n N implies xn x 12 . Since there are arbitrarily large even numbers, there is always some n N (eg n = 2N ) such that xn = 1, so 1 x 12 , in particular, x 12 . On the other hand, since there are arbitrarily large odd numbers, there is always some n N (eg n = 2N + 1) such that xn = 1, so 1 x 12 , hence x 12 . The inequality 12 x 12 is impossible, so there is no limit.
∈
| − |
| − |
−
−
| − − |
−
14
CHAPTER 2. SEQUENCES
√
Example 2.7. ( n)n tends to
∞ as n → ∞.
Proof. Let K > 0. Then set N N to be greater than K 2 . Hence n implies x n = n N > K , as required.
∈
√ √ 1
−
Example 2.8.
n
n
n
N
tends to
−∞ as n → ∞.
∈ R. Then set N ∈ N so that N > 1 − K . So n N implies
Proof. Let K
xn =
1 n
− n 1 − N < K,
as required. Example 2.9. (Standard sequences) Let a
∈ R.
1. If a < 1 then (an )n converges to 0.
| |
2. If a > 1 then (an )n tends to
∞.
Proof. We assume Bernoulli’s Inequality: for x (1 + x)n
0 and n
∈ N,
∗
1 + nx
( ).
(Proof is an easy exercise in induction.) 1 |a| − 1 > 0, for N ∈
| |
1) Using the fact that a < 1 to deduce that n N , n
n
|a −0| = |a|
=
1
n
1
1
| | − | | − | | − 1+
1
a
1
∈ N such that N
so choosing N
1 + n
1
1 ε |a |
−1
1
a
1
1 + N
, we are finished.
∈ R. Then for n ∈ N and n N , a = (1 + (a − 1)) 1 + n(a − 1) N (a − 1).
2) Let K
n
So if N
K , a 1
N and
− we are finished.
n
1
a
1
,
15
2.2. LIMIT THEOREMS
2.2
Limit Theorems
In this section we’ll consider uniqueness and algebraic properties of limits. Theorem 2.1. If a sequence is convergent, its limit is unique.
Proof. Let (xn )n be a sequence that converges to both s and t. Let ε > 0. Then since xn s, there exists N 1 N such that n N 1 implies
→
∈ |x − s| ε. Similarly, since x → t, there exists N 2 ∈ N such that n N 2 implies |x − t| ε. Therefore taking n max{N 1 , N 2 }, |s − t| = |s − x + x − t| |s − x | + |x − t| ε + ε = 2ε. n
n
n
n
n
n
n
Since this holds for any ε > 0, this means s = t. If a sequence can have very large values for large n, this can cause problems for the algebraic properties of that sequence (eg see next theorem). The following definition deals with that. Definition 2.3. A sequence (xn )n is called bounded if there exists M 0 such that xn M for all n N.
| |
∈
Theorem 2.2. Every convergent sequence is bounded.
Proof. Suppose that (xn )n is a convergent sequence and denote its limit by x R. Then there exists N N such that xn x 1 for all n N . So n N implies xn = xn x + x xn x + x .
∈
∈ | − | | | | − | | − | | | So defining M := max {|x1 |, |x2 |, . . . , |x |, |x| + 1 }, we deduce that |x | M for all n ∈ N. N
n
The next theorem simplifies many questions involving combinations of more than one sequence.
16
CHAPTER 2. SEQUENCES
R and (xn )n , (yn )n be convergent sequences with Theorem 2.3. Let a . Then xn x and yn y as n
→
∈
→
→∞
1. xn + yn
→ x + y ; 2. x y → xy ; 3. If y = 0 for all n ∈ N and y = 0 then 1 → 1 ; → ; = 0 for all n ∈ N and y = 0 then 4. If y 5. ax → ax and a + x → a + x. n n n
yn
y
n
xn yn
x y
n
n
Proof. 1) Let ε > 0. Since x n
→ x, there exists N 1 ∈ N such that |x − x| 2ε ∀n N 1. Similarly, since y → y, there exists N 2 ∈ N such that |y − y| 2ε ∀n N 2. Set N := max{N 1 , N 2 }. Then n N implies |(x + y ) − (x + y)| = |(x − x) + (y − y)| |x −x|+|y −y| 2ε + 2ε = ε, n
n
n
n
n
n
n
n
n
as required. 2) Let ε > 0. First note that
|x y − xy| = |x y − x y + x y − xy| |x y − x y | + |x y − xy | |x ||y − y | + |y ||x − x| (∗) → x, there exists N 1 ∈ N such that |x − x | 2(| |+1) for all n n
n n
n
n n
n
n
Since xn n N 1 .
n
n
n
n
ε y
n
| | ∗
By Theorem 2.2, (xn )n is bounded, i.e., there exists M 0 such that xn M for all n N . Since also yn y, there exists N 2 N such that ε yn y 2(M +1) for all n N 2 . Let N := max N 1 , N 2 . Then by ( ), n N implies
| − |
∈
→
{
∈ }
|x y − xy| |x ||y − y| + |y||x − x| M · 2(M ε+ 1) + |y|· 2(|y|ε + 1) ε, n n
n
n
n
17
2.2. LIMIT THEOREMS
as required. 3) Let ε > 0. First observe that
1 yn
−
−
1 y yn y yn = = y yn y y yn
| − | (∗∗). | || |
→ y, there exists N 1 ∈ N such that n N 1 implies |y − y | |y| ∀n N 1.
Since yn
n
2
Therefore,
|y| = |y + (y − y )| |y | + |y − y | |y | + |y2| n
n
n
n
n
| | |2| for all n N 1. Moreover, there exists N 2 ∈ N such that n N 2 implies ε|y |2 y y | − | ∀n N 2. y
which implies yn
n
2
So setting N := max N 1 , N 2 , ( ) implies
{
as required. 4) Since
xn yn
} ∗∗
= x n
·
1 yn
1 yn
−
1 y
| |2 2|y | |2| εy
y
= ε,
, this follows by 2) and 3).
5) Exercise. Example 2.10. Let p
∈ N. Then (x )
n n =
1
2+
np n
converges to 2.
Proof. Let (an )n be the constant sequence 2 and let (bn )n be the sequence given by bn = n1p for all n N. Then x n = an + bn .
∈
Further, let cn = n1 for all n N. We know that n1 0 as n . So by p Theorem 2.3, cn cn cn = c n 0 as n . Hence a n 2 and b n 0 as n , so by Theorem 2.3, xn 2 + 0 = 2 as n .
→ ∞
· ··
∈ → →
→ ∞
→ → ∞ → → →∞
18
CHAPTER 2. SEQUENCES
(Above we used the constant sequence 2 and the sequence (1/n)n as building blocks for which we knew the limiting behaviour. We’ll be able to use this type of idea from here on, unless we are asked to prove ‘from first principles’ (or a similar phrase) that a sequence converges.)
− →
Example 2.11. The sequence
n2 +4n n3 3
n2 + 4n n3 3
n
is convergent, in fact
0 as n
−
n
→ ∞.
Proof. 1 1 + n42 + 4 n2 + 4n n n = = 3 n3 3 1 n3 1 3
−
Since
1 n
−
→ 0, by Theorem 2.3, 1 +4 n2 + 4n n = n3 3 1 3
−
2.3
1 2
−
1 2
→ − n
1 3
n
n
1 3
.
n
0 + 4 02 = 0. 1 3 03
· − ·
Monotone Sequences and Subsequences
In most of our examples of convergent sequences so far, proving convergence has involved guessing a limit before taking any further steps. In this section we’ll develop tools which can overcome this problem: even when we don’t have a candidate for a limit, in some cases we can use the completeness of the reals to guarantee that one exists. Definition 2.4. Let (xn )n be a sequence of real numbers.
• We say that a sequence is increasing if it satisfies x 1 x2 x3 ·· · . • We say that a sequence is decreasing if it satisfies x 1 x2 x3 ·· · . • We say that a sequence is monotone if it is either increasing or decreasing.
Example 2.12.
a) (an ) = (n)n is increasing;
19
2.3. MONOTONE SEQUENCES AND SUBSEQUENCES
b) (bn ) = (4n )n is increasing; 1
−
c) (cn ) = 3
n n
is increasing;
d) (dn ) = (1, 1, 2, 2, 3, 3, 4, 4, . . .)n is increasing (despite some adjacent terms being equal); e) (en ) =
3
n n
is decreasing;
f) (f n ) = ( 2n)n is decreasing;
−
g) (gn ) = (k , k , k , k , . . .) for some k
∈ R is both increasing and decreasing;
h) (hn ) = (( 1)n )n is not monotone; i) (in ) =
−− ( 1)n n
n
is not monotone.
Note that (cn )n , (en )n , (gn )n , (hn )n , (in )n are all bounded, while (an )n , (bn )n , (dn )n are unbounded. Also note that the bounded monotone sequences ( (cn )n , (en )n , (gn )n ) are convergent, while the unbounded sequences are not (as in Theorem 2.2 ). These are examples of a broader phenomenon: Theorem 2.4 (Monotone Convergence Theorem). Let (xn )n be a monotone sequence. Then the following are equivalent:
1. (xn )n is convergent; 2. (xn )n is bounded. Proof. (1
⇒ 2): This is true by Theorem 2.2.
(2 1): Assume first that (xn )n is bounded and increasing. Let A := xn : n N . Since we have assumed that A is bounded, A has a supremum x = sup A R by the completeness of the reals.
⇒ ∈ }
{
∈ Claim. x → x as n → ∞. n
Proof of Claim. Let ε > 0. Since x = sup A, x ε is not an upper bound for A. Hence there exists N N such that xN x ε. Since (xn )n is increasing, n N implies
−
∈
−
x ε xN xn so the claim is proved.
−
| − x| ε ∀n N,
x, i.e., xn
20
CHAPTER 2. SEQUENCES
The proof where (xn )n is decreasing is similar. Corollary 2.5. Suppose that (xn )n is a bounded sequence.
1. If (xn )n is increasing then limn→∞ xn = supn xn . 2. If (xn )n is decreasing then limn→∞ xn = inf n xn . Proof. This follows immediately from the proof of Theorem 2.4. Example 2.13. If 0 < a < 1 then an 0 as n already proved this, but here’s an alternative proof.)
→ ∞ .
→
(Note that we’ve
Proof. Let xn = an . Since an+1 < an for all n N, (xn )n is a decreasing sequence bounded above by a and below by 0. So MCT implies there is a limit x R. Therefore (see eg TS2 Q11) x n+1 x also. So x n+1 = an+1 = axn x. But since xn x, Theorem 2.3 implies that axn ax. Hence ax = x. Since a = 1, the only solution is x = 0.
→
∈
∈ →
→
→
| |
Exercise: extend this to a < 1.
√
Example 2.14. Define (xn )n by x1 = 1 and xn+1 = 1 + xn for n The sequence is bounded and increasing, thus convergent: indeed
→
xn
∈ N.
√
1+ 5 . 2
Proof. Claim 1. xn
| | 2 for all n ∈ N.
Proof of Claim 1. We prove this by induction. For n = 1, x1 = 1 Suppose that the claim is true for n N. Then xn+1 =
√ 1 + x
∈
n
2.
√ 1 + 2 = √ 3 2,
so the claim is true for n + 1. Hence by induction, the claim holds. Claim 2. (xn )n is increasing, i.e., for all n
√ √
∈ N, x
n
xn+1 .
