Solutions Manual for Joseph Taylor’s
Foundations of Analysis Michael Senter
2
2
Part I Single Variable - 3210
3
Chapter 1 The Real Numbers 1.1
Sets and Functions
Exercise 1.1.1. If a, b R and a < b, give a description in set theory notation for each of the intervals (a, b), [a, b], [a, b), and (a, b] (see Example 1.1.1).
∈
{x ∈ R : a < x < b } {x ∈ R : a ≤ x ≤ b} {x ∈ R : a ≤ x < b} {x ∈ R : a < x ≤ b}. Theorem 1.1.2. If A, B , and C are sets, then A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ). Proof. If x ∈ A ∩ (B ∪ C ), then x ∈ A and x ∈ (B ∪ C ). Thus, either x ∈ B or x ∈ C . Thus, x ∈ A ∩ B or in x ∈ A ∩ C . Thus, if x ∈ A ∩ (B ∪ C ), then x ∈ (A ∩ B) ∪ (A ∩ C ). If an x ∈ (A ∩ B) ∪ (A ∩ C ), then either x ∈ (A ∩ B) or x ∈ (A ∩ C ). This means that surely x ∈ A, and also that x ∈ B ∪ C . Hence, if x ∈ (A ∩ B) ∪ (A ∩ C ), then x ∈ A ∩ (B ∪ C ). Therefore, A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ). (a, b) = [a, b] = [a, b) = (a, b] =
Exercise 1.1.3. Solved in Class Question 1.1.5. What is the intersection of all the closed intervals containing the open interval (0, 1)? Justify your answer.
A
Let denote the set of all sets such that (0, 1) intervals is denoted . It is defined as
A
A
{
= x : x
∈ A.
The intersection of all closed
∈ A ∀A ∈ A}.
In other words, we are looking for some set A such that A is a subset of every other set in . This set is A = x : 0 < x < 1 . Consider any other subset C of . If C = A, then necessarily there must exist an element x such that x C and x / A, showing that A C but C A. Since x is not in every subset of , x / .
A
{
}
A 5
∈ ∈ A
∈
A
⊂
6
CHAPTER 1. THE REAL NUMBERS
Question 1.1.6. What is the union of all of the closed intervals contained in the open interval (0, 1)? Justify your answer. Let be the set containing all sets containing (0, 1) as a subset. The union of all these sets is denoted by . An object x is an element of if there exists some set C such that x C . We need to consider two cases: either the object x 0 or 1 x. The case of 0 < x < 1 is trivial. For any x such that 1 x we can create a set C such that . The case of x 0 is C = y : 0 < y < x . This since 1 x, it is guaranteed that C analogous. Hence, = ( , ).
A
A } A
∈
{
A
−∞ ∞
≤
≤
≤
⊂ A
≤
⊂ A
≤
Problem 1.1.7. If A is a collection of subsets of a set X, formulate and prove a theorem like Theorem 1.1.5 ( from book numbering ) for the intersection and union of A. Theorem 1.1.7. Let ( )c = A c1 Ac2 ...
A be a collection of subsets A , A ,...,A ∩ ∩ ∩ A and ( A) = A ∪ A ∪ ... ∪ A .
A
1
c n
c
c 1
n of
2 c n
c 2
some set X . Then
Proof. This is a generalization of DeMorgan’s law, proved in the book. We begin with the statement ( )c = A c1 Ac2 ... Acn . We can rewrite ( )c as (A1 A2 ... An )c . We can then sub-partition this collection of unions into a collection of two unions, as such:
A
A
∩ ∩ ∩
A
)c = [A1
(
∪ ∪ ∪
c
∪ (A ∪ ... ∪ A )] 2
n
Then we will refer to A2 ... An as B. We can then rewrite the above as (A1 B)c , for which DeMorgans laws apply. Thus, we write (A1 B)c = A c1 B c = A c1 (A2 ... An )c . As next step, we sub partition B into two sets, as such
∪ ∪ (A2
∪
c
∪ ... ∪ A ) n
= [A2
∩
∪ (A ∪ .... ∪ A )] 3
n
∪
∩
∪ ∪
c
Then DeMorgan’s laws apply again as above, and we can write [ A2 (A3 .... An )]c = Ac2 (A3 ... An )c . Since intersections and unions are associative, we can then write
∩
∪ ∪ ∪
∪ ∪
A
(
)c = (Ac1
c 2
c
c 1
c 2
c
∩ (A ∩ (A ∪ ... ∪ A ) )) = A ∩ A ∩ (A ∪ ... ∪ A ) 3
n
3
n
We continue an inductive application of DeMorgan’s laws as outlined above, until we see that ( )c = A c1 Ac2 ... Acn
A
∩ ∩ ∩
The other proof is analogous, requiring a sub-partition of the collection of intersections and rewriting them into series of intersections of two sets to which DeMorgan’s laws apply.
R are one to one and which ones Problem 1.1.8. Which of the following functions f : R are onto. Justify your answer. (a) f (x) = x2 ; This function is neither onto, nor one-to-one. It is not onto, since there is no x such that f (x) < 0. It is not one-to-one since f (x) = f ( x) for all x R. (b) f (x) = x3 ; This function is both one-to-one and onto. It is one-to-one since there f (x) = f (y) for all x, y such that x = y. It is onto, as for any y R , there exists an x R such that f (x) = y. (c) f (x) = e x This function is one-to-one, but not onto. It is one-to-one, for f (x) = f (y) for all x, y R such that x = y. It fails to be onto since there exists no x such that f (x) < 0 for any x R.
→
−
∈
∈
∈
∈
∈
7
1.1. SETS AND FUNCTIONS
Theorem 1.1.9. If f : A F ) = f −1 (E ) f −1 (F ).
∩
1
→ B is a function and E and F are subsets of B, then f − (E ∩
Proof. If x f −1 (E F ), then f (x) E F . This means that f (x) is both in E as well as in F . If f (x) E , then x f −1 (E ) . If f (x) F , then x f −1 (F ). Since f (x) is in both E and F , x is in f −1 (E F ). Assume x is in f −1 (E ) f −1 (F ). Then, x f −1 (E ) as well as x f −1 (F ). If x f −1 (E ), then f (x) E . If x f −1 (F ), then f (x) F . Since x is both in f −1 (E ) as well as f −1 (F ), we know that f (x) E F . This implies that x f −1 (E F ). Since every x f −1 (E F ) implies that x f −1 (E ) f −1 (F ) and vice versa, it is true that f −1 (E F ) = f −1 (E ) f −1 (F ).
∈
∩ ∈ ∈ ∩ ∩ ∈ ∈ ∈ ∩ ∈ ∩ ∩ ∩
∈ ∩
∈ ∈ ∈ ∈ ∈ ∩ ∈ ∩
∈
∈
→
Theorem 1.1.10. If f : A B is a function and E and F are subsets of B, then 1 1 1 − − − f (E F ) = f (E ) f (F ) if F E .
\
\ ⊂ Proof. If x ∈ f − (E \F ), then f (x) ∈ E \F . Thus f (x) ∈ E but f (x) ∈ / F . This means that − − − − x ∈ f (E ) and but also x ∈ / f (F ). In other words, x ∈ f (E )\f (F ). − Assume now that x ∈ f (E )\f − (F ). Then x ∈ f − (E ) but x ∈ / f − (F ). This means that f (x) ∈ E \F , and hence x ∈ f − (E \F ). It follows that f − (E )\f − (F ) = f − (E \F ). Theorem 1.1.11. If f : A → B is a function and E and F are subsets of A, then f (E ∪ F ) = f (E ) ∪ f (F ). Proof. If y ∈ f (E ∪ F ), then y = f (x) for some x ∈ E or x ∈ F . If x ∈ E , then y ∈ f (E ). If x ∈ F , then y ∈ f (F ). Since x is in either one of these, we know that y ∈ f (E ) ∪ f (F ). Assume now that y ∈ f (E ) ∪ f (F ). This implies that y = f (x) for some x ∈ E or x ∈ F . Thus we can write x ∈ E ∪ F . Then y ∈ f (E ∪ F ). Since any element of f (E ∪ F ) is in f (E ) ∪ f (F ) and vice versa, we conclude that f (E ∪ F ) = f (E ) ∪ f (F ). Theorem 1.1.12. If f : A → B is a function and E and F are subsets of A, then f (E ∩ F ) ⊂ f (E ) ∩ f (F ). Proof. Assume that y ∈ f (E ∩ F ). Then y = f (x) for some x ∈ E ∩ F . This means that both x ∈ E as well as x ∈ F . Then, f (x) ∈ f (E ) and f (x) ∈ f (F ), showing that f (x) ∈ f (E ) ∩ f (F ), or - equivalently - that y ∈ f (E ) ∩ f (F ). This proves that f (E ∩ F ) ⊂ f (E ) ∩ f (F ). Question 1.1.13. Give an example of a function f : A → B and subsets F ⊂ E of A for which f (E )\f (F ) = f (E \F ). 1
1
1
1
1
1
1
1
1
1
1
1
1
The above conditions are fulfilled for a function f (x) = x with A = B = [0, 10], and the subsets E = [1, 6] and F = [1, 2] E .
⊂
Exercise 1.1.14. Solved in Class Exercise 1.1.15. Solved in Class
8
CHAPTER 1. THE REAL NUMBERS
1.2
The Natural Numbers
1.3
Integers and Rational Numbers
Exercise 1.3.1. Given that N has an operation of addition which is commutative and associative, how would you define such an addition operation in Z ? Assume a, b Z. We differentiate three cases: a) Both a, b N. In this case, proceed as in N . b) Both a,b < 0. In this case a + b = (c + d) where c = a and d = b. As such, we may proceed as in the case of natural numbers. c) assume a > 0, b < 0. We now distinguish two cases. Provided a > b, we may reduce a to a sum of two integers m, n N such that n = b. We then define a + b = m + n + b = m, since n + b = 0 by definition of the additive inverse. Assume now instead that a < b. In this case, we can split b into an addition problem, with two integers m, n N such that m + n = b and a = m. Then
∈
∈
−
−
∈
− −
−
−
−
∈
−
−
−
a + b = a + ( (m + n)) = a + ( m) + ( n) =
−
−n.
This defines addition in Z .
1.4
The Real Numbers
Exercise 1.4.1. For each of the following sets, describe the set of all upper bounds for the set: (a) the set of odd integers; The integers are unbounded. (b) 1 1/n : n N ; The set of all upper bounds for this set is x N : x 1 . (c) r Q : r 3 < 8 ; The set of all upper bounds for this set is x Q : x 2 . (d) sin x : x R ; The set of all upper bounds for this set is x R : x 1 .
{ − { ∈ {
∈ } } ∈ }
{ ∈ ≥ } { ∈ ≥ } { ∈ ≥ }
Exercise 1.4.2. For each of the sets in (a), (b), (c) of the preceding exercise, find the least upper bound of the set, if it exists. (a) There is no upper bound, and hence no least upper bound. (b) The least upper bound is 1. (c) The least upper bound is 2.
Theorem 1.4.3. If a subset A of R is bounded above, then the set of all upper bounds for A is a set of the form [x, ). What is x?
∞
Proof. Let B denote the set of all upper bounds of A. By definition, a number m R is considered an upper bound for the set A if z m for all z A. If the set A has a largest number, then this largest number - y - will be in the set B . In that case, it is obvious that all numbers m > y will also be upper bounds, since we assumed that x y for all x A, and that m > y , it follows that x y < m. Therefore, the set [y , ) would be the set of
≤
≤
∈
∈
∞
≤
∈
9
1.5. SUP AND INF
all upper bounds of A. Assume now that A does not have a largest number. By the completeness theorem we know that any subset A of an ordered field - such as R - is indeed bounded above. Specifically, according to theorem 1.4.4 of the book we know that any subset of R not only is bounded above, but has a least upper bound. By definition, a number c is a least upper bound if and only if it is a number such that x c for all x A, and for every k R , if k is an upper bound of A, then k c. It is obvious then that the set of all upper bounds of A will be the set [c, ) where c is the least upper bound of A.
≤
≥
∞
Exercise 1.4.4. If x 2 < 1
2
so x
∈
− − √ − √ 5
2 upper bound.
,
1+ 2
∈
− x, then x + x − 1 < 0, hence √ √ 1+ 5 1− 5 − − x− x−
1
∈
5
2
2
< 0,
√ − 1+ 5 , hence A is bounded above by , which is its least 2
Exercise 1.4.6. For the forward direction, observe that for each pair x, y F with x > 0, y/x in F and since F is archimedean, we have that for some integral n, n > y/x so nx > y as desired. For the reverse direction, we have for each pair x, y F, x > 0, there exists integral n with nx > y. In particular, let x = 1 > 0, so n > y . Hence F is archimedean.
∈
∈
−
Exercise 1.4.7. By the archimedean property, there exists natural n with y x > 1/n. Let k be an integer with k/n < x < (k + 1)/n, then adding the inequalities y x > 1/n and x > k/n gives us y > (k + 1)/n, so (k + 1)/n is a rational between x and y. This proves the claim if both x, y are positive. If both are negative, the proof works essentially the same way. If one is positive and the other negative, then take 0 as the rational number between the two.
−
Exercise 1.4.8. . Suppose not, so that x + r = s Q and xr = s and x = s /r Q, both of which lead us to a contradiction.
∈
∈
1.5
∈ Q, then x = s − r ∈ Q
Sup and Inf
Exercise 1.5.1. For each of the following sets, find the set of all extended real numbers x that are greater than or equal to every element of the set. Then find the sup of the set. Does the set have a maximum? (a) ( 10, 10); The set of all numbers greater than this set is the set [10 , + ). The supremum of the set in question is 10. The set does not have a maximum. (b) n2 : n N ; In the extended set of real numbers, the only element greater than or equal to all the elements in the set in question is + , which thereby must also be its supremum. The set does not have a maximum. (c) 2n+1 ; The set of all real numbers greater than the set in question is the set [2 , ). The n+1 supremum is 2 and the set does not have a maximum.
− {
{
∞
∈ }
}
∞
∞
10
CHAPTER 1. THE REAL NUMBERS
Exercise 1.5.2. Find the sup and inf of the following sets. Tell whether each set has a maximum or a minimum. (a) ( 2, 8]; The infimum of the set is 2 and the supremum is 8. The has a maximum, but not a minimum. (b) nn+2 2 +1 ; The infimum of the set is 0, and the supremum is 2. The set has a maximum, but no minimum. (c) n/m : n, m Z , n2 < 5m2 ; The infimum of the set is 5, and the maximum is 5. Seeing that 5 is not a rational number, the set has neither a maximum nor a minimum.
−
−
{
√ ∈
−√
}
√
∞, then for each n ∈ N there is an element a ∈ A
Exercise 1.5.3. Prove that if sup A < such that sup A 1/n < an sup A.
−
≤
n
Proof. This is true since we can easily construct an element an such that this equality holds. We assume that A is defined for all m/n with m, n Z within A. In this case, we constructs our term to be an = sup A 1/(n + 1). It is obvious that since 1/(n + 1) < 1/n, that sup A 1/n < sup A 1/(n + 1) sup A. Alternatively, we may also note that sup A 1/n < sup A for all n N by definition, so the inequality holds in the trivial case of a n = sup A.
−
∈
−
−
≤
Exercise 1.5.4. Prove that if sup A = such that an > n.
−
∈
∞, then for each n ∈ N there is an element a ∈ A n
∞
∀ ∈
∞
Proof. Assume some set A whose supremum is + . In that case, x A, x < . Both from the Archimedean property and from the Peano Axioms we know that for every n N, there is a successor element n which is also in N, such that n < n . Since there a such that a = , and n < , this implies that an such that a n = n and a n A, showing that n < an < .
∞ ∞
∞
∃
∈
∈
Exercise 1.5.5. Formulate and prove the analog of Theorem 1.5.4 for inf. Theorem. Let A be a non-empty subset of R and let x be an element of R . Then (a) inf A x if and only if a x for every a A; (b) x > inf A if and only if x > a for some some a A.
≥
≥
∈
∈
≥
Proof. By definition, a x if and only if x is a lower bound for A. If x is a lower bound for A, then A is bounded below. This implies that its inf is its greatest lower bound, which is necessarily greater than or equal to x. Conversely, if inf A x, then inf A is finite and is the greatest lower bound for A. Since inf A x, x is also a lower bound for A. Thus, inf A x if and only if a x for every a A. If x > inf A, then x is not a lower bound for A, which means x > a for some a A. Conversely, if x > a for some a A, then x > inf A, since a inf A. Thus, x > inf A if and only if x > a for some a A.
≥
∈
∈ ∈
≥
≥
≥
≥ ∈
Exercise 1.5.6. Prove part (d) of Theorem 1.5.7. Theorem. Let A, B be non-empty subsets of R . Then sup(A
− B) = sup A − inf B.
11
1.5. SUP AND INF
Proof. According to the book, sup(A + B) = sup A + sup B (proof on p. 30). We can then write sup(A + ( B)) = sup A + sup( B). We then apply Theorem 1.5.7b, to rewrite sup( B) as inf A. From this it follows that
−
−
−
−
−
sup(A + ( B)) = sup(A
− B) = sup A + (− inf B) = sup A − inf B
.
Exercise 1.5.7. Prove (e) of Theorem 1.5.7. Theorem. Let A, B be non-empty subsets of R. If A inf B inf A.