Proof. For n = 1 we have x 1 = 1 1 + 1 = x 2 . Assume the claim is true for n N. Then x n+1 = 1 + xn 1 + xn+1 = x n+2 , so the claim is true for n + 1. Hence by induction the claim holds.
∈
√
21
2.3. MONOTONE SEQUENCES AND SUBSEQUENCES
∈ R such that x → x. By Theorem 2.3, 1 + x → 1 + x, and by TS2 Q5, setting y = 1 + x and √ √ √ √ y√ = 1 + x, we have y → y, i.e., 1 + x → 1 + x. Since x +1 = √ 1 + x → x, this means that x = 1 + x and hence x 2 = 1 + x. Therefore √ √ 1+ 5 1− 5 Combining these claims and the MCT there exists x n
n
n
n
n
n
n
n
x =
or x =
2
2
.
∈ N, only the positive solution for x is possible,
But since x√ n 0 for all n 1+ 5 i.e., x = 2 .
Example 2.15. Let (xn )n be given by xn =
1 1 1 + + + 12 22 32
·· · + n12 .
The sequence is bounded and increasing, and hence convergent (in fact the 2 limit is π6 , but we won’t show that here). Proof. Claim. 0 < xn
2 for all n
∈ N.
Proof of Claim. We write 1 1 1 1 1 1 1 1 + 2 + 2 + 2 + + 2 1 + + + + 2 1 2 3 4 n 1 2 2 3 3 4 1 1 1 1 1 1 1 1 =1 + + + + + 1 2 2 3 3 4 n 1 n 1 =1+1 2. n
0
···
− − − · · · · ·
· − −
·· · + (n −1 1)n
−
So (xn )n is bounded. Clearly (xn )n is increasing since xn =
1 1 1 + 2 + 2 + 2 1 2 3
·· · + n12 112 + 212 + 312 + · ·· + n12 + (n +1 1)2 = x +1, n
so the proof is complete. Definition 2.5. Let (xn )n be a sequence and (mk )k be a sequence of natural numbers m1 < m2 < m3 < . then the sequence (xmk )k is called a subsequence of (xn )n .
· · ·
22
CHAPTER 2. SEQUENCES
Note that in this definition, we must have m k Example 2.16. The sequence (xn )n =
1
n n
k for all k
∈ N.
has various subsequences, eg
• (s ) = (x2 ) = 21 = 12 , 14 , 16 , . . . ; • (t ) = (x +3) = 14 , 15 , 16 , . . . ; • the sequence (x ) is a subsequence of itself. k k
k k
k k
k
k k
k
n n
Note that eg 13 , 12 , 14 , 15 , 16 , . . . is not a subsequence of (xn )n : the order of terms must be preserved.
Theorem 2.6. Suppose that (xn )n converges to x sequence also converges to x R.
∈
∈ R.
Then any sub-
Proof. Let ε > 0. Since xn x then there exists N N such that n N implies xn x ε. Since (xmk )k is a subsequence of (xn )n . Let K N be such that k N implies mk N . Then
| − |
→
∈
∈
|x − x| ε for all k K, mk
as required. N.B. that we could have made the statement of the theorem into an ‘if and only if’ since if any subsequence of (xn )n converges to x, then certainly (xn )n converges to x - since (xn )n is a subsequence of itself. Theorem 2.7 (Monotone Subsequence Theorem). Any sequence of real numbers (xn )n has a monotone subsequence.
Proof. In this proof, we say that p x p
∈ N is a ‘peak’ if
xn for all n
p.
Case 1: (xn )n has infinitely many peaks. We list the peaks in the order in which they occur: p1 < p2 < p3 < . From the definition of peaks, we have x p1 x p2 x p3 ,
·· ·
·· ·
so (x pk )k is the monotone (decreasing) subsequence we require.
23
2.3. MONOTONE MONOTONE SEQUEN SEQUENCES CES AND SUBSEQU SUBSEQUENCES ENCES
Case Case 2: (xn )n has finite finitely ly many many peaks peaks.. Ag Agai ain n we list list all all the the peaks peaks in the order in which they occur: p1 < p2 < < pN . Let t1 > pN . Sinc Sincee t1 is not not a peak, peak, ther theree exis exists ts t2 > t1 such such that xt2 > xt1 . Since t2 is also not a peak, there exists t3 > t2 such that xt3 > xt2 . Proce Proceed edin ing g in this this way way, we obtain obtain an infinit infinitee sequen sequence ce t1 < t2 < t3 < t4 < such that xt1 < xt2 < xt3 < xt4 < . So (x (xtk )k is the monotone (increasing) subsequence we require.
· · ·
···
· · ·
Theorem 2.8 (Bolzano-Weierstrass 2.8 (Bolzano-Weierstrass Theorem). Theorem). Let (xn )n be a bounded sequence quence of real real numbers. numbers. Then there exists a subsequenc subsequence e (xmk )k and a real number x such that xmk x as n .
→
→∞
Proof. The Monotone Subsequence Theorem guarantees the existence of a monotone monotone subsequen subsequence ce (xmk )k . The Then sin since (x (xmk )k is also also bounded bounded,, the Monotone Convergence Theorem implies that (x (xmk )k converges to some limit x. Example Example 2.17. Let (xn )n = (( ( ( 1)n )n . Then this is a bounded sequence, eg bounded above by 2 and below by 2. So the Bolzano-Weierstrass Theorem implies implies there exists a convergent onvergent subsequenc subsequence. e. In fact in this case case we can can (1, 1, 1, . . .) .) converges to 1. check this by hand: eg the subsequence (x4n )n = (1, On the other hand, the subsequence (x6n+1 )n = ( 1, 1, 1, . . .) .) converges to 1.
− − −
− − −
−
Example 2.18. Consider (sin (sin n)n . Note that this sequence is bounded above by 1 and below by 1. Hence the Bolzano-Weierstrass Theorem implies that there exists (mk )k such that (xmk )k is a conver convergent gent subsequenc subsequence. e. (In fact, x [ 1, 1] there exists a subsequence while we won’t show this here, for any x which converges to x.)
− −
∈ −
24
CHAPTER CHAPTER 2. SEQUENCES SEQUENCES
Chapter 3
Series Outside maths, ‘sequence’ and ‘series’ are often understood to mean the same thing. However, in maths they are distinct notions. Notation: Given a sequence (x (xn )n , for n write
m , we ∈ Z and m ∈ N with n m,
xn + x + xn+1 + x + xn+2 + Definition 3.1. x1 + x2 + x3 + by (xn )n .
m
+ x · · · + x
m =
xk .
k =n
• Given∞a sequence (x ) , we can formally write · · · as =1. This is called the (infinite) series generated n n
k
• For each n ∈ N, we write
n
sn = x 1 + x + x2 +
+ x · · · + x
n =
the nth partial sum of our series.
• We say that the series
∞
n=1 xn converges if
xk ,
k=1
the partial sums converge, lim n→∞ sn exists. In this case, ∞ i.e., the limit lim n=1 xn is the (infinite) sum of our sequence (xn )n .
(s • If (s
n )n does
or sn
s n + not converge, we say that the series diverges. If s ∞ ∞ , we write n=1 xn = + , n=1 xn = + respectively.
→ −∞
∞
N.B. If (s (sn )n doesn’t converge or tend to + ∞ of n=1 xn as a sum.
25
∞
→ ∞
∞ or −∞, then we don’t think
26
CHAPTER 3. SERIES
Example 3.1. Consider the sequence (xn )n where xn = N. The associated series is
1 n(n+1)
for all n
∈
∞ ∞ xn =
n=1
1 . n(n + 1) n=1
The nth partial sum is n
sn =
k=1
1 1 1 1 = + + + k(k + 1) 1 2 2 3 3 4 =
· − · −· − 1 1
=1
Since
1 n+1 n
1 · ·· + n(n + 1)
−
1 + 2 1 . n + 1
1 2
1 3
1 3
+
1 4
+
· · · +
− 1 n
−
converges to 0 and (1)n converges to 1, then (sn )n = 1
converges to 1. Hence
1 n + 1
∞
1
n+1
n=1 xn is a convergent series with sum equal to 1.
Example 3.2. Let (xn )n = (( 1)n )n . Then the series this generates is ∞ ( 1)n. Here the nth partial sum is n=1
−
−
n
sn =
− − ( 1)k =
k=1
1
0
if n is odd , if n is even .
Hence (sn )n is a divergent sequence (check), so (xn )n = (( 1)n )n is divergent.
−
In the first example, but not the second, the sequence which generated the sequence converged to zero. In fact this is necessary for a sequence to converge: Theorem 3.1. Let .
∞
∞
n=1 xn be
a convergent series. Then xn
→ 0 as n →
∞ xn being convergent means that the partial sums (sn )n conProof. n=1 verge to some limit s. Note that
− s −1 for all n 2. Since s → s and s −1 → s, this means that x → s − s = 0, as required. xn = s n
n
n
n
n
n
27
→
It might be nice if the converse of this theorem were true (i.e., xn 0 implies ∞ xn converges), but as in the next example, this isn’t true, which leads n=1 to some fundamentally important theory.
Example 3.3. The harmonic series is defined as
∞
n=1
1 1 1 = 1 + + + n 2 3
∞ ∞ ∞ · ·· 1
Theorem 3.2.
n=1 n
=
··· .
.
Proof. Letting s n = k=1 k1 be the nth partial sum, we’ll show s n grouping summands in an appropriate way. For k N, we write s2k = 1 + +
1 2
∈
1 1 1 1 1 1 + + + + + + 3 4 5 6 7 8 1 1 1 + + + 2k−1 + 1 2k−1 + 2 2k
+
→ ∞ by
···
+
1 1 1 + + + 9 10 11
···
1 + 16
.
So the j th bracketed expression is t j =
1
1
1
− + 1 + 2 j−1 + 2 + · · · + 2 j . There are 2 j − 2 j −1 = 2 j −1 (2 − 1) = 2 j −1 terms in this sum, each of which is 1 . So t j 2 j −1 1 = 1 for any j ∈ N. Hence 2j
2 j 1
2
2j
s2k = 1 + t1 + t2 +
· · · + t
k
1+
k . 2 M
∈ R, let M M be an integer. Then n 22
So for any M
sn
s22M
1+
implies that
2M = 1 + M > M. 2
→ ∞, as required. Example 3.4. Given a ∈ R and r ∈ R\{0}, the sequence (ar −1 ) generates ∞ So sn
n
the geometric series
n
n−1 . n=1 ar
n−1 converges if and only if Theorem 3.3. The geometric series ∞ n=1 ar ∞ a n−1 r < 1 . When r < 1 then n=1 ar = 1−r .
||
| |
28
CHAPTER 3. SERIES
Proof. We start by noting that the nth partial sum is sn = a + ar + ar 2 + So if r < 1 then r n
·· · + ar −1 = a(11 −− r r n
| |
n
)
(Exercise). a
→ 0 (see Example 2.9), so s → 1− by Theorem 2.3. If r = 1 then s = na which tends to +∞ if a > 0 and −∞ if a < 0. If r = − 1, then s is zero if n is even and a if n is odd, so the series is n
r
n
n
divergent.
| |
If r > 1 then (sn )n diverges (Exercise). So a particular case is the sequence
∞ −
n=1
3.1
1 2
n 1
= 1+
−
1 n 1 2 n
1 1 1 + + + 2 4 8
generating the series
· · · = 1 −1 1 = 2. 2
The Comparison Test
As we’ve seen before, its common to try to understand complicated mathematical examples in terms of some basic building blocks. The Comparison Test is another example of this approach: given a new series ∞ n=1 xn , we can try to investigate its convergence in terms of some known series ∞ n=1 an which we’ve studied before.