≤
⊂ ≥
∈ ∈ − ∈ ≤
∈
⊂ B, then sup A ≤ sup B and ∈
∈ ∈ − ∈ − ∈
Proof. If A B , then a A implies that a B for all a. Then, if sup A A, sup A B . Since sup B b for all b B, it is obvious that sup A sup B. Assume now that sup A / A. In that case, sup A A for all > 0. Thus, sup A A and sup A B. By definition, sup B is greater than or equal to all b B. This means that if sup A B implies that sup A sup B. The proof for the infimum is analogous.
≤
∈
− ∈
Exercise 1.5.10. Prove (a) of Theorem 1.5.10.
c
∈
Theorem. Let f and g be functions defined on a set containing A as a subset, and let R be a positive constant. Then sup A cf = c supA f and inf A cf = c inf A f .
→
Proof. Let f be function f : A B. Then sup f is the supremum of B provided that f is surjective. Let M be an arbitrary upper bound of cx for some x B. We say that cx M if and only if x M/c. This shows that M is an upper bound of cx if and only if M/c is an upper bound of B. Hence, sup cB = c sup B and similarily sup cf = c sup f . The result for the infimum follows similarily.
∈
≤
≤
Exercise 1.5.8. [(a)] supI f = 1, inf I f = 0. These are the max and min respectively. 2. 1. supI f =
∞, inf f = 3 at x = 2. Neither are the max nor min of f on I . I
3. inf I f = 0, supI f = 1. The inf of f on I is actually a minimum, but the sup is not a maximum.
Exercise 1.5.9. Solved in Class Exercise 1.5.11. Prove (b) of Theorem 1.5.10.
c
∈
Theorem. Let f and g be functions defined on a set containing A as a subset, and let R be a positive constant. Then sup A ( f ) = inf A f .
−
−
12
CHAPTER 1. THE REAL NUMBERS
→
Proof. We have a function f : A B. A number x is a lower bound for f (a) for all a A if and only if x is an upper bound for the set f (a). Let L be the set of all lower bounds for f (a). Then L is the set of all upper bounds for f (a). Furthermore, the largest member of L and the smallest member of L are negatives of each other. That is, inf f (a) = sup(f (a)), or equivalently inf f = sup( f ).
∈
−
−
−
−
−
−
−
−
Exercise 1.5.12. Prove (c) of Theorem 1.5.10. Theorem. Let f and g be functions defined on a set containing A as a subset, and let c R be a positive constant. Then supA (f + g) supA f + supA g and inf A f + inf A g inf A (f + g).
∈
≤
≤
≤
∈
≤
∈
Proof. By definition, f (a) sup f for all a A and g(a) sup g for all a A. Therefore, f (a) + g(a) sup f + sup g. Let c denote the supremum of f +g. We know that sup f + sup g is an upper bound for f (a) + g(a). Since the supremum is always less than or equal to an upper bound, we find that c sup f +sup g. This implies that sup(f +g) sup f +sup g.
≤
≤
≤
Exercise 1.5.13. Prove (d) of Theorem 1.5.10.
c
∈
Theorem. Let f and g be functions defined on a set containing A as a subset, and let R be a positive constant. Then sup f (x) f (y) : x, y A = supA f inf A f .
{
−
∈ }
−
∈
Proof. This appears somewhat obvious. The function f is defined on A, i.e., for every a A, f maps to some value f (a) in some set, let’s call it B. The value sup f is defined as to be the least upper bound of f (a), i.e. x such that f (x) > sup f for some x A. The infimum is defined as the value such that there is no value x A such that x < inf f . The value defined by f (x) f (y) for all x, y A is a measure of the distance between these two values. Since sup f and inf f are defined as above, we can see that there cannot be a greater distance between any other two points in B than the distance between sup f and inf f . Therefore, for any collection of distances between points in B reached by f (x) for all points x A, the supremum of this collection - namely, the largest value of this set such that no other value is larger - cannot be any other than the distance between the supremum and the infimum of the function itself.
−
∈
∈
∈
∈
Chapter 2 Sequences 2.1
Limits of Sequences
Exercise 2.1.1. Show that (a) if x 5 < 1, then x is a number greater than 4 and less than 6.; This is equivalent to saying 1 < x 5 < 1. We add 5 to the inequality, and we get 4 < x < 6. (b) if x 3 < 1/2 and y 3 < 1/2, then x y < 1 ; We add the inequalities, such that we see x 3 + y 3 < 1/2 + 1/2 = 1. We notice that y 3 = 3 y . We rewrite using the triangle inequality:
| − | − − | − | | − | | − | | − |
| − |
| − | | − |
|(x − 3) + (3 − y)| ≤ |x − 3| + |3 − y| < 1 |x − y| ≤ |x − 3| + |3 − y| < 1. (c) if |x − a| < 1/2 and |y − b| < 1/2, then |x + y − (a + b)| < 1 . We add the inequalities and get |x − a| + |y − b| < 1/2 + 1/2 = 1. We can then rewrite using the triangle inequality as above
|(x − a) + (y − b)| ≤ |x − a| + |y − b| |x + y − a − b| ≤ |x − a| + |y − b| |x + y − (a − b)| ≤ |x − a| + |y − b|
< 1 < 1 < 1.
Exercise 2.1.2. Adding the two inequalities gives
|x − 1| + |x − 2| < 1, hence by the triangle inequality | 2x − 3 | ≤ |x − 1 | + |x − 2 | < 1, so |x − | < . Hence x ∈ (1, 2). Combining this with |x − 1 | < 1/2 and |x − 2 | < 1/2 we have x ∈ (1, ) and x ∈ ( , 2), a contradiction. 3 2
3 2
1 2
3 2
Exercise 2.1.3. Put each of the following sequences in the form a1 , a2 , a3 , . . . , an . This requires that you compute the first 3 terms and find an expression for the nth term. (a) the sequence of positive odd integers; This is a sequence of the form 1, 3, 5, . . .. To find the n th term, we express this sequence as a n = 2n 1, with n N. (b) the sequence defined inductively by a1 = 1 and an+1 = a2n ; The sequence begins
−
−
13
−
∈
14
CHAPTER 2. SEQUENCES
with 1, 1/2, 1/4, . . .. The nth term will be something like an = (( 1)n−1 )/(2n−1 ) for n N. an . This is the series (c) the sequence defined inductively by a1 = 1 and an+1 = n+1 2 1, 1/3, 1/12, 1/60, . . .. The nth term is: an = (n+1)! .
−
−
∈
Exercise 2.1.4. Find lim 1/n2 . The larger n become, the smaller 1/n2 will become. We guess the limit to be 0. For any > 0, we need an N such that whenever n > N , 1/n2 < . We find that this is true whenever 1/ < n2 , or in other words - whenever 1/ < n.
−1 . Exercise 2.1.5. Find lim 2n 3n+1 We guess the limit to be 2/3.
− 1 − 2 | = | 3(2n − 1) − 2(3n + 1) | = | 6n − 3 − 6n − 2 | | 2n 3n + 1 3 3(3n + 1) 9n + 3 5 5 5 =| < | | < | | 9n + 3 9n n We must choose an n > N such that N > 5 so that this will be true.
Exercise 2.1.6. Find lim( 1)n /n
−
n
We guess the limit to be 0. We see (−n1) = such that N > 1 for this inequality to be true.
|
Exercise 2.1.7. The limit is zero. Let N =
n = 0. n→∞ n3 + 4
Hence lim
n n3 + 4
−
0 =
Hence we need to choose an n > N
√ 1 . Then if n > N ,
1 1 n n = = . < < n3 + 4 n3 n2 N 2
− − ⇒ √ | −
Exercise 2.1.8. The limit is zero. Let N =
√ n >
1 n
| | |.
2
1
2
2
1
2
. Then for n > N ,
2
2 n > 1
2 ,
|
hence adding 2 + n to both sides yields
√ | − | +
2 + n + 2 n > 1
2
2
+ n
≥ 1 + n,
where the last step follows from the triangle inequality. Factoring the left side gives
√
( n + )2 > n + 1, and since both sides are positive we may take the square root which yields hence
√ n + > √ n + 1 ⇒ > √ n + 1 − √ n = |√ n + 1 − √ n − 0|, √ √ lim n + 1 − n = 0. →∞
n
15
2.2. USING THE DEFINITION OF LIMIT
Exercise 2.1.9. The limit is zero. We know for any > 0, there exists N 1 such that for all n > N , 1 < /2, n since /2 is another positive real and we have shown the limit of 1 /n as n goes to infinity is zero in problem 4. Similarly, we may find N 2 such that for all n > N 2 ,
−
( 1)n < /2, n2
since we found the limit as n of ( 1)n /n earlier to be 0, and it can be shown in much the same way that lim ( 1)n /n2 = 0. Hence, for n > max N 1 , N 2 , both inequalities hold n→∞ and we may add the two and apply the triangle inequalityto obtain
−
hence the limit is zero.
→∞
−
1 ( 1)n + n n2
− −0 ≤
{
−
}
1 ( 1)n + < , n n2
Exercise 2.1.10. Prove that lim 2−n = 0. Hint: prove first that 2 n numbers n.
≥ n for all natural
Proof. We wish to show that 2 n > n for all n. Proof by induction. The base case, 2 1 > 1 is obviously true, since 2 1 = 2. We assume now that 2 n > n for some n. Then we wish to check 2n+1 . But, we can rewrite this simply as 2 n 21 . Let k = 2n . Since we know that k > n, it is obvious that 2k > n + 1. Thus, 2n > n for all n N. We note that 2−n = 21n . Thus, lim2−n = lim 21n . Since 2n increases until infinity, we see that 1/2n will grow smaller and smaller, since 1/2n > 1/2n+1 for all n. We see that for any > 0, we need to simply pick n such that 1/ < 2 n . As such, the limit is 0.
∈
Exercise 2.1.11. Prove that if a n
→ 0 and k is any constant, then ka → 0. n
→ 0, this means that a < for any > 0. We multiply by k and find that ka
If an
2.2
n
n <
k.
Using the Definition of Limit
Exercise 2.2.1. Make an educated guess as to what you think the limit is, then use the definition of limit to prove that your guess is correct. 2
2
2
−2 . I assume the limit will be 3. We note that 3n2 −2 < 3n2 = 3. Hence, the limit lim 3n n2 +1 n +1 n is 3. 1 Exercise 2.2.2. The limit is zero. Let N = , then 1 1 n n n 0 = = = . < < n2 + 2 n2 + 2 n2 n N
−
16
CHAPTER 2. SEQUENCES
Exercise 2.2.3. Make an educated guess as to what you think the limit is, then use the definition of limit to prove that your guess is correct. lim √ 1n I assume the limit will be 0. We see √ 1n =
√ we choose an n > N such that N >
1 n1/2
| | | |; therefore, this is true whenever
1 .
2 Exercise 2.2.4. The limit is 1. Let N = , then
− − − − 2
n n + 1
2n 1 2n + 1 2 1 = 2 = 2 < . n + 2n + 1 n + 2n + 1 n
This last step is true because for any positive integer n, we have 3n + 2 > 0, hence 2n2 + 4n + 2 > 2n2 + n, and so dividing by n(n2 + 2n + 1) on both sides yields the desired inequality. Hence, 2 2 2 n 1 < < = . n + 1 n N
Exercise 2.2.5. Make an educated guess as to what you think the limit is, then use the definition of limit to prove that your guess is correct. The limit is
1 . Let A = n 2
{ ∈ N : |a − 12 | ≥ } for a given . This set is finite since
√
n2
n
1 2
− n − ≥ ⇔ n
+ n
⇔ 0
2
+ n
≥
1 n + + 2 2
≥
2
1 + 2
2
+ n(2 + 1)
+ 2n.
But this last inequality is never true, since each of the terms is strictly greater than zero. Of course, we might also have 1 n + 2
−
√
n2 + n
2
2
≥ ⇔ n + +n(1 − 2) + ≥ n ⇔ −2n + ≥ 0 ⇔ 2 ≥ n,
2
+ n
2
which is certainly only true for finite n. Hence for a given the set A is in fact finite, so the 1 limit is . 2
Exercise 2.2.6. Make an educated guess as to what you think the limit is, then use the definition of limit to prove that your guess is correct. lim(1 + 1/n)3 = 1. Proof:
|(1 + n1 ) − 1| = | n1 3
3
+
3 3 + + 1 n2 n
− 1| = | n1
3
+
3 3 + n2 n
|
17
2.2. USING THE DEFINITION OF LIMIT
We note that each term is of the form c/n or multiples thereof for some constant c. It has already been shown that each such term tends can be made smaller than any . This also holds for the sum.
Exercise 2.2.7. Suppose the sequence converges to a, then for all > 0, there is an N so that when n > N , an a < a < an < a + . In particular, let = 1, then certainly for all n > N , the sequence is bounded above by a + 1 and below by a 1. For n < N, notice that the a1 ,...,aN is bounded above by e = max a1 ,..., an and below by f = min a1 ,...,aN . Hence the entire sequence is bounded above by max e, a + 1 and min f, a 1 .
| − |
≤ { { − }
⇒ −
−
{| | | |} {
}
}
Exercise 2.2.8. Prove that if lim an = a, then lim a3n = a3 . 3 n
3
2 n
2
|a − a | = |(a − a)(a + a a + a )| We then note that we are given that |a − a| < . From this we see that |(a − a)(a + a a + a )| < (a + a a + a ). Exercise 2.2.9. Does the sequence {cos(nπ/3)} have a limit? Justify your answer. No. The sequence {cos(nπ/3)} oscillates between −1 and 1; a limit cannot converge to n
n
n
2 n
n
2
n
2 n
2
n
two different values. Hence, this sequence does not have a limit.
Exercise 2.2.10. For a similar reason as in the previous problem, the sequence an = cos(πn) does not converge but the sequence an = cos(πn) does converge since an = 1 for all n.
| | |
{ }
|
| |
{ }
| | ≤ b
Exercise 2.2.11. Prove that if an and bn are sequences with an lim bn = 0, then lim an = 0 also.
n for
all n and if
| | ≤
| |≤ | |≤ | | ≤ | | ≤ | | Exercise 2.2.12. Prove the following partial converse to Theorem 2.2.3: Suppose {a } is a convergent sequence. If there is an N such that a ≤ c for all n > N , then lim a ≤ c. Also, if there is an N such that b ≤ a for all n > N , then b ≤ lim a .
We are given that an lim bn . We bn for all n. Therefore, we know that lim an know that lim bn = 0. Hence we can write - equivalently - that lim an 0. We notice that lim an 0, from an is defined to be greater than or equal to zero. Hence we have 0 which it follows by the squeeze theorem (proof on p. 43 of the book) lim an = 0. n
n
n
n
n
Note that an is bounded by c according to the premise. In this case, we can say that sup an sup an , and an c for all n. Let a = lim an . We know by definition that a therefore we can write that lim an sup an c. Likewise, we can say that b is a lower bound for an such that b inf an . We know that by definition inf an a, allowing us to write b lim an .
≤
≤
≤
≤
≤ ≤
≤
≤
18
CHAPTE CHAPTER R 2. SEQUEN SEQUENCES CES
Exercise Exercise 2.2.15. 2.2.15. Suppose this sequence converges to zero. Then for all > 0, there exists N consider n = = + 1 106 . This is a multiple N such N such that for all n > N , N , an < . But . But consider n 6 10 of one million, so a so a n . And, . And, it is larger than N N since
·
| |
≥
N N +1 < 106 106 by definition, 106
2.3 2.3
N + 1 > N. Hence we have a contradiction. 106
Limi Limitt Theo Theore rems ms
Exercise Exercise 2.3.1. 2.3.1. Solved in Class Exercise Exercise 2.3.2. 2.3.2. Use the Main Limit Theorem to find lim n2 5 lim 3 = 2 n2 + 5 n + 2n 1/n 5/n3 lim = 1 + 2/n 2 /n + + 5/n 5 /n3 lim(1/n lim(1/n 5/n3) 0 = = 0. lim(1 + 2/n 2/n + + 5/n 5/n3) 1
−
n2 5
−
n3 +2n +2n2 +5
.
(dividing top and bottom by n 3 )
−
−
Exercise Exercise 2.3.3. 2.3.3. By dividing numerator and denominator by 2 n, in a similar fashion to problem 1, we can see that the limit is 1. Exercise Exercise 2.3.4. 2.3.4. Let an = 1/n, /n, then lim an = 0, and let bn = sin n, then bn sin n theorem 2.3.2, the sequence = a n bn has limit zero. n
| | ≤ 1.
By
Exercise Exercise 2.3.5. 2.3.5. Prove Theorem 2.3.2. Proof. We know that lim an = 0, hence we know that for all > 0, there exists an N N such that whenever n > N , Likewise, e, we know know that bn is bounded, such that we can N , an < . Likewis state that q b b q . We can then also write an < |q | . We guess that the limit limit of (a ( an )(b )(bn ) is zero, so we write:
− ≤ ≤ ≤
| |
| |
|a b − 0| = |a b | ≤ |a q | |a b | ≤ |q | |q | |a b | ≤ . n n
n n
n
n n n n
Thus, the lim an bn = 0, since we there is an N an N such that the above ab ove inequality inequality is true whenever we pick an n an n > N . N .