Theorem 3.4 (The Comparison Test). Suppose that (xn )n and (an )n are sequences with no negative terms.
1. If ∞ n=1 an is convergent and xn convergent.
∞
2. If n=1 an is divergent and xn divergent.
an for all n
∞ ∞
∈ N, then
n=1 xn
is
∈ N, then
n=1 xn
is
an for all n
Proof. 1) Denote the partial sums by An = nk=1 an and X n = nk=1 xn . By assumption (An )n is convergent: let A denote the limit. Since all summands an are positive, (An )n is monotone increasing with An A. Since xk ak for all k N, we have X n An A for all n N, so (X n )n is bounded.
∈
∈
29
3.1. THE COMPARISON TEST
Hence the Monotone Convergence Theorem implies that (X n )n is convergent (i.e., ∞ n=1 xn is convergent).
2) Suppose that ∞ n=1 xn is convergent. Then, swapping the roles of (an )n ∞ an converges. This contradiction and (xn )n in 1), we deduce that n=1 shows that ∞ x divverges. n n=1
With minor changes, the proof implies a stronger result which we will also refer to as the Comparison Test: Corollary 3.5. Suppose that (xn )n and (an )n are sequences with no negative terms.
1. If xn 2. If xn
∞ ∞
n=1 an is
∈
convergent and there exists N N and c > 0 such that ∞ xn is convergent. can for all n N , then n=1 n=1 an is
∞
divergent and there exists N N and c > 0 such that can for all n N, then n=1 xn is divergent.
∈
Example 3.5. Show that
Proof. For n
1, n2
∞
n(n+1)
2
1
n=1 n2 is
∈
convergent.
, so 1 n2
2 . n(n + 1)
1 Now since ∞ n=1 n(n+1) is convergent by Example 3.1, the Comparison Test 1 (in particular Corollary 3.5 part 1 with c = 2) implies that ∞ n=1 n2 is convergent.
More generally: Theorem 3.6. Let α α > 1 .
∈ R.
Then
∞
1
n=1 nα
is convergent if and only if
Proof. Suppose that α 2. Then n1α n12 for all n N, so convergent using Example 3.5 and the Comparison Test.
∈
∞ ∞
1
n=1 nα
is
Now suppose that α 1. Then n1α n1 for all n N. So using the fact that the harmonic series is divergent, the comparison test implies that n=1 n1α is divergent.
∈
30
CHAPTER 3. SERIES
We omit the proof of convergence in the case α uses the ‘Integral Test’). Example 3.6. Given xn =
n+1 , n 1
5n3
−−
is
Yes: Since n + 1 2n 2n3 for all n
∈ (1, 2) (the standard proof
∞
n=1 xn convergent?
∈ N,
n + 1 n 1
5n3
− −
5n3
2n n
− − 1
<
5n3
2n 2n 2 1 = 3 = . 3 2n 3n 3 n2
−
So using the Comparison Test, with comparison series ∞ xn is convergent. n=1
∞
1
n=1 n2 ,
we see that
∈
Note that given a sequence (xn )n , for any m N, we can ask the same ∞ question about the convergence of n=m xn (actually by Corollary 3.5, convergence/divergence of this series is equivalent to that of ∞ n=1 xn ).
Example 3.7. Investigate the convergence of
∞
n=1 xn where x n
n 1 . n2 +3n+4
−
=
For all n N, n3 + 3n + 4 n 2 + 3n2 + 4n2 = 8n2 . Moreover, n for all n 3. So for n 3,
∈
−
n 1 2 n + 3n + 4
n
2
8n2
=
n
− 1 2
1 16n
So we can use the Comparison Test, comparing our series with the harmonic 1 series ( xn c n1 for n N in Corollary 3.5 with N = 3 and c = 16 ) to see ∞ that n=1 xn = .
∞
Example 3.8. Investigate the convergence of
√
3
∞
n=1 xn where x n =
√ 2n3+2 n3 +3
.
The dominant term on the top is n3 = n 2 and on the bottom is n 3 . So we 3
will try to compare with nn23 = 1
3
xn =
n2
1
√ 2n3 + 2
3 n2
Since ∞ n=1 converges.
1
3 n2
(n3 + 3)
1
3
n2
1
=
√ 3
.
√ 2n3 + 2
n n3 3 n2
+
3
3 n2
=
2+
3
n2 +
2
n3
3
3 n2
√ 4 3
n2
=
2 3
n2
is convergent, the Comparison Test implies that
.
∞
n=1 xn
Theorem 3.7 (The Ratio Test). Let (xn )n be a sequence of positive terms.
31
3.1. THE COMPARISON TEST
i) If there exists N N and r < 1 such that n then ∞ n=1 xn is convergent.
∞
∈
N implies that xxn+1 n
ii) If there exists N then n=1 xn =
∈ N and r > 1 such that n N implies that ∞.
r
xn+1 xn
r
n−1 Note that for the geometric series ∞ , the (n + 1)st term over the n=1 ar n ar nth term is arn−1 = r, so the ratio test agrees with Theorem 3.3:
• 0 < r < 1 implies ∞=1 ar −1 is convergent. • r > 1 implies ∞=1 ar −1 is divergent.
n
n
n
n
Proof of the Ratio Test. By Corollary 3.5, we only need to apply the comparison test to terms x n for n N . i) xN +1 xN and xN +2 rxN +1 r2 xN and so on, so xN +k rk xN . So in the Comparison Test, we use (an )n = (xN r−N r n )n as our sequence to compare (xn )n with. Since (an )n is a convergent geometric series and for n N we have xn = xN +(n−N ) rn−N xN = an , the comparison test implies that ∞ n=1 xn converges.
·
ii) Exercise.
Note that if we know that limn→∞ xxn+1 exists, then we can take this as our n value of r in the comparison test, so we just check if this limit is > 1 or < 1. It’s important to note that in the case that a series has xxn+1 1, the Ratio n ∞ 1 Test gives us no information at all. For example, n=1 n is divergent, but
∞
n=1
xn+1 xn
1
is convergent, but in the former case, n2
while in the latter case,
xn+1 xn
=
1 (n+1)2 1 n2
=
n2
(n+1)2
Example 3.9. Given a fixed x > 0 , does
xn+1 = xn
xn+1
(n+1)! xn n!
=
1
=
n+1
1
n
=
n n+1
→ 1.
xn n=1 n! converge?
∞
Solution: The sequence of terms here is xn =
→
xn . n!
xn+1 n! x = xn (n + 1)! n + 1
Then
→ 0 as n → ∞.
→ 1,
32
CHAPTER 3. SERIES
Hence the ratio test shows that of x we had in the beginning.
xn n=1 n! converges,
∞
regardless of the value n
0 as n Note that one consequence of this is that by Theorem 3.1, xn! , independently of the value of x (this can be extended to negative x also, as we’ll see later). Example 3.10. Does the series
n n=1 3n converge?
n n = 3n for
n+1 3n+1
n
3n
→ ∞
∞
• Solution 1: Letting x xn+1 = xn
→
∈ N, we consider
all n
3n (n + 1) 1 = = 3n+1 n 3
→ n n + 1
1 as n 3
→ ∞.
So by the Ratio Test, the sequence is convergent.
• Solution 2: Since x ∞ 1
1
all n N, and we know that the series ∞ nn n=1 2n is convergent, the Comparison Test implies that n=1 3 converges. n 2n for
Example 3.11. Does the series
∈
∞
1
n=1 n2 +1 converge?
Solution: To apply the Ratio Test, let xn = xn+1 = xn
1 (n+1)2 +1
1 n2 +1
1
n2 +1
. Then
1 + n12 n2 + 1 n2 + 1 = = 2 = (n + 1)2 + 1 n + 2n + 1 1 + n2 + n12
→ 1 as n → ∞,
so the Ratio Test tells us nothing in this case. On the other, hand n21+1 n12 for all n the Comparison Test implies that ∞ n=1 Example 3.12. Does the series
Solution: Letting xn = xn+1 = xn
(n+1)! (n+1)4 +3
n!
n4 +3
n! , n4 +3
∞
∈ N and since 1
n2 +1
∞
1
n=1 n2 converges,
is convergent.
n! n=1 n4 +3 converge?
(n + 1)! n4 + 3 = = (n+1) n! (n + 1)4 + 3
n4 + 3 (n + 1)4 + 3
→ ∞
so the Ratio Test (with r any number > 1 ) implies that the series diverges.
as n
→ ∞,
33
3.2. SUMS OF POSITIVE AND NEGATIVE TERMS
3.2
Sums of positive and negative terms
So far, we’ve mostly restricted ourselves to series ∞ n=1 xn where all x n (−1)n ∞ But consider, for example n=1 n2 . Does it converge? Definition 3.2. A series is convergent.
∞
n=1 xn is
called absolutely convergent if
0.
∞ |
xn
n=1
Theorem 3.8. Every absolutely convergent series converges. Proof is omitted.
∞ ∞ − ∞ ∞ Example 3.13. n
( 1)
n=1
n2
=
( 1)n
−
n=1
is convergent since it is absolutely convergent:
n2
1
n=1 n2 ,
which is convergent.
Geometric series n=1 ar n−1 give examples of series with positive and neg1 n−1 ative terms. For example for a = 1 and r = 12 , we consider ∞ . n=1 2 1 We already know that this is convergent since r = 2 < 1. Moreover note
− ∞ ||− − ∞ − −
that it’s also absolutely convergent since which is also convergent. n
1 n 1 2
n=1
=
n=1
1 n 1 , 2
(−1) 1 1 Example 3.14. The series ∞ = 1 + 12 is actun=1 3 + 4 n ally convergent (which we won’t prove), but it is not absolutely convergent, (−1)n ∞ 1 which is the harmonic series, which is di= since ∞ =1 n n=1 n n vergent (when a series converges, but doesn’t absolutely converge, it’s called conditionally convergent).
−
−
− ·· ·
Another test for convergence: Theorem 3.9 (The Root Test). Let
∞
n=1 xn be
a series. 1
1. If there exists N N and r < 1 such that n then ∞ n=1 xn converges.
N implies xn
n
2. If there exists N N and r > 1 such that n then n=1 xn diverges.
N implies xn
n
∞
∈ ∈
1
| | | |
1
r,
r,
Proof. 1) n N implies xn n r so xn rn . Since an = rn generates a convergent geometric series, by the Comparison Test ∞ n=1 xn converges ∞ so n=1 xn converges by Theorem 3.8.
| |
| |
| |
|
34
CHAPTER 3. SERIES
2) n N implies xn r n . So in particular, xn does not converge to zero as n , so by Theorem 3.1, ∞ n=1 xn diverges.
| |
→ ∞
| |
1
N.B. If limn→∞ xn n exists, then we can take this limit as the value of r in the Root Test (so we get convergence if this limit is < 1 and divergence if it’s > 1). Note that if this limit is equal to zero, then we can take any r < 1 in case 1) of the Root Test (eg r = 1/2), so the series is convergent. 1
N.B. Similarly to the Ratio Test, limn→∞ xn n = 1 tells us nothing about convergence.
| |
Useful tools here: 1
1. limn→∞ n n = 1.
Theorem 3.10.
1
2. If r > 0 then limn→∞ r n = 1. 3. Suppose that (an )n is a sequence of positive numbers and 1
Then an
an+1 an
| | → r .