19
2.3. 2.3. LIMIT LIMIT THEO THEOREM REMS S
{ } {| |}
Exercise Exercise 2.3.6. 2.3.6. Prove that a sequence an is both bounded above and bounded below if and only if its sequence of absolute values an is bounded above. such that an M Proof. By definition, if an is bounded above, then there exists some M such for all n. This is equivalen equivalentt to saying saying M a n M , M , which proves that an is bounded above and below.
{| |}
− ≤ ≤ ≤
{ }
| |≤
Exercise Exercise 2.3.7. 2.3.7. Prove part(b) of Theorem 2.3.6. Proof. Since both an and bn have a limit, we can write an a < 2 and bn b < 2 . For all all ,, we have an N an N such such that if we choose n > N , N , these inequalities are true. We know add them together and find
| − |
| − |
|a − a| + |b − b| < |(a ) + (b(b − b)| ≤ |a − a| + |b − b| < |(a + b + b ) − (a + b + b))| ≤ |a − a| + |b − b| < . n
n
n
n
{ { }
n
n
n
n
n
n
Exercise Exercise 2.3.8. 2.3.8. Prove that if bn is a sequence of positive terms and bn there is a number m > 0 such that b that b n m for all n.. m for all n
≥
→ b > 0, then
This is true by virtue of the definition of R . The statement above is equivalent to saying that we are looking for some m such that 0 < m definitio tion n R is full, such that bn . By defini between any two numbers, there are infinitely more numbers.
≤
Prove part (d) (d) of Theorem 2.3.6. 2.3.6. Hint: Hint: Use the previous previous exercise. exercise. I.e, that Exercise Exercise 2.3.9. 2.3.9. Prove if a and b n = 0 for all n. a n a and a and b n b, b , a n /bn a/b, a/b, if b = 0 and b
→
→
→
Proof. 1 1 − a 1b | ≤ |a = a | = a || | + |a|| b b −b We know that { 1/b } is bounded, and hence {| 1/b |} is bounde bounded d abov above. We also also have have |a − a| → 0. Therefore, |a − a||1/b | → 0. Also, |a||1/b − 1/b| → 0. By (b) we know that |a − a||1/b | + |a||1/b − 1/b| → 0, proving that a that a /b → a/b. a/b.
|a
n
1 bn
− a 1b | = |a
n
1 bn
− a 1b
+ a
n
1 bn
n
n
n
n
n n
n
n
n
n
n
n
n
n
Exercise Exercise 2.3.10. 2.3.10. Prove part (f) of theorem 2.3.6. Hint: use the identity: xk 1/k
with x = a = an
−y
k
= (x
k 1
)(x − − y)(x
1/k
and y = a = a1/k , to show that a that a n
+ xk−2 y + . + . . . + y + y k−1 )
→ a
1/k
≥ 0 for all n.
if a a n
Proof. We notice that an1/k
1/K
−a
= (an
− a)(a )(a
1/K 1 n
− + a1/K −2a1/k + . . . + a + a1/K −1 ) = (an − a)bn n
20
CHAPTE CHAPTER R 2. SEQUEN SEQUENCES CES
where = an1/K −1 + an1/K −2 a1/k + . . . + a + a1/K −1 bn = a
{ } {|a |} is bounde b ounded d above. We choose an upper bound m for {|a |} which satisfies that |a | ≤ m. Th Then en b ≤ sowing that {|b |} is bounded above. According to theorem 2.3.2 we conclude that |a − a||b | → − and find from theorem 2.3.1 that a → a . We know that an converges, and hence that n
n
n
1/k n
1/k 1/k m
n
n
n
n
1/k
Exercise Exercise 2.3.12. 2.3.12. Prove that lim a1/n = 1. Hint: use the result of the previous exercise. We notice that n1/n > 1 for all n N .We can therefore write that we are looking for a solution to n to n 1/n 1 < .We can rearrange and raise both sides to the nth nth power, resulting in n the equation n equation n < (1 + + )) . We can expand the right hand side using the binomial theorem:
∈
−
1 + n + + n(n n < 1 + n 2
2
− 1) 1)
+ . . .
As long as n < 12 n(n 1) 1)2 this inequality holds, requiring that n > 1 + any any > 0 there exists an N such N such that whenever n > N , N , n1/n 1 < .
−
2.4 2.4
|
− |
2 . 2
Therefor Therefore, e, for
Mono Monoto tone ne Sequ Sequen ence cess
Exercise Exercise 2.4.1. 2.4.1. Tell which of the following sequences are non-increasing, non-decreasing, bounded? Justify your answers. (a) n2 ; for n is N, this sequence is non-decreasing since n2 < (n + 1)2 for all n. It is bounded below by 1. (b) √ 1n ; this sequence is non-increasing, since n = n = n 1/2 < (n ( n + 1)1/2 for all n all n.. This then implies √ 1n > √ n1+1 . The sequence is bounded by 0 and 1.
{ } { }
n
∈
√
(c) (−n1) ; this sequence is neither non-increasing, nor non-decreasing as the sign of the value of the sequence fluctuates due to the term ( 1)n . It is, however, bounded by 1 and 1/2. (d) 2nn ; this is the sequence 12 , 24 , 38 , . . . which . which is clearly non-increasing. It is bounded by 0 and 1. n (e) n+1 ; this is the sequence 12 , 23 , 34 , 45 , . . . which . which is clearly non-decreasing and tending to 1. It is bounded by 1/ 1/2 and 1.
{
}
−
−
{ } { }
Exercise Exercise 2.4.2. 2.4.2. Prove that the sequence xn with x1 = 1 and xn+1 = and decide what number it converges to.
√ x + 1 converges n
√ √ √ √
Proof. The first first few terms of the sequence sequence:: that x that x n is both increasing an bounded: xn+1 =
1+
1 + . + . . .
n terms
1+
√ 1 , 2,
2>
1+
1+
2,
1+
1 + . + . . .
n
− 1 terms
1+
1+
2, . . .. see .. We see
2 = x n
21
2.4. MONOTONE SEQUENCES
Then x n is increasing for all n N. We prove that xn < 2 by induction. We have x 1 = 1 < 2. Then, if xk < 2, xk+1 = 1 + xk < 1 + 2 = 3 < 2. Thus by the monotone convergence theorem, xn converges and is bounded by 2. Finding the√ limit: We solve a = 1 + a, which implies a 2 a 1 = 0. The solutions to this √ are a 1 = 1+2 5 , a 2 = 1−2 5 . The correct limit is a 1 , since a 2 < 0.
√ ∈
√
√
√
− −
Exercise 2.4.3. If a 1 = 1 and an+1 = (1
n
− 2− )a , prove that {a } converges. n
n
Proof. We notice that 2−n is monotone and converges to 0. Therefore we see that 1 2−n is also monotone, converging to 1. The whole term then is monotone and non-increasing. It is also bounded by 0 and 1. Therefore, by the monotone convergence theorem, an converges.
−
Exercise 2.4.4. 5
3
+2 Exercise 2.4.8. Prove that lim nn+3n 4 −n+1 =
∞.
n5 + 3n3 + 2 n5 (1 + 3n3 /n5 + 2/n5 ) lim 4 = lim 4 n n + 1 n (1 n/n4 + 1/n4 ) 1 + 3/n2 + 2/n5 = lim n ( ) 1 1/n3 + 1/n4
−
−
− →∞
5
3
+2 And hence, lim nn+3n 4 −n+1 =
∞.
→1
Exercise 2.4.10. If the sequence is bounded, then we are done by theorem 2.4.1. Otherwise there is no M R such that M a n for all n. This is equivalent to saying any M R, there exists N such that aN > M . Since the sequence is monotone, for all n > N , an > aN > M . Hence it has limit equal to infinity.
∈
≥
∈
Exercise 2.4.11. Prove Part (c) of Theorem 2.4.7. Theorem lim an = iff lim( an ) =
∞
−
−∞
Proof. If an , there exists some value of a n such that a n > M for any possible M R. If we consider the sequence an ( 1), we see clearly that an < M for any M R. But if that is true, then lim an = .
→ ∞
−
− −∞
−
∈
∈
Exercise 2.4.12. Suppose lim bn < . Then bn is bounded by theorem 2.2.3, so there exists M with b n M for all n. Since a n b n , we have a n M so lim an < , a contradiction.
∞ ≤
≤
≤
∞
0, M/k R. Hence there exists N , for all Exercise 2.4.13. For all M R, given k n > N , an > M/k. Since b n k, we have a n bn > k M/k = M, hence lim an bn = .
≥
∈
Exercise 2.4.14. Solved in Class
≥
·
∈
∞
22
CHAPTER 2. SEQUENCES
2.5
Cauchy Sequences
Exercise 2.5.1. Give an example of a nested sequence of bounded open intervals that does not have a point in its intersection. Exercise 2.5.2. Let I n = [n, ) which are nested, closed and unbounded. Suppose there is an x R with x I i . But there exists an integer m,m > x by the archimedean property.