→ r.
n
1
N. By Theorem 2.3, it suffices Proof. 1) Let bn = n n 1 for all n to show that bn 0. First note that bn 0 for all n N. Moreover 1 n 1 + b n = n n implies n = (1 + bn ) . Using the first three terms from the binomial expansion of (1 + bn )n ,
−
→
∈
∈
− 1)b2 > 21 n(n − 1)b2 Simplifying and rearranging n > 12 n(n − 1)b2 to b2 < 2/(n − 1) whenever 2 n 2, we obtain b −1 → 0, as required. n = (1 + bn )n
1 1 + nbn + n(n 2
n
n
n
n
n
n
2) Suppose first that r 1. Then for n r we have 1 1 this gives limn→∞ r n = 1.
1
3) Let ε > 0. Then there exists N
Since n
an+1 an
N implies
− r
1
(an ) n =
ε 2
∈ N such that n N implies ε a +1 ε i.e., r − r + . 2 a 2
an an−1 an−1 an−2
n
n
· ··
aN +1 aN aN
1 n
1
r n n n .
,
By 1),
35
3.3. POWER SERIES
we obtain 1
− − − − r
ε 2
n N
n
aN
so
ε 2
r
1
1
(an ) n =
N n
1
Since N 0 and (aN ) n n n M implies
→
r
1
(aN ) n
an an−1 an−1 an−2 1
(an ) n
· ··
1
−
aN +1 aN aN
ε r + 2
1
−
N n
n
ε r + 2
n N
1
(aN ) n .
→ 1 as n → ∞, there exists M N such that
− ε (a
1
n) n
| − r| ε,
r + ε, i.e., an
as required. Example 3.15. Consider the series
∞
2(−1)
n=0 n
Setting xn = 2(−1)
n
−n = 2 + 1 + 1 + 1 + 1 + 1 + ·· · 4
2
16
8
64
−n , the Ratio Test gives n+1
xn+1 2(−1) −(n+1) 1 (−1)n+1 −(−1)n = = 2 = n xn 2 2(−1) −n
2 1 8
if n is odd , if n is even .
So the Ratio Test tells us nothing. However, for the Root Test,
|x | n
1 n
= (2(−1)
n
−n) 1 = 2 (−1) −1 = 1 · 2 (−1)
1
Since for n even, we have 2 n 1
1
n
n
→
n
n
2
n
1 (see Theorem 3.10 ) and for n odd
2− n = 12 n 1 (again, see Theorem 3.10 ), Theorem 2.3 implies that 1 limn→∞ xn n = 12 , so the Root Test implies that the series converges.
| |
3.3
→
Power Series
Definition 3.3. Given a sequence (an )n , the series power series
∞
n n=1 an x
is called a
aN
1 n
,
36
CHAPTER 3. SERIES
n In a power series there is a variable x. So if, for a particular x, ∞ n=1 an x converges, then the series can be interpreted as a sum and so the power series is a function of x. These come up in different areas of maths and physics, for example probability and combinatorics, as well as lots of areas of applied maths.
Determining convergence can be a difficult problem, but what always happens is either: i) the power series converges for all x
∈ R;
ii) the power series converges only at x = 0; iii) the power series converges in some interval of x-values with centre zero. n Theorem 3.11. Let ∞ n=1 an x be a power series and suppose lim n→∞ an β and R = β1 (if β = 0, then set R = ; if β = , then set R = 0).
| |
∞ ∞ 1. The power series converges for |x| < R. 2. The power series diverges for |x| > R.
1 n
R is called the radius of convergence of the power series. Note that 1) is vacuous if R = 0 and 2) is vacuous if R = .
∞
Proof. Let b n = a n xn for all n N, so our series is that generated by (bn )n . Then in order to apply the Root Test, we compute
∈
1
1 n n
1
|b | = |a x | = |a | |x| → β |x| as n → ∞. We will input the value β |x| into the Root Test. Case 1: a) β = 0 (so R = ∞, and necessarily |x| < R). Then the Root Test ∞ n
implies that
n
n
n n=1 an x
||
n
n
is convergent.
| |
b) β > 0 and β x < 1 (so x < R). Again the Root Test implies that ∞ an xn is convergent. n=1
Case 2: a) β = is divergent.
∞ (so R = 0). Then the Root Test implies that ||
| |
∞
n n=1 an x
b) β is finite and β x > 1 (so x > R). Again the Root Test implies that ∞ an xn is divergent. n=1
=
37
3.3. POWER SERIES 1
| exists, then it equals β and lim Note that if limn→∞ |a|na+1 n→∞ an n = β n| (Exercise), so it’s often easier to check this ratio, rather than finding the root directly.
| |
Example 3.16. Consider
Then an =
1
n!
. Hence
∞
1 n n=1 n! x .
1 |a +1| = ( +1)! n! 1 → 0 = = 1 |a | (n + 1)! n + 1 ! Hence β = 0 and R = ∞, so the radius of convergence is ∞, i.e., the series converges for all x ∈ R (actually to e ). n
n
n
n
x
Example 3.17. Consider
Letting an =
1 n
,
∞
1 n n=1 n x .
|a +1| = |a | n
1
n+1
n
1
n n + 1
=
n
→ 1,
so β = 1 and R = 1 - the radius of convergence is 1. This means that the , 1) power series converges for all x ( 1, 1), and diverges for x ( (1, ). In fact, Theorem 3.2 and Example 3.13 imply that the series converges if and only if x [ 1, 1).
∞
Example 3.18. For
∈−
∈ −∞ − ∪
∈ −
2n n n=1 n2 x ,
∞ 2n n2
1
1 n
=
an = 2 2
nn
2n n2
=
. We calculate 2 1
(n n )2
→ 2
(Note that we use limn→∞ n n = 1 from Theorem 3.10 here.) So β = 2 and the radius of convergence is 12 .
38
CHAPTER 3. SERIES
Chapter 4
Cauchy Sequences It’s important to know if a sequence has a limit or not. The Monotone Convergence Theorem is an example of a result which guarantees the existence of a limit, but its main drawback is that the sequence has to be monotone. We will show that any ‘Cauchy sequence’ has a limit. Definition 4.1. A sequence (xn )n of real numbers is called a Cauchy sequence if for all ε > 0 , there exists N N such that n, m N, n, m N implies
∈
∀
∈
|x − x | ε. n
Example 4.1. (xn )n =
1
n n
m
is Cauchy.
Proof. Let ε > 0. Then setting N N such that N 2ε , n, m N implies that 1 1 1 1 1 1 2 xm xn = + + = ε, n m n m N N N
∈
| − |
so (xn )n is Cauchy. Example 4.2. (xn )n =
−
n
1+n n
is Cauchy.
∈ N such that N 2 .
Proof. Let ε > 0. Then set N
ε
39
Then for n, m
N ,
40
CHAPTER 4. CAUCHY SEQUENCES
assuming n m implies that
−
n m n(1 + m) m(1 + n) xn = = 1 + n 1 + m (1 + n)(1 + m) n m n m m n 1 = = = (1 + n)(1 + m) nm nm n 1 1 1 1 2 + + = ε. n m N N N
|x − | m
− −
| − |
The case where n > m follows similarly. So Example 4.3. (xn )n = 1 + 12 +
· ·· 1
n n
−
n
1+n
n
− m1
is Cauchy.
is not Cauchy.
Proof. We’ll show that (xn )n satisfies the negation of the definition of Cauchy sequence. That is, we must show that
∃ε > 0 : ∀N ∈ N, ∃m, n ∈ N s.t. n, m N and |x − x | > ε. (∗) We start by fixing some N ∈ N. Put n = N and m = 2N , so clearly n
m
n, m N . Then
|x − x | = |x − x2 | 1 1 − = 1 + + · · · + 2 N m
n
N
=
N
1 + N + 1
1 1 + + 2
1 1 ·· · + 2N +··· 2N
1 1 1 + + + + N N + 1 2N 1 1 1 1 + = N = > . 2N 2N 2 4
···
···
Since N was arbitrary here, this calculation implies that we can put ε = 1 into ( ) to derive that 1 + 12 + is not Cauchy. n n
∗
·· ·
1 4
Note that in the first two examples here the sequence was convergent as well as being Cauchy, while in the third example, the sequence was neither convergent nor Cauchy. This suggests that being convergent and Cauchy are related, an idea which we develop below. Proposition 4.1. Every convergent sequence is Cauchy.
Proof. Suppose that (xn )n is a convergent sequence with limit x. Let ε > 0. Since xn x there exists N N such that n N implies xn x 2ε . So for n, m N, n, m N implies ε ε xn xm = xn x + x xm xn x + x xm + = ε, 2 2
→ ∈ | − | | −
∈
− | | − | | − |
| − |
41 as required. Lemma 4.2. Any Cauchy sequence is bounded.
Proof. Let (xn )n be a Cauchy sequence. Then there exists N n, m N implies xn xm 1. Hence n N implies
∈ N such that
| − | |x | = |x − x + x | |x − x | + |x | 1 + |x |. n
n
N
N
n
N
N
N
Therefore, for
{| | | |
| || | }
M := max x1 , x2 , . . . , xN , xN + 1 ,
|x | M for all n ∈ N. n
Lemma 4.3. Let (xn )n be a Cauchy sequence. If (xn )n has a subsequence , (xmk )k that converges to some real number x R, i.e., x mk x as k then the original sequence (xn )n also converges to x , i.e., x n x as n .
∈
→ →
→ ∞ →∞
N such that Proof. Let ε > 0. Since (xn )n is Cauchy, there exists N 1 n N 1 implies ε xn xm . 2 Since, moreover, x mk x, there exists N 2 N such that k N 2 implies
∈
| − |
→
∈
|x − x| 2ε . Let N := max{N 1 , N 2 }. Then for n ∈ N, n N implies |x − x| = |x − x + x − x| |x − x | + |x − x| 2ε + 2ε = ε. (Here we used that m n N to estimate |x − x |.) mk
n
n
mn
n
mn
n
mn
mn
n
mn
Theorem 4.4 (The General Principle of Convergence). Let (xn )n be a sequence of real numbers. The following are equivalent.
1. (xn )n is convergent. 2. (xn )n is Cauchy.
42
CHAPTER 4. CAUCHY SEQUENCES
Proof. (1
⇒ 2): This is Proposition 4.1.
(2 1): Let (xn )n be a Cauchy sequence. Lemma 4.2 implies that (xn )n is bounded. So the Bolzano-Weierstrass Theorem implies that there exists a subsequence (xmk )k which converges to some limit x. Finally Lemma 4.3 implies that (xn )n itself is also convergent (converging to x).
⇒
N.B. Note that this theorem is also known as the Cauchy Convergence Criterion. Example 4.4. Let (xn )n be a sequence and assume there is A > 0 such that
|x +1 − x | nA2 for all n ∈ N. n
n
Then (xn )n is convergent. Proof. By the General Principle of Convergence, it suffices to show that (xn )n is a Cauchy sequence. Let ε > 0. Choose N N such that N 1−1 Aε . Then n m N implies
∈
|x − x | |x − x −1| + |x −1 − x −2| + · · · + |x +2 − x +1| + |x 1 1 1 1 ·· · A + + + + n2 (n + 1)2 (m + 1)2 m2 n
m
n
A A
n
n
−
n
m
···
···
= A
− x |
1 1 1 1 + + + + (n 1)n n(n + 1) m(m + 1) (m 1)m 1 1 1 1 + + n 2 n 1 n 3 n 2 1 1 1 1 + + m m + 1 m 1 m 1 1 A A ε. m 1 n 1 n 1 N 1
− − −
m+1
m
− − −
· ·· − − −
− − −
−
−
−
Hence (xn )n is a Cauchy, so (xn )n is convergent. Example 4.5. The sequence (xn )n where 1 1 xn = 1 + + + 2 3
· ·· + n1 − log n is convergent. The number γ = lim →∞ 1 + 12 + 13 + · · · + 1 − log n is called n
n
Euler’s constant. Whether it’s rational or not is a long-standing question in maths.
m
43
Proof. By Example 4.4, convergence of (xn )n will follow if xn+1 xn x for all n N. We compute this using the fact that log x = 1 1t dt:
|
∈
n2
1 1 1 1 1 1 1 log(n + 1) 1 + + + + xn = 1 + + + + + 2 3 n n + 1 2 3 n 1 1 n + 1 = log(n + 1) log n = log n + 1 n + 1 n 1 1 1 1 1 1 1 1 = + log 1 + + log 1 + n + 1 n n n n + 1 n n n
|x +1 − | n
2
− | ··· − − · · · − − − − − − − − − −
− −
1 = + n(n + 1) 1 = 2+ n
as required.
1 + n2
1 1+ n
1
1 1+ n
1
1 1+ n
1 dt
1
1
t 1 dt t 1 dt n
1 1+ n
1 + n2
1 dt t
1 1+ n
1 + n2
1 1+ n
1
1
1 dt t
t 1 dt
1
1 1 2 + = , n2 n2 n2
N.B. We could also have used the theory of Cauchy sequences to get further information on series via Cauchy properties of partial sums (this leads to proofs of Theorem 3.8 and the statement in Example 3.14).
log n
44
CHAPTER 4. CAUCHY SEQUENCES
Chapter 5
Continuous Functions →
∈
Recall that given sets A and B, f : A B is a function if for each x A there exists a unique value y inB such that f (x) = y (A is the domain of f and f (A) = y B : x A s.t. f (x) = y is the range of f ).
{ ∈
∃ ∈ } Given an interval I ⊆ R , a rough description of a function f : I → R being continuous is that it can be graphed without removing the pen from the paper. Here we’ll give a more rigorous treatment.
Definition 5.1. Let A R and f : A R be a function on A. Then f is said to be continuous at a point x 0 A if
⊆
→
∈ ∀ε > 0∃δ > 0 s.t. ∀ x ∈ A, |x − x0| δ ⇒ |f (x) − f (x0)| ε.
• We say that f is discontinuous at x0 ∈ A if f is not continuous at x 0. • We say that f is continuous if it is continuous at all x0 ∈ A. • We say that f is discontinuous if it is not continuous. Theorem 5.1. Let A ⊆ R and f : A → R be a function. Then the following are equivalent:
1. f is continuous at x0 . 2. If (xn )n is any sequence in A with xn
→ x0, then f (x ) → f (x0).
45
n
46
CHAPTER 5. CONTINUOUS FUNCTIONS
Proof. (1) 2)): Suppose that (xn )n A has xn x0 . To show that f (xn ) f (x0 ), let ε > 0. Since f is continuous at x0 , there exists δ > 0 such that x x 0 δ implies f (x) f (x0 ) ε. Since xn x0 , there exists N N such that xn x0 δ for all n N . So if n N has n N , then f (xn ) f (x0 ) ε. Therefore 2) follows.
⇒ ⊆ → → | − | | − | → ∈ | − | ∈ | − | (2) ⇒ 1)): To obtain a contradiction, we assume 2), but assume that 1) is false, i.e., [negating ∀ ε > 0 ∃ δ > 0 s.t. ∀x ∈ A, | x − x 0 | δ ⇒ |f (x) − f (x0 )| ε], ∃ε > 0 s.t. ∀δ > 0 ∃x ∈ A s.t. |x − x0| δ and |f (x) − f (x0)| > ε. (∗) Suppose that ε > 0 is such that (∗) holds, i.e., ∀δ > 0 ∃x ∈ A s.t. |x − x0| δ and |f (x) − f (x0)| > ε. In particular, for each n ∈ N, setting δ = 1 gives some x ∈ A such that n
n
|x − x0| n1 and |f (x ) − f (x0)| > ε. n
n
→
But this implies that we have a sequence (xn )n such that xn x0 , but (f (xn ))n doesn’t converge to f (x0 ), which contradicts 2). So f must in fact be continuous at x0 . This theorem means that we now have two equivalent definitions of continuity:
• the original one, Definition 5.1, which is called the ε-δ definition of continuity ;
• the condition “for all sequences (x ) ⊂ A such that x → x0, we have f (x ) → f (x0 )”, which is called the sequential definition of continuity . Example 5.1. Let c ∈ R and define f : R → R by f (x) = c for all x ∈ R. n n
n
n
Then f is continuous.
Proof using the ε-δ definition of continuity. Let x0 δ = 300. Then x R and x x0 δ implies
∈
∈ R. Let ε > 0 and set
| − | |f (x) − f (x0)| = |c − c| = 0 < ε, so f is continuous at x 0 . Since x 0 ∈ R was arbitrary, f is continuous.
47 R be the identity map, i.e., f (x) = x for all Example 5.2. Let f : R x R. Then f is continuous.
∈
→
Proof using the ε-δ definition of continuity. Let x0 for δ = , for all x R, x x0 δ implies
∈ R.
Let ε > 0. Then
∈ | − | |f (x) − f (x0)| = |x − x0| δ = ε, so f is continuous at x 0 . Since x0 ∈ R was arbitrary, f is continuous. Example 5.3. Define f : R → R by f (x) = 2x2 + 1.
Then f is continuous. Proof using the ε-δ definition of continuity. Let x0 R. Let ε > 0. Note that if x x0 1 implies x x0 +1. Now choose δ = min 1, 2(2|xε0 |+1) . Then x R and x x0 δ implies
∈
|− | || | | ∈ | − | |f (x) − f (x0)| = |(2x2 + 1) − (2x20 + 1)| = 2|x2 − x20| = 2|(x − x0)(x + x0)| = 2|x − x0 ||x + x0 | |x − x0 |2 (|x| + |x0 |) |x − x0 |2 (|x0 | + 1 + |x0 |) δ 2 (2|x0 | + 1) ε, so f is continuous at x 0 . Since x0 ∈ R was arbitrary, f is continuous.
R. Let (xn )n Proof using the sequential definition of continuity. Let x0 be a sequence in R such that xn x 0 . Then by Theorem 2.3,
∈
→
f (xn ) = 2x2n + 1
→ 2x20 + 1,
so f is continuous. Example 5.4. Let f : (0,
∞) → R be given by f (x) =
Then f is continuous.
1 . x2
48
CHAPTER 5. CONTINUOUS FUNCTIONS
Proof using the ε-δ definition of continuity. Let x0 First note that
|
− | −
1 f (x) f (x0 ) = 2 x
δ = min δ , so
3 x0 x 0 ε ,
2
10
∈ (0, ∞).
1 x2 x20 x x0 x + x0 = = x20 x2 x20 x2 x20
. Then x
−
| − ||
Let ε > 0.
| |x − x0| (|x| + |x0|) . (∗) x2 x2 0
∈ (0, ∞) and |x − x0| δ implies −δ x − x0
x x0
− δ x0 − x20 = x20
and x x0 + δ x0 +
x 0 3x0 = . 2 2
So for all x (0,
∈ ∞), by (∗), |x − x0| δ implies δ 32 + x0 | x − x0 | (x + x0 ) |f (x) − f (x0)| x0
5x0 2
x2 x20
x20
x20
2
=
δ
x40
4
10 = δ 3 x0
ε.
∈ (0, ∞) and hence continuous.
so f is continuous at each x 0
N.B. The sequential definition can also be used in conjunction with Theorem 2.3 here (Exercise). Example 5.5. Let f : R
→ R be defined by −1 if x < 0, f (x) =
Then f is continuous at any x0
1
if x 0.
∈ R \ {0}, but discontinuous at 0.
1 . Proof. We’ll first show that f is discontinuous at 0. Let (xn )n = n n Then xn 0. Also since xn < 0, f (xn ) = 1 for all n N: but this means that f (xn ) does not tend to 1 = f (x0 ), so f is not continuous at 0.
→
−
∈
−
Now let x0 R 0 . Let ε > 0. Then since x0 = 0, δ := |x20 | > 0. So for | which implies all x R, if x x0 δ = |x20 | then |x20 | x x0 |x20 ,
∈
∈ \ { } | − |
− − |x0| x |x0| . x0 − 2
2
49 Hence
x0
⇒ |x0| = x0 and x 2 > 0, so f (x) = 1 = f (x0), ⇒ |x0| = −x0 and x − 2 < 0, so f (x) = −1 = f (x0), so in either case |f (x) − f (x0 )| = 0 < ε. Example 5.6. Let f : R → R be defined by x if x ∈ Q, f (x) = 0 if x ∈ R \ Q. x0 > 0 x0 < 0
x0
Then f is continuous only at 0. Proof. We’ll first show that f is continuous at 0. Let ε > 0. Then for δ = ε, for all x R with x x0 δ , we have
∈
| − | |f (x) − f (x0)| = |f (x) − 0| = |f (x)| |x| δ ε,
so f is continuous at 0. We next show that f is discontinuous at x0 = 0.
∈ \ →
Case 1: x0 R Q. Then f (x0 ) = 0. Since the rationals are dense in R there exists xn x0 such that xn Q for all n N. Hence f (xn ) = xn , but limn→∞ f (xn ) = x 0 = f (x0 ) = 0. Hence f is not continuous at x 0 .
∈
∈
√
Case 2: x0 Q . Then f (x0 ) = x 0 . Since for xn = x + n2 , xn is irrational and f (xn ) = 0 for all n N. But since xn x0 and (f (xn ))n does not converge to f (x0 ) = x 0 , f is not continuous at x 0 .
∈
∈
→
As we’ve seen before, it can be useful to break complicated mathematical objects down into more basic building blocks and thus understand the more complicated object. For example, taking f (x) = x 2
g(x) = x 3
as our building blocks (which we can fairly easily prove are continuous), we can make f +g : x
→ x2 +x3, f − g : x → x2 − x3, f · g : x → f (x)g(x) = x2x3 = x5, f ◦ g : x → f (g(x)) = f (x3 ) = (x3 )2 = x 6 .
50
CHAPTER 5. CONTINUOUS FUNCTIONS
As one might expect, continuity is preserved by the operations above (namely addition, subtraction, multiplication, composition), which is the content of the next two results.
⊆
∈
Theorem 5.2. Let A A. Suppose that f, g : A R and x0 functions which are continuous at x0 and let λ R. Define
∈
• f + g by (f + g)(x) = f (x) + g(x), • fg by (fg)(x) = f (x)g(x), • λf by (λf )(x) = λf (x), • min(f, g) by min(f, g)(x) = min{f (x), g(x)}, • max(f, g) by max(f, g)(x) = max{f (x), g(x)}, • |f | by |f |(x) = |f (x)|, • If g(x) = 0 for all x ∈ A then is given by f g
f g
(x) =
→ R are
f (x) . g (x)
Then 1. f + g is continuous at x0 , 2. fg is continuous at x0 , 3. λf is continuous at x0 , 4. min(f, g) is continuous at x0 , 5. max(f, g) is continuous at x0 , 6. f is continuous at x0 ,
||
7. If g(x) = 0 for all x A then f g is continuous at x0 .
∈
Proof. 1) Let (xn )n A have xn x 0 . Then since f and g are continuous at x0 , f (xn ) f (x0 ) and g(xn ) g(x0 ). So by Theorem 2.3,
⊆
→ → → (f + g)(x ) = f (x ) + g(x ) → f (x0 ) + g(x0 ) = (f + g)(x0 ), n
n
so f + g is continuous at x0 .
n
51 2), 3), 6) and 7) follow similarly. 5) Since
1 1 max(f, g) = (f + g) + f g , 2 2 the result follows by 1), 3) and 6).
| − |
4) Since
1 min(f, g) = (f + g) 2 the result follows by 1), 3) and 6).