∈
∈
∞
i
Hence I m does not contain x, a contradiction.
Exercise 2.5.4. Prove by induction that if nk is an increasing sequence of natural numbers, then nk k for all k.
≥
{ }
Proof. Assume the base case nk = n, which is the series 1, 2, 3, 4, 5, . . .. Since k is the counter index, i.e. k N, it is obvious that n k = k = 1, 2, 3, 4, 5, . . .. We generalize to the n + 1 case, i.e. nk = n + 1. In that case we have n k = n + 1 = k + 1 > k.
∈
Exercise 2.5.5. Which of the following sequences an have a convergent subsequence? Justify your answer. (a) an = ( 2)n ; None of the subsequences are convergent, as they either tend to + or . 5+(−1)n n (b) an = 2+3n ; This sequence is convergent for all n such that n mod 2 = 0, which is the sequence starting with 0.875, 0.6428, 0.55, 0.5, 0.46875, 0.4473, 0.4318, 0.42, 0.41071, . . . n (c) a n = 2(−1) This sequence has convergent subsequences for all n such that n mod 2 = 0 and for n mod 2 = 1.
{ }
−
∞ −∞
Exercise 2.5.6. [(a)] The sequence is 1, 1, 1, 1, 1, 1,.... The subsequence ani , ni = 2i 1.
− − − { } − 1 converges to − √ √ √ √ 2. 1. The sequence is 2/2, 1, 2/2, − 2/2, −1, − 2/2, 0,.... Since the sequence is periodic, we can find infinitely many terms equal to 0, which gives a convergent subsequence (convergent to 0).
3. Let ank = 1/2k which happens when n = 2k + 1. Then by a previous homework this has limit 0. Hence it is a convergent subsequence.
Exercise 2.5.7. For each of the following sequences, determine how many different limits of subsequences there are. Justify your answer. (a) 1 + ( 1)n ; This sequence is 0, 2, 0, 2, 0, 2, . . . and as such has two different limits: 0 and 2. (b) cos(nπ/3) ; There are two different limits. The first approaches 1 for the sequence of all n where n mod 6 = 0. The second limit is attained at 1 for all n such that n mod 6 = 3. (c) 1, 12 , 1, 12 , 13 , 1, 12 , 13 , 14 , 1. 12 , 13 , 14 , 15 , . . . The terms a 1 , a3 , a6 , a10 , a1 5, . . . are convergent to 1.
{ {
− } }
−
Exercise 2.5.8. Does the sequence sin n have a convergent subsequence? Why? Yes, it has three convergent subsequences, provided n convergent.
∈ R.
If n
∈ N, then it is not
23
2.6. LIM INF AND LIM SUP
Exercise 2.5.9. Prove that a sequence which satisfies an+1 an < 2 −n for all n is a Cauchy sequence.
|
− |
Proof. We notice that the sequences defined by the above condition are non-increasing and covergent. We notice the following pattern:
|a − a | = |a − a − a | = |a − a + a − a n+2
|a
n+3
n
n
n+3
n+2
n+2
n+1 + an+1
n+2
n+1 + an+1
n
n+2
n+1
n+1
n
n
n+3
n+2
n+2
n+1
∈ N .
− + 2−(m−2) + . . . + 2−n = 2−n (1 + 2−1 + 2−2 + . . . + 2−(m−1)+n )
|a − a | < 2− n
n+1
< 2 −(n+2) + 2−(n+1) + 2−n
Inductively, we see that this pattern continues for all patterns an and an + k with k Now, we assume two indices m and n such that m > n. We find m
n+1
(m 1)
geometric series
We rewrite and solve the geometric series: 0
2−n (
2k ) = 2−n (2
k= m+n+1
−
m+n+1
− 2−
) = 21−n
1 m
−2 −
We want to to prove 21−n 21−m 2 1−n + 21−m < . We solve the equations 2 1−n < log() 21−m < 2 . The solution to this is 2 log(2) < n, m
|
−
|≤ −
2
and
Exercise 2.5.10. Solved in class
2.6
n
− a | ≤ |a − a | + |a − a | < 2− + 2− − a | ≤ |a − a | + |a − a | + |a − a |
lim inf and lim sup
Exercise 2.6.1. Solved in Class Exercise 2.6.2. Find lim inf and lim sup for the sequence an = largest integer k so that 2k n.
≤
n 2kn
− 1 with k being the n
This is the sequence 0, 0, 12 , 0, 14 , 24 , 34 , 0, 18 , 28 , 38 , 48 , 58 , 68 , 78 , 0, . . .. It is clear that lim inf = 0 and lim sup = 1.
Exercise 2.6.3. Find lim inf and lim sup for the sequence 1 , 12 , 1, 12 , 13 , 1, 12 , 13 , 14 , 1, . . .. We find that lim inf = 0 and lim sup = 1.
Exercise 2.6.4. Solved in class
−
Exercise 2.6.5. If limsup an is finite, prove that lim inf( an ) =
− lim sup a . n
n
24
CHAPTER 2. SEQUENCES
≥
∈ { } − ≤ − −
Proof. By assumption, lim supan is equal to some a, such that a an for all an an . We multiply this by 1 to find the inverse sequence an . Then we have a an for all an an . By definition, this means a = liminf( an ). Therefore, lim inf( an ) = lim sup an .
−
∈{ }
−
−
{− } −
Exercise 2.6.7. Solved in Class
{ }
{ }
{ }
Exercise 2.6.8. If an and bn are non-negative sequences and bn converges, prove that lim sup an bn = (lim sup an )(lim bn ).
{ }
Proof. We need to consider two cases. First, assume an is not bounded above. Then lim sup an = . It then doesn’t matter what we multply a n with, we will always get infinity provided that bn = 0.Then lim sup(an bn ) = limsup an lim bn = . We now consider case 2, where a n is bounded above. By Bolzano-Weierstrass we know that an then has at least one convergent subsequence. Let a be the subsequential limit of a n , and let M be the upper bound of an . We know then that lim sup an exists and lim sup an M . MORE WORK NEEDED ON THIS . We note that according to the main limit theorem, if an a and b n b, a n bn ab. Thus lim sup(an bn ) = limsup an lim bn .
∞
∞
≤
→
→
→
Exercise 2.6.9. Solved in class Exercise 2.6.12. Which numbers do you think are subsequential limits of sin n ∞ n=1 ? Can you prove that your guess is correct?
{
All x
∈ R with |x| ≤ 1 are limits for sin.
}
Chapter 3 Continuous Functions 3.1
Continuity
Exercise 3.1.1. If f is a function with domain [0, 1], what is the domain of f (x2 1)? ======= If f is a function with domain [0, 1], what is the domain of f (x2 1)?
−
−
g is defined at point x iff x 2
∈ −√ 2, −1] ∪ [1, √ 2].
2
− 1 ∈ [0, 1], 0 ≤ x − 1 ≤ 1. x − 1 x ∈ ( −∞, −1] ∪ [1, ∞] √ √ x ≤ 2 x ∈ [ − 2, 2] 2 2
Thus x [
Exercise 3.1.2. What is the natural domain of the function is this function continuous? Why?
x2 +1 ? x2 1
−
With this as its domain,
The domain is R 1, 1 . The function is continuous everywhere except for the points not part of the domain.
\{− }
Exercise 3.1.3. For any real x, x2 + 1 is defined and nonzero. And, 1 + x2 is continuous on R , so by theorem 3.1.9(d), it is continuous. Exercise 3.1.4. Show that the function f (x) = x is continuous on all of R .
||
Proof. We need to find a δ such that for any > 0, we have x a < δ .
| − |
Exercise 3.1.5. Assuming sin is continous, prove that sin(x3
|| x| − |a|| < whenever
− 4x) is continuous.
Proof. We know that sin(x) < 1 for all x.
|
|
Exercise 3.1.6. Since f (x), g(x) are continuous, and f (xn ) converges to f (a) and g(xn ) converges to g(a), when xn converges to a. By theorem 2.3.6(d), f (xn )/g(xn ) converges to f (a)/g(a) since g(a) = 0 by assumption and g(xn ) = 0 in the domain of f (x)/g(x) by definition.
{ }
{ }
25
{
}
{
}
26
CHAPTER 3. CONTINUOUS FUNCTIONS
{ }
∈ ∈
→ { } { }
→
Exercise 3.1.7. Let xn by a sequence in Df ◦g . Then x n D g so if xn a, g(xn ) g(a). But all the xn , a D f ◦g implies xn , a D g and g(xn ), g(a) D f . Hence g(xn ) is a sequence in Df which converges to g(a) when xn converges to a, hence f (g(xn )) converges to f (g(a)) , so by theorem 3.1.6 it is continuous.