− 12 |f − g|, →
One consequence of this theorem is that since f : R R defined by f (x) = x is continuous (Exercise), any polynomial p : R R is continuous.
→
Theorem 5.3. Let A, B A. Suppose that f : A R and x0 R and R be functions with f (x0 ) g:B B . If f is continuous at x0 and g is continuous at f (x0 ), then g f is continuous at x0 .
⊆
→
∈
◦
∈
→
Proof. Let (xn )n A be a sequence with xn x0 . Since f is continuous at x0 , the sequence (f (xn ))n has f (xn ) f (x0 ). Since g is continuous at f (x0 ), this means
⊆
→
→
◦ → g(f (x0)) = (g ◦ f )(x0), i.e., (g ◦ f )(x ) → (g ◦ f )(x0 ), so g ◦ f is continuous at x0 . (g f )(xn ) = g(f (xn ))
n
We’ll next derive some major theorems on continuous functions which are intuitively obvious, but require quite sophisticated proofs. We start with a definition.
→
Definition 5.2. A function f : A R is called bounded if there exists M > 0 such that f (x) M for all x A.
|
|
∈
Theorem 5.4 (The Extreme Value Theorem). Given a < b, let f : [a, b] R be a continuous function. Then
→
1. f is bounded;
∈ [a, b] such that f (x) f (x0 ) for all x ∈ [a, b];
2. f attains its maximum, i.e., there exists x0
52
CHAPTER 5. CONTINUOUS FUNCTIONS
3. f attains its minimum, i.e., there exists y0
∈ [a, b] such that f (x) f (y0 ) for all x ∈ [a, b].
Proof. 1) Assume, for a contradiction, that f is unbounded above. So for each n N there exists x n [a, b] such that f (xn ) > n. Since (xn )n [a, b], it is a bounded sequence, so the Bolzano-Weierstrass Theorem implies that there is a subsequence (xmk )k and c R such that xmk c as k . Since xmk [a, b] for all k N, this means that c [a, b]. Since f is continuous in [a, b], in particular it’s continuous at c, so xmk c implies f (xmk ) f (c). Hence (f (xmk ))k is bounded sequence by Theorem 2.2. But this contradicts the fact that f (xmk ) > mk for all k, so f is bounded above. The proof that f is bounded below follows similarly.
∈
∈
∈
∈
∈
∈
⊆ → ∞ →
→
→
2) Let M := sup f (x) : x [a, b] . We’ll show that there exists x0 [a, b] such that f (x0 ) = M . By definition of sup, for each n N there exists xn [a, b] such that 1 M f (xn ) M. n (see eg TS1 Q9). This implies that
{
∈
∈
}
∈
∈
−
f (xn )
→ M as n → ∞ (∗).
The Bolzano-Weierstrass Theorem implies that for the sequence (xn )n there exists a convergent subsequence (xmk )k with limit, say x0 [a.b]. Continuity of f means that x mk x 0 implies that
∈
→
f (xmk )
∗
→ f (x0) as k → ∞ (∗∗).
∗∗
Combining ( ) and ( ) shows that f (x0 ) = M , so 2) is proved.
−f (Exercise).
3) Apply 2) to the function
N.B. If we allowed f to be defined on an open interval (a, b) (or half-open), then the conclusions of the Extreme Value Theorem may fail, as the next examples show. Example 5.7. Define f : (0, 1)
→ R by f (x) =
1 . x
Then f is continuous (Exercise), but not bounded (eg given any M there exists n N such that n > M , so f n1 = n > M .).
∈
0,
53 R by f (x) = x for all x Example 5.8. Define f : (0, 1) (0, 1). Then f is bounded and continuous, but it does not have a maximum since f (x) : x (0, 1) = (0, 1), so it certainly can’t attain its maximum.
→
∈
∈
}
{
The following is the main result of this chapter. Theorem 5.5 (The Intermediate Value Theorem). Let I R be an interval and let f : I R be a continuous function. If a, b I with a < b and y lies between f (a) and f (b) (i.e., either f (a) < y < f (b) or f (b) < y < f (a)), then there exists x (a, b) such that
→
⊆
∈
∈
f (x) = y.
Proof. Assume that f (a) < y < f (b) (the other case follows similarly). Let
{ ∈
}
S := x [a, b] : f (x) < y .
∈
∅
Since a S , S = . Set x0 := sup S . The idea is that f (x0 should be y.
• Proof1that f (x0) y: for each n ∈ N, since x0 − 1 < x0, we know that x0 − is not an upper bound on S , so there exists s ∈ S such that n
n
n
x0
− n1 < s
n
x0 .
→ x0 and the continuity of f implies that f (s ) → f (x0 ).
This means that sn
n
∈
Since sn S , f (sn ) < y for each n, so any limit of (f (sn ))n must be y, i.e., f (x0 ) y.
• Proof that f (x0) y: For each n ∈ N, let t := min b, x0 + 1 . So t ∈ [a, b], t → x 0 and by continuity, f (t ) → f (x0 ). (∗). By definition for each n ∈ N, t ∈ / S , so f (t ) y. Hence this inequality passes to the limit, i.e., (∗) implies f (x0 ) y.
n
n
n
n
n
n
In conclusion, f (x0 ) = y, as required. Note that x0 f (x0 ) < f (b), so in fact x0 (a, b).
∈
n
∈ [a, b], but f (a) <
54
CHAPTER 5. CONTINUOUS FUNCTIONS
R be an interval and f : I Corollary 5.6. Let I function. Then the range of f : i.e., f (I ) = f (x) : x interval or a single point.
⊆
→ R a continuous ∈ I } is either an
{
Proof. If there are two distinct points in f (I ) (say f (a) = f (b)), the IVT guarantees that all points between these are also in f (I ) (eg f (a), f (b) f (I ) and f (a) > f (b) implies all y (f (b), f (a)), y f (I )).
∈
∈
∈
Note that if I is closed then f (I ) = [inf f (I ), sup f (I )]. Example 5.9. Suppose that f : [0, 1] [0, 1] is a continuous function. [0, 1] such that Then there exists a ‘fixed point’, i.e., there exists x0 f (x0 ) = x 0 .
→
∈
−
Proof. Let g(x) = f (x) x. By Theorem 5.2, g is also continuous on [0, 1]. Now notice that
− g(1) = f (1) − 1 1 − 1 = 0 g(0) = f (0) 0 = f (0) 0
so 0 [g(1), g(0)].
∈
If g(0) = 0 or g(1) = 0 then 0 or 1 are fixed points, respectively. Assume that g(0) > 0 > g(1). Then the IVT implies that there exists x0 (0, 1) such that g(x0 ) = 0, i.e., f (x0 ) x0 = 0, so f (x0 ) = x 0 , as required.
∈
−
Example 5.10. If y > 0 and m
∈ N then y has a positive mth root.
Proof. Suppose y = 1, since then the required root is 1. By Theorem 5.2, f (x) = x m is continuous. Now let
b =
1 y
if y < 1, if y > 1.
Then 0 < y < b m , i.e., f (0) < y < f (b). So the IVT implies that there exists x (0, b) such that f (x) = y, i.e., x m = y, i.e., xm = y, so x is an mth root of y.
∈
55
5.1. LIMITS
5.1
Limits
⊆
Suppose that A R and f : A to use the notation, for a R,
∈
→ R is a function. It will be convenient later lim f (x).
→a
x
∈
This means that there exists R such that as x becomes arbitrarily close to, but not equal to, a, f (x) becomes arbitrarily close to . More formally: Definition 5.3. limx→a f (x) exists and equals R if for all ε > 0 there exists δ > 0 such that x A a with x a δ implies
∈
∈ \ { } | − | |f (x) − | ε. Example 5.11. Let f : [0, ∞) \ {1} → R be defined by x2 − 1 f (x) = 2 . x + x − 2 Then f is not defined at 1, but limx→1 f (x) exists. [Note that a standard technique here would be to write
x2 1 (x + 1)(x 1) x + 1 = = , 2 x + x 2 (x + 2)(x 1) x + 2
−
− −
−
but strictly speaking, the function on the LHS is not equal to that on the RHS since LHS is not defined at 1.] Proof. We make a guess that = 23 . Let ε > 0. Since x + 2 ε 0, for all x [0, ) 1 ,
∈ ∞ \{ } 2 x2 − 1 2 x + 1 2 − − f (x) − = 2 = 3 x + x − 2 3 x + 2 3 |x − 1| 3(x + 1) − 2(x + 2) = =
3(x + 2)
3(x + 2)
2,
for all
|x − 1| . 6
∈ [0, ∞) \{1} and |x − 1| δ implies f (x) − 23 ε. Theorem 5.7. Suppose that f : A → R is a function. Given a point x 0 ∈ A, So setting δ = 6ε, x
then limx→x0 f (x) exists and equals f (x0 ) if and only if f is continuous at f (x0 ).
56
CHAPTER 5. CONTINUOUS FUNCTIONS
Proof. Exercise. For general functions, it’s possible for limx→x0 f (x) to exist, but not equal f (x0 ):
→ R be defined by x2 if x = 3, f (x) =
Example 5.12. Let f : R
0
if x = 3.
Then limx→3 f (x) = 9 ( = f (3)).
| − 3| 1 implies 2 x 4, so also if
Proof. Let ε > 0. First note that x x = 3 then
|f (x) − 9| = |x2 − 9| = |x − 3||x + 3| 7|x − 3|. So if δ := min 1, 7 , then x ∈ R \ {3} and |x − 3| δ implies |f (x) − 9| 7|x − 3| ε. ε
→
Theorem 5.8. Suppose that f, g : R R are functions and a limx→a f (x) = , limx→a g(x) = m . Then
∈ R has
1. limx→a(f + g)(x) = + m; 2. limx→a(fg)(x) = m;
→
3. limx
a
f g
(x) =
so m
long as m = 0;
4. limx→a f (x) =
||
|| ∈ \ { }
a , R Proof. 1) Let ε > 0. Then there exists δ 1 > 0 such that x x a δ 1 implies ε f (x) . 2 Similarly, there exists δ 2 > 0 such that x R a , x a δ 2 implies
| − |
|
−|
∈ \ { } | − |
|g(x) − | 2ε .
57
5.1. LIMITS
{ } ∈ R \ {a}, |x − a| δ implies |(f + g)(x) − ( + m)| = |(f (x) − ) + (g(x) − m)| ε ε |f (x) − | + |g(x) − m| + = ε. 2 2
So setting δ := min δ 1 , δ 2 , x
The proofs of the 2), 3), 4) are exercises. It’s sometimes useful to be able to distinguish between the limit of a function as we approach from the left, and the limit when we approach from the right. Definition 5.4. Given A
⊂ R and f : A → R, for a ∈ A, →
a exists and equals and 1. We say that the left-limit of f (x) as x write lim f (x) = ,
→a−
x
if for all ε > 0 there exists δ > 0 such that if x x a δ then f (x) ε.
| − |
|
∈ A has x < a and
−|
→
2. We say that the right-limit of f (x) as x a exists and equals and write lim f (x) = ,
→a+
x
if for all ε > 0 there exists δ > 0 such that if x x a δ then f (x) ε.
| − |
|
∈ A has x > a and
−|
Example 5.13. Define f : R
→ R by |x| + | | f (x) =
Then limx→0− f (x) =
x x
0
if x = 0,
if x = 0.
−1 and lim →0 f (x) = 1 . Theorem 5.9. Let A ⊆ R, f : A → R and a ∈ A. x
+
Then limx→a f (x) exists if and only if lim x→a− f (x) and lim x→a+ f (x) both exist and are equal. Moreover, if limx→a− f (x) = limx→a+ f (x) = then limx→a f (x) = .
58
CHAPTER 5. CONTINUOUS FUNCTIONS
Proof. Suppose that limx→a f (x) = . Then for all ε > 0 there exists δ > 0 such that x A a , x a δ implies f (x) ε. So in particular if x A and x < a then x a δ implies f (x) ε, so limx→a− f (x) = ; and if x A and x > a then x a δ implies f (x) ε, so limx→a+ f (x) = .
∈
∈ \ { } | − | | − | ∈
| | | − |
−| −|
|
− |
Now suppose that limx→a− f (x) = limx→a+ f (x) and denote this value by . Set ε > 0. Then there exists δ 1 > 0 such that if x A and x < a then x a δ 1 implies f (x) ε. Also there exists δ 2 > 0 such that if x A and x > a then x a δ 2 implies f (x) ε. So setting δ := min δ 1 , δ 2 , if x A a then x a δ implies f (x) ε, i.e., limx→a f (x) = .
| − | ∈ { }
| − | | − | | ∈ \ { } | − |
∈ − | | − |
Note that in Example 5.13, limx→0 f (x) does not exist since limx→0− f (x) = limx→0+ f (x).
Definition 5.5. Given A
⊆ R, f : A → R and a ∈ A,
1. we say that limx→a f (x) = if for all M R, there exists δ > 0 such that if x A a then x a δ implies f (x) M .
∞ ∈ ∈ \ { } | − | 2. we say that lim → f (x) = −∞ if for all M ∈ R, there exists δ > 0 such that if x ∈ A \ {a} then |x − a| δ implies f (x) M . x
a
Chapter 6
Differentiation Differentiation is a crucial topic in any area of science where a system evolves in time. In this chapter we’ll give the fundamental ideas of differentiation a rigorous treatment, aided by our knowledge of limits. Definition 6.1. Let I R be an open interval, f : I c I . Then f is differentiable at c if
→ R a function and
⊆
∈
− −
f (x) f (c) x→c x c lim
exists,
in which case we denote this limit by f (c), the derivative of f at c. Letting c vary over the whole of I , if f is differentiable at every such c, then f : I R can be thought of as a function f (x). This can also be denoted
→
d df f (x), , Dx f (x). dx dx
→ c if and only if h → 0, so we can also use f (c + h) − f (c) lim
Note that letting x = c + h, x
h
→0
h
existing as the definition of differentiability and the value of f (c). Example 6.1. Let f : R since x = 2 implies
f (x) f (c) x2 = x 2 x
− −
→ R be defined by f (x) = x2.
Then f (2) = 4,
− 22 = x2 − 4 = (x − 2)(x + 2) = x + 2, − 2 x − 2 x − 2 59
60
CHAPTER 6. DIFFERENTIATION
so since g(x) = x is continuous at 2, Theorems 5.7 and 5.8 imply
− − = c , Similarly, for c ∈ R, for x f (x) − f (c) x2 − c2 x2 − c (x − c)(x + c) = = = = x + c. x − c x − c x − c x − c f (x) f (c) = lim x + 2 = 2 + 2 = 4. x→2 x→2 x c lim
Hence Theorems 5.7 and 5.8 imply
− −
f (x) f (c) = lim x + c = c + c = 2c. x→c x→c x c lim
∈ R, and f (x) = 2x. √ Example 6.2. Set f : [0, ∞) → R to be f (x) = x. Then for c > 0 , x = c , √ x − √ c √ x − √ c √ x + √ c f (x) − f (c) √ x + √ c = = x − c x − c x − c x − c 1 √ √ = = √ √ . (x − c)( x + c) x + c So f (x) − f (c) 1 1 lim = lim √ √ = √ → → x + c 2 c x − c So f is differentiable at every x
x
c
x
c
by Theorems 5.7 and 5.8 and TS6 Q3. So for x > 0 , f (x) = Example 6.3. Given n f (x) = nx n−1 .
∈
N, define f : R
→ R by f (x)
√ = 12 x− 12 .
1 2 x
= xn . Then
Proof. Let c
∈ R. Then for x ∈ R we observe that f (x) − f (c) = x − c = (x − c)(x −1 + cx −2 + c2 x −3 + · ·· + c −2 x + c −1 ). So if x = c, f (x) − f (c) = x −1 + cx −2 + c2 x −3 + ·· · + c −2 x + c −1 x − c n
n
n
n
n
n
n
n
n
n
n
n
and
− −
f (x) f (c) = lim xn−1 + cxn−2 + c2 xn−3 + + cn−2 x + cn−1 x→c x→c x c = c n−1 + ccn−2 + c2 cn−3 + + cn−2 c + cn−1 = ncn−1 .
f (c) = lim
···
· ··
61 Our first main theorem of this chapter shows that differentiability is stronger than continuity. Theorem 6.1. Let I R be an open interval. Then for c differentiable at c implies f is continuous at c.
⊆
f (x) f (c) x c
− −
Proof. Since limx→c implies that lim f (x)
x
→c
−
= f (c) and limx→c x
− −
f (x) f (c) f (c) = lim (x c) = x→c x c = f (c) 0 = 0,
·
−
∈ I , f being
− c = 0, Theorem 5.8
− −
f (x) f (c) lim x→c x c
−
lim (x c)
x
→c
i.e., limx→c f (x) = f (c). So by Theorem 5.7, f is continuous at c. It is not true that f being continuous at c implies f is differentiable at c, as the following example shows. Example 6.4. Define f : R R by f (x) = x . Then f is continuous (see Theorem 5.2 plus the fact that g(x) = x is continuous), but f is not differentiable at 0 since the left limit is
→
lim
→0−
x
| |
f (x) f (0) x = lim = x 0 x→0− x
− −
−
−1,
but the right limit is lim
→0+
x
so limx→0
f (x) f (0) x 0
− −
f (x) x
− f (0) = − 0
lim
x
→0−
x = 1, x
doesn’t exist by Theorem 5.9.
We’ve seen that, intuitively, a function is continuous if it has ‘no gaps’ and the above example suggests that intuitively a function is differentiable if it has ‘no corners’. However, a function like f (x) = x 2 sin x1 (see TS10) eludes such intuition. The next theorem gives the standard rules of differential calculus. Theorem 6.2. Suppose that I R is an open interval, f, g : I and g are differentiable at c I .
∈
⊆
→ R and f
62
CHAPTER 6. DIFFERENTIATION
1. Given a constant λ (λf ) = λf .
∈
R, the function λf is differentiable at c and
2. f + g is differentiable at c and (f + g) = f + g . 3. f g is differentiable at c and (f g) = f g + f g .
·
·
·
·
1
4. If g(c) = 0, then g is differentiable at c and
g
= −gg2 .
− −
1
Proof. 1)
− −
(λf )(x) (λf )(c) f (x) f (c) lim = lim λ , x→c x→c x c x c so as in Theorem 5.8, this limit exist and equals λf (c). 2) Exercise. 3) In this case (fg)(x) lim x→c x
− (fg)(c) = lim → − c
f (x)g(x) f (x)g(c) + f (x)g(c) f (c)g(c) x c x c g(x) g(c) f (x) f (c) = lim f (x) + g(c) . x→c x c x c
−
−
− −
−
− −
Since f is differentiable at c, it is continuous at c (Theorem 6.1), so Theorems 5.7 and 5.8 imply that this limit exists and equals f (c)g (c) + g(c)f (c).
4) First note that g(c) = 0 and g being continuous at c (Theorem 6.1) means that there exists δ > 0 such that x (c δ, c + δ ) implies g(x) = 0. Then
− 1
lim
→c
x
∈ −
g
1
(x)
g
x c
−
(c)
1
1 1 = lim x→c x c g(x) g(c) 1 g(c) g(x) = lim x→c x c g(x)g(c) g(x) g(c) 1 = lim . x→c x c g(x)g(c)
− −
−
− −
− −
So since g is continuous at c and g(c) = 0, Theorems 5.7 and 5.8 imply that this limit exists and is equal to gg(c(c)2) .
−
Corollary 6.3. For an open interval I that f and g are differentiable at c,
⊆ R, f, g : I → R and c ∈ I such
63
1. f g is differentiable at c and (f g) = f
−
−
− g
2. if also g(c) = 0, then f g is differentiable at c and
f g
=
f g f g . g2
−
Proof. 1) f g = f + ( 1)g, so the proof is immediate by parts 1) and 2) of Theorem 6.2.
−
f g
−
·
2) Since = f 1g , part 4) of Theorem 6.2 implies that 1g is differentiable at c, so part 3) of Theorem 6.2 implies f g
= f
1 g
+ f
1 = g
−fgg2
+
f f g f g = , g g2
−
as required.
→ }
One consequence of the last two results is that if p, q : R R are polynop mials, then p and q are differentiable. Moreover, q (which is known as a rational function ) is differentiable on the set x R : q (x) = 0 .
{ ∈ Example 6.5. Let n ∈ N and let f : R \ {0} → R be given by 1 f (x) = for x ∈ R \ {0}. x n
Then f (x) =
−
n . xn+1
Proof. Let g(x) = x n . Then Theorem 6.2 implies
1 f (x) = g
g −nx −1 − −n . = = = n
(g)2
x2n
xn+1
−nx− −1.
Note we could also have phrased this as f (x) = x −n and f (x) =
n
The final rule of differentiability we’ll give is about composition of functions.
⊆ ∈ ◦
→
Theorem 6.4 (Chain Rule). Let I , J R be open intervals and f : I R, g : J R be functions. Suppose that c I , f (c) J , f is differentiable at c and g is differentiable at f (c). Then g f is differentiable at c and
→
∈
(g f ) (c) = (g (f (c))f (c).
◦
64
CHAPTER 6. DIFFERENTIATION
‘Proof’: We wish to show
◦
(g f )(x) x
− (g ◦ f )(c) → (g(f (c))f (c), − c
so we could write
◦
− (g ◦ f )(c) = g(f (x)) − g(f (c)) = g(f (x)) − g(f (c)) f (x) − f (c) − c x − c f (x) − f (c) x − c and then argue that as x → c, f (x) → f (c), so g(f (x)) − g(f (c)) g(u) − g(f (c)) lim = lim = g (f (c)). → f (x) − f (c) → ( ) u − f (c) The problem here is that f (x) − f (c) could be zero for many values of x, so (g f )(x) x
x
c
u
f c
the above argument is invalid.
We omit a genuine proof of the Chain Rule. Example 6.6. Given the function h : R compute h (x).