∈
{
}
∈ { }
√
≥
Exercise 3.1.8. We know x is continuous at all a 0 by theorem 3.1.7. Give another proof of this fact by using only the definition of continuity. Proof. We need to distinguish between two cases: Case 1 - a = 0: 0 = x < iff 0 x < 2 , δ = 2 . Whenever x < 2 we find that x x < and therefore x is continuous at a = 0. |x√ −a| < a Case 2 - a > 0: x a = x a x + a . This implies x a = √ xx− a −√ a if we have x a < a, δ = a.
| √ − √ √ | √ ≤ √ ||√ √ | | − √ | |√ −√ | − |
√
|√ − √ |
≤
Exercise 3.1.9. Consider the function: f (x) =
1
≥
: x 0, 1 : x < 0
−
Is this function continuous if its domain is R ? Is it continous if its domain is cut down to x R : x 0 ? How about if its domain is x R : x 0 ?
{ ∈
≥ }
{ ∈
≤ }
Exercise 3.1.10. Let f be a function with domain D and suppose f is continuous at some point a D. Prove that, for each > 0, there is a δ > 0 such that
∈
|f (x) − f (y)| < whenever x, y ∈ D ∩ (a − δ, a + δ ) Exercise 3.1.11. Prove that the function f (x) =
sin(1/x) x = 0 0 x = 0
is not continuous at 0.
→
→
Proof. Whenever xn 0 we have f (xn ) f (0) = 0. We are looking for a sequence 1 0 but where f (xn ) . This goes to 0 but xn f (0) = 0. We choose xn = π/2+2πn sin( π1 + 2n) = 1.
→
→
Exercise 3.1.12. Prove that the function f (x) =
x sin(1/x) x = 0 0 x = 0
is continuous at 0.
|
− f (0)|. |f (x) − f (0)| = |x sin x1 | = |x|| sin
Proof. We need to estimate f (x)
| − 0| < for δ = .
Thus x
| ≤ | | 1 x
x <
27
3.2. PROPERTIES OF CONTINUOUS FUNCTIONS
√
≥
Exercise 3.1.8. We know x is continuous at all a 0, by Theorem 3.1.7. Give another proof of this fact using only the definition of continuity (Def. 3.1.1). solved in class
Exercise 3.1.11. Prove that the function (piecewise function) is not continuous at 0. solved in class
Exercise 3.1.12. Prove that the function (piecewise function) is continuous at 0. solved in class
3.2
Properties of Continuous Functions
Exercise 3.2.1. The minimum of f on [0, 3) is sup f [0,3) = 6 which is achieved at x = 3 [0, 3).
∈
−1.
The maximum does not exist, since
Exercise 3.2.2. Prove that if f is a continuous function on a closed bounded interval I and if f (x) is never 0 for x I , then there is a number m > 0 such that f (x) m for all x I or f (x) m for all x I .
∈ ∈
≤−
≥
∈
Proof. Assume f (a) > 0. We have that f ([a, b]) = [m, M ] . We know that m = min f , M = max f . Let’s prove that m > 0. By contradiction: assume m < 0. Value 0 is taken (non-legible) an intermediate value [m, f (a)] which contradicts f (x) = 0. Prove for case 2 is analogous (show that M < 0).
Exercise 3.2.3. Prove that if f is a continuous function on a closed bounded interval [a, b] and if (x0 , y0 ) is any point in the plane, then there is a closest point to ( x0 , y0 ) on the graph of f . Proof. Pick any point x [a, b]. Then the distance to x0 , y0 is dist((x0 , yo ), (x, f (x))) = 1 ((x x0 )2 + (y0 f (x))2 )) 2 . We must prove that this function attains its minimum value 1 in [a, b] and that if f is continuous, then ((x x0 )2 + (y0 f (x))2 )) 2 is also continuous on [a, b]. Then distance takes its minimum value there.
−
−
∈
−
−
Exercise 3.2.4. Find an example of a function which is continuous on a bounded (but not closed) interval I , but is not bounded. Then find an example of a function which is continuous and bounded on a bounded interval I , but does not have a maximum value. The function f : (0, 1) R with f (x) = x1 fulfills the first condition. The second condition cannot be fulfilled; according to theorem 3.2.1 (p. 65): “If f is a continuous function on a closed bounded interval I , then f is bounded on I and in fact, it assumes both a minimum and a maximum value on I .” The only way to create a function
→
28
CHAPTER 3. CONTINUOUS FUNCTIONS
which would not assume a maximum on such an interval would be by violating the continuity. For example, the function f (x) =
2x x < 1/2 0 x 1/2
≥
fails to achieve its maximum on a bounded interval [0 , 1]. However, it does so by having a discontinuity at x = 1/2.
Exercise 3.2.5. Let f (x) = ex , I = [1, ). Then f is continuous on a closed (but not bounded) interval, but is not bounded. If f (x) = 1/(1 + e −x ) and I = [0, ), then f is continuous and bounded by 1 on I which is closed but does not have a maximum.
∞
∞
Exercise 3.2.7. Give an example of a function defined on the interval [0 , 1] which does not take on every value between f (0) and f (1). In other words, we are looking for a function with a discontinuity between [0, 1]. One example would be: x : 0 x < 12 f (x) = 2x : 12 x 1
≤ ≤ ≤
Exercise 3.2.8. Show that if f and g are continuous functions on the interval [a, b] such that f (a) < g(a) and g(b) < f (b), then there is a number c (a, b) such that f (c) = g(c).
∈
−
Proof. We create a function h(x) = f (x) g(x). This is continuous since it is a linear combination of continuous functions, and it is defined on [a, b]. We know that h(a) = f (a) g(a) < 0 and h(b) = f (b) g(b) > 0. Bt the intermediate value theorem there exists a c such that h(c) = f (c) g(c) = 0, which implies f (c) = g(c).
−
−
−
Exercise 3.2.9. Let f be a continuous function from [0, 1] to [0, 1].Prove that there is a point c [0, 1] such that f (c) = c - that is, show that f has a fixed point . Hint: Apply the Intermediate Value Theorem to the function g(x) = f (x) x.
∈
−
− ≤
Proof. Let g(x) = f (x) x. Since f (x) is continuous, we know that g(x) is also continuous. Then g(a) 0 and g(b) 0. By the intermediate value theorem we know that there exists some x [0, 1] such that g(x) = 0, which implies that f (x) = x.
∈
≥
Exercise 3.2.10. Use the intermediate value theorem to prove that if n is a natural number, then every positive number a has a positive n-th root. Proof. We write the function f (x) = x n which is continuous on [0, ) since it is a polynomial. We notice f (0) = 0 < a. We know that there is a number m N such that m > a which implies f (m) = mn m > a. Thus we have f (0) < a and f (m) > a and since f is continuous on [0, m], the intermediate value theorem states that there exists a c such that f (c) = c n = a.
≥
∈
∞
Exercise 3.2.11. Prove that a polynomial of odd degree has at least one real root.
29
3.3. UNIFORM CONTINUITY
Proof. Assume g(x) : R R is an odd degree polynomial. Then g is of the form nk=0 ak xk , where a k is the k-th coefficient of the polynomial an n N such that n mod 2 = 1. We can n−1 −1 a xk ). We note that lim xk then factor g to be of the form g(x) = x n (an + nk=0 = a k xn x→±∞ k k=0 xn n n n 0. We then consider lim x→±∞ x an . We note that since n is odd, x 0 if x 0 and x 0 n n if x 0. Therefore, lim x→+∞ x an = + and limx→−∞ x an = , provided that an > 0. Hence, we find that lim x→−∞ g(x) = and limx→∞ g(x) = . In the case of an < 0, we find that limx→−∞ g(x) = and limx→∞ g(x) = . We know that any polynomial is continous, and the above shows that there are some a, b R such that g(a) < 0 and g(b) > 0. We now consider the interval [a, b]. By the Intermediate value theorem, we find that for every c [g(a), g(b)] there exists some x [a, b] such that g(x) = c, implying that there exists at least one x such that g(x) = 0.
→
≥
∞ −∞
∞
∈
≤ −∞ ∞
−∞
≤
≥
∈
∈
∈
Exercise 3.2.12. Use the Intermediate Value Theorem to prove that f is a continuous function on an interval [a, b] and if f (x) m for every x [a, b), then f (b) m.
≤
∈
≤
≤
∈
Proof. Assume that m < f (b), such that f (x) m < f (b) for all x [a, b). We then know that m = f (b) δ for some δ > 0 R . But, by properties of the real numbers, we would also have m = f (b) δ < f (b) < f (b) some some , such as = 2δ . But f (b) [a, b) - contradiction: by the intermediate value theorem, since f is continuous, we know that there exists some x such that f (x) = f (b) , and thus we require m f (b) . Hence, f (b) m.