→ R where h(x) = (x3 + x2 + 3)6,
Solution: We can write h(x) = g f (x) where f (x) = x3 + x 2 + 3 and g(u) = u6 . So since f (x) = 3x2 + 2x and g (u) = 6u5 , the Chain Rule implies that
◦
h (x) = (g f ) (x) = g (f (x))f (x) = 6(f (x))5 (3x2 + 2x)
◦
= 6(x3 + x2 + 3)5 (3x2 + 2x). Example 6.7. Suppose that h : R R is defined by h(x) = sin(x3 + 7x). Then h (x) = (3x2 + 7) cos(x3 + 7x).
→
Proof. We can write h(x) = g f (x) where f (x) = x 3 + 7x and g(u) = sin u. Then since f (x) = 3x2 + 7 and g (u) = cos u, the Chain Rule implies that
◦
h (x) = (g f ) (x) = g (f (x))f (x) = (cos(x3 + 7x))(3x2 + 7).
◦
Our next results extract information about maxima/minima and average slopes using the definition of derivative.
65 Theorem 6.5. Let I function.
⊆ R be an open interval and f : I → R a differentiable
1. If f attains its maximum at c I then f (c) = 0.
∈ 2. If f attains its minimum at c ∈ I then f (c) = 0. Proof. 1): By assumption,
∈
f (x) f (c) for all x I. Hence f (x)
− f (c) 0 for all x ∈ I .
∈ I has x < c, then x − c < 0, so f (x) − f (c) f (x) − f (c) 0, hence lim 0. x − c x − c → On the other hand, if x ∈ I has x > c, then x − c > 0, so f (x) − f (c) f (x) − f (c) 0, hence lim 0. x − c x − c → If x
x
c−
x
c+
Since f is differentiable at c, the left and right limits must exist and be equal, which is only possible if f (x) x→c x lim
as in Theorem 5.9.
− f (c) = 0 − c
2) follows similarly. The main Theorem of this chapter is the Mean Value Theorem. This gives information on the derivative of a function between two given points. The following result is a simpler version of this, and roughly states that if a graph is differentiable and f (a) = f (b) then there is a point between a and b where the graph is flat.
∈
Theorem 6.6 (Rolle’s Theorem). Suppose that a, b R with a < b. Suppose further that f : [a, b] R is continuous and is differentiable on (a, b) with f (a) = f (b). Then there exists θ (a, b) such that
→
∈
f (θ) = 0.
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CHAPTER 6. DIFFERENTIATION
Proof. Case 1: f (x) = f (a) for all x x (a, b), so we are finished.
∈
•
∈ [a, b]. Then f (x) = 0 for all
• Case 2: there exists y ∈ (a, b) such that f (y) > f (a). Since f : [a, b] → R is continuous, the Extreme Value Theorem implies that there exists θ ∈ [a, b] such that f (x) f (θ) for all x ∈ [a, b]. In particular, since there exists y ∈ (a, b) such that f (y) > f (a), this means that θ = a, b, so θ ∈ (a, b). So f has a maximum at θ ∈ (a, b). Since f is differentiable on (a, b), Theorem ?? implies that f (θ) = 0.
• Case 3: there exists y ∈ (a, b) such that f (y) < f (a).
This follows
similarly to case 2, with minimum replacing maximum.
One immediate consequence of Rolle’s Theorem is the following.
∈
Theorem 6.7 (Mean Value Theorem). Suppose that a, b R with a < b, f : [a, b] R is continuous and f is differentiable on (a, b). Then there exists θ (a, b) such that
→ ∈
f (θ) =
Proof. Set g(x) = f (x)
− −
f (b) f (a) . b a
− λx where we choose λ such that g(a) = g(b). Thus f (a) − λa = f (b) − λb,
so λ =
− −
(f (b) f (a) . b a
We can then apply Rolle’s Theorem to g to obtain θ g (θ) = 0, which means that f (θ) = λ = as required.
− −
f (b) f (a) , b a
∈ (a, b) such that
67
6.1. HIGHER DERIVATIVES
Note that one way of phrasing the Mean Value Theorem is that there exists θ (a, b) such that f (b) = f (a) + (b a)f (θ).
∈
−
(This is useful in itself, but also tees up Taylor’s Theorem.) Theorem 6.8. Let f : [a, b] R be continuous and f be differentiable on (a, b). If f (x) = 0 for all x (a, b) then there exists k R such that f (x) = k for all x [a, b].
→ ∈
∈
∈
∈ −
∈
Proof. Let k = f (a) and c (a, b]. Then MVT implies there exists θ (a, c) such that f (c) = f (a) + (c a)f (θ). Since f (θ) = 0 then f (c) = f (a) = k, as required. Example 6.8. Let f : R actually differentiating!)
→ R be given by f (x) = x2.
Prove (without
∈ R such that f (θ) = 1; 2. there exists θ ∈ R such that f (θ) = −2; 1. there exists θ
Proof. 1) Since f (0) = 0 and f (1) = 1, MVT implies there exists θ such that f (1) f (0) f (θ) = = 1. 1 0
∈ (0, 1)
− −
−
2) Since f (0) = 0 and f ( 2) = 4, MVT implies there exists θ that f (0) f ( 2) 0 4 f (0) = = = 2. 0 2 2
− − −−
6.1
−
∈ (−2, 0) such
−
Higher Derivatives
Differentiating a function once yields another function f . If this function is differentiable, differentiating again yields yet another function f , the second derivative of f . Similarly we can often find the third derivative etc.
68
CHAPTER 6. DIFFERENTIATION
Repeating this process n times, if this is possible, yields the nth derivative f (n) . This can also be denoted (Dxn f )(x),
dn f (x). dxn
Supposing that f and g are n 2 times differentiable, 1. Since we know (f + g) = f + g , then (f + g) = (f + g ) = f + g . (Clearly (f + g)(n) = f (n) + g (n) .) 2. If λ R, then (λf )(n) = λf (n) .
∈
3. (fg) is more difficult. By Theorem 6.2, (fg) = f g + f g , so (fg) = (f g + f g ) = (f g) + (f g ) = (f g + f g ) + (f g + f g ) = f g + 2f g + f g . This generalises further for (fg)(n) (Leibniz’s Theorem), but we omit that here. The following theorem, a fundamental theorem used throughout mathematics, is a generalisation of the MVT Theorem 6.9 (Taylor’s Theorem). Let n N and suppose that f : [a, b] R is a continuous function which is n times differentiable on (a, b). Then there exists θ (a, b) such that
∈
→
∈
2
− a)f (a) + (b −2! a) f (a) + · ·· (b − a) −1 ( −1) ·· · + (n − 1)! f (a) + (b −n! a)
f (b) = f (a) + (b
n
n
Proof. We define the function F n : [a, b]
n
f (n) (θ).
→ R by
(b − x)2 (b − x)n−1 (n−1) F n (x) = f (b) −f (x)− (b− x)f (x)− f (x)−···− f (x). 2!
(n
− 1)!
69
6.1. HIGHER DERIVATIVES
Then F n (x) = f (x) + [f (x) (b x)f (x)] 1 + (b x)f (x) (b x)2 f (x) + 2! (b x)n−2 (n−1) (b x)n−1 (n) + f (x) f (x) (n 2)! (n 1)! (b x)n−1 (n) = f (x). (n 1)!
−
Now set Gn
− − − −
−
− −
∗
( )
− −−
· ·· − −− : [a, b] → R to be
Gn (x) = F n (x)
· ··
n
− − b x b a
F n (a).
−
−
−·
Then Gn (a) = F n (a) F n (a) = 0 and Gn (b) = F n (b) 0 F n (a) = 0. So since Gn is differentiable, Rolle’s Theorem implies that there exists θ (a, b) such that G n (θ) = 0. I.e., using ( ),
∗ −n(b − θ) −1 F (a) 0 = G (θ) = F (θ) − (b − a) 1 − (b − θ) n(b − θ) −1 ( ) =− f (θ) + F (a) (n − 1)! (b − a) (b − θ) −1 (b − θ) −1 ( ) = n F (a) − f (θ) (b − a) (n − 1)! (b − θ) −1 = n f (b) − f (a) − (b − a)f (a) − ·· · (b − a) −1 ·· · − (b(n−− a) 1)! f ( −1)(a) − (b −n! a) n
n
n
n
n
n
n
n
n
n
n
n
n
n
∈
n
n
n
n
n
n
f (n) (θ) .
Hence n 1
(b − a) f (b) = f (a) + (b − a)f (a) + · · · + (n
−
− 1)!
n
(b − a) (n) f − (a) + f (θ), (n 1)
n!
as required. A more Scottish version of this result comes from setting a = 0 and b = x to obtain Maclaurin’s Theorem:
xn−1 (n−1) 2 f (0) f (x) = f (0) + xf (x) + x +··· f (0) + Rn 2!
(n
− 1)!
70
CHAPTER 6. DIFFERENTIATION n
where Rn = xn! f (n) (y) for some y (0, x). (In fact, there is evidence that Gregory http://www-history.mcs.st-andrews.ac.uk/history/Biograph ies/ Gregory.html ‘discovered Taylor’s Theorem’ before Taylor, while working at St Andrews.)
∈
Example 6.9. Let f ”R R be given by f (x) = ex . Then f (x) = ex , f (x) = ex , , f (n) (x) = ex . So since e0 = 1, the expression from Maclaurin’s Theorem gives
→
· ··
f (x) = 1 + x +
for some y
x 2 x3 + + 2! 3!
n 1
−
n
· ·· + (nx− 1)! + xn! e
y
∈ (0, x).
Example 6.10. If f (x) = sin x, then f (0) = 0; f (x) = cos x, so f (0) = 1; f (x) = sin x, so f (0) = 0; f (x) = cos x, so f (0) = 1; f (x) = sin x, so there is y (0, x) such that
−
−
∈
2
3
−
4
3
4
x x x x x f (x) = 0 + 1 · x + 0 · − 1 + sin y = x − + sin y. 2! 3! 4! 3! 4! Example 6.11. If f (x) = log(1 + x) (here we mean log to base e), then −1 2 , so f (0) = 1; f (0) = log 1 = 0; f (x) = 1+1 x , so f (0) = 1; f (x) = (1+ x) f (x) = f (n) (x) =
2 , so f (1+x)3 ( 1)n+1 (n 1)! , (1+x)n
−
(0) = 2. In fact we can show by induction that
−
−
so
f (x) = x
−
x2 x3 + 2! 3
−
x 4 x 5 + 4 5
− ·· · .
We can actually show that this is a convergent series (this goes slightly beyond this course). Note that in particular, log 2 = 1
− 12 + 13 − 41 + 51 − . . . .
Appendix A
Some notation for MT2502 and beyond Important note: notation varies from mathematician to mathematician (within reason, usually), so a definitive list is impossible.
A.1
∀
Logic
for all
A.2
s.t. there exists such that
∃
⇒
implies
⇐
is implied by
⇐⇒
if and only if
Numbers
• N denotes the set of natural numbers. In MT2502 this means 1, 2, 3, . . .
(. . . means ‘and so on forever’). (Note that for many other mathematicians, and courses at St Andrews, N is 0, 1, 2, . . .: there is no universally, or even locally, agreed convention.)
• Z denotes the integers:
the set of whole numbers, both positive and negative, and 0, i.e., . . . , 2, 1, 0, 1, 2, . . .
− −
• Q denotes the rational numbers, i.e, numbers which can be expressed p where q
p, q are integers (it’s common to assume that one of these is 71
iff if and only if