−
−
−
∈
−
≤
3.3
− ∈ −
≥
Uniform Continuity
Exercise 3.3.1. Is the function f (x) = x2 uniformly continuous on (0, 1)? Justify your answer. Yes, it is. According to Theorem 3.3.4, if a function is continuous on a closed bounded interval I , it is uniformly continuous there. Assume I = [0, 1]. Then by theorem 3.3.4, f is uniformly continuous on I . By theorem 3.3.6, f is then also uniformly continuous on (0, 1).
Exercise 3.3.2. Is the function f (x) = 1/x2 uniformly continuous on (0, + ) (actual text printed asks only about interval up to 1 )? Justify your answer.
∞
Assume f were uniformly continuous. Then it it is uniformly continuous on a subinterval, such as (0, 1). But f (x) is not bounded on the interval (0, 1). Therefore, it is not uniformly continuous.
Exercise 3.3.3. Is the function f (x) = x 2 uniformly continuous on (0, + )? Justify your answer.
∞
No, it its not. As x we find that the distance between y, y gets bigger and bigger, such that x, x need to be closer and closer for y, y to still be within of each other. This means that δ does depend on a, so it is not uniformly continuous.
→∞
30
CHAPTER 3. CONTINUOUS FUNCTIONS
−
Exercise 3.3.4. Using only the δ definition of uniform continuity, prove that the function x is uniformly continuous on [0, ). f (x) = x+1
∞
Proof. 1) + y(x + 1) |f (x) − f (y)| = | x +x 1 − y +y 1 | = | x(y + | (x + 1)(y + 1) xy + x − xy − y =| | = |x − y| ≤ |x − y| (x + 1)(y + 1)
| −
− |
Estimate f (x) f (y) in f (x) f (y) < .
|
(x + 1)(y + 1)
| ≤ |x − y|. Then for all > 0, δ = implies |x − y| < δ = will result
Exercise 3.3.5. In example 3.3.8 we showed that Show that it is also uniformly continuous on [0, 1].
√ x is uniformly continuous on [1, ∞).
√
By theorem 3.3.4: if x is continuous on [0, 1], it is uniformly continuous there.
Exercise 3.3.6. Prove that if I and J are overlapping intervals in R (I J = and f is a function, defined on I J , which is uniformly continuous on I and uniformly continuous on J , then it is also uniformly continuous on I J . Use this and the previous exercise to prove that x is uniformly continuous on [0, + ).
∩ ∅
∪
√
∞
∩ ∅ −
∪
Proof. By assumption I J = . We shall assume that the interval I is the “lower” one of the two. Then there exists an x such that x I J . Since I is uniformly continuous by assumption, we know that [x a, x] is uniformly continous for all (x a) I . Likewise we know that [x, x + b] is uniformly continous for all (x + b) J since J is uniformly continous by assumption. This implies that the whole interval [x a, x + b] is uniformly continous. To prove that x is uniformly continuous, we assume we are given some > 0. For this, we pick δ = 2 . We note that x y x + y . If x y < δ = 2 , we find:
∈ ∩
− ∈
∈
− |√ − √ | ≤ |√ √ | | − | |√ x − √ y| ≤ |√ x − √ y||√ x + √ y| = |x − y| < √ √ √ This guarantees that | x − y| < , proving that x is uniformly continous on (0, ∞). √
2
2
Exercise 3.3.7. Suppose f is not continuous on I , then since uniform continuity implies continuity, f is not uniform continuous on I . Suppose f is continuous on I . Then f is ¯. But a continuous function on a closed, uniform continuous on I if f is continuouss on I bounded interval is bounded, but f is unbounded. This is a contradiction so f cannot be uniformly continuous. Exercise 3.3.8. Let f be a function defined on an interval I and suppose that there are positive constants K and r such that r
|f (x) − f (y)| ≤ K |x − y| Prove that f is uniformly continuous.
∈ I .
for all x, y
31
3.4. UNIFORM CONVERGENCE
Proof. According to assumption, we find that f (x) f (y) K x y r for all x, y I . r This implies that if K x y < , f (x) f (y) < . Thus we find that we need to solve . Since δ K x y r < δ , and find that for any given , we pick a δ such that δ = r K does not depend on where x, y are in the interval, f is uniformly continuous.
| − |
≤
| − |
|
−
|
|
−
|≤ | − |
∈
Exercise 3.3.9. Is the function f (x) = sin( x1 ) continuous on (0, 1)? Is it uniformly continuous on (0, 1)? Justify your answers. Proof. Since sin is a trigonometric function, it is continuous on its whole domain. Likewise, sin(1/x) is continous since it is merely a composition of two elementary functions. However, sin(1/x) is not uniformly continous. The reason for this is that the as x 0 the function oscillates between 1 and 1. Thus, a δ that would work at one point in the function will can produce potentially a difference f (x) f (y) = 2 for x, y sufficiently close to 0. Hence, the functions is not uniformly continous.
→
−
|
−
|
Exercise 3.3.10. Is the function f (x) = x sin(1/x) uniformly continuous on (0, 1)? Justify your answer. Proof. Method 1: f (1) = sin(1). It is still uniformly continuous. lim x→0 f (x) = limx→0 x sin(1/x) = 0. By squeeze theorem:
→ 0 by squeeze thrm.
|| ≤ | | ≤ 0
x sin
→ 0
1 x
x
→ 0
If we define f ( 0) = 0, f (1) = sin(1), then f (x) becomes continuous on [0, 1]. then by theorem 3.3.4, f is uniformly continuous on [0, 1] = f is uniformly continuous on (0, 1). Method 2:
⇒
|f (x) − f (y)| = |x sin(1/x) − y sin(1/y)| ≤ |x sin(1/x)| + |y sin(1/y)| ≤ |x| + |y| Then for all > 0 we have |f (x) − f (y)| < if x, y ∈ (0, ]. If now x,y > /3, then there exists a δ > 0 such that |f (x) − f (y)| < whenever |x − y | < δ : |f (x) − f (y)| ≤ |x| + |y| < 3 + |x| + |y − x| < 3 + 3 + δ < 2
if δ < 3 . Then we choose δ = min( 3 , δ 0 ).
3.4
Uniform Convergence
32
CHAPTER 3. CONTINUOUS FUNCTIONS
Chapter 4 The Derivative 4.1
Limits of Functions
4.2
The Derivative
4.3
The Mean Value Theorem
Exercise 4.3.1. Exercise 4.3.1. If we apply the MVT to the intervals [ 1, 1], [0, 1] and [ 1, 0] we find that there exist points where the derivative equals 1 /2, 1, and 0. Exercise 4.3.2. Observe that f (x) = sin x satisfies the hypotheses of theorem 4.3.9 where we let (a, b) = ( , ) and so it must satisfy f (x) f (y) M x y where M is a bound for the derivative. In this case f (x) = cos x 1 for all x, hence we may set M = 1. Then sin x sin y = f (x) f (y) M x y = x y as desired.
−
|
−
−∞ ∞ | | −
| − ≤ |≤ | − | | − |
−
|≤ | − |
Exercise 4.3.3. Exercise 4.3.4. We know that there is some c (0, ) with f (x) f (y) = f (c)(x y) for x, y (0, ). Then f (x) f (y) f (c) x y M x y . If we take the limit of both sides as y 0, then since f (x) f (y) is continuous, lim f (x) f (y) = f (x) f (0) = f (x) .
∈ ∞ | − |≤ → | − | Hence |f (x)| ≤ M x.
∈ ∞ | − |≤ | − | | − → y
0
− | |
−
− | |
|
Exercise 4.3.5. Exercise 4.3.6. We know that f (x) = 6x2 + 6x 12 = 6(x +2)(x 1). Hence f (x) 0 on [ 2, 1] and is positive elsewhere. Hence f (x) is decreasing on [ 2, 1] and increasing elsewhere.
−
−
Exercise 4.3.7. Exercise 4.3.8. 33
− −
≤
34
CHAPTER 4. THE DERIVATIVE
Exercise 4.3.9. Exercise 4.3.10. Exercise 4.3.11. Exercise 4.3.12. Exercise 4.3.13. Exercise 4.3.14. Exercise 4.3.15. Exercise 4.3.16.
4.4
L’Hopital’s Rule
Exercise 4.4.1. By Cauchy’s form of the MVT for the interval [1, x] and f (x) = ln x and g(x) = x r , we have ln x 1/c 1 = = , xr 1 rcr−1 rcr hence xr 1 ln x = . rcr We know moreover that c > 1 and r > 0, hence r log c 1, or cr 1 hence 1/cr 1. Then
−
−
≥
ln x =
xr 1 rcr
r
− ≤ x − 1, r
as desired.
Exercise 4.4.2. Exercise 4.4.3. Exercise 4.4.4. Exercise 4.4.5. Exercise 4.4.6. The limit is of the form
∞/∞, hence
ln x 1/x = lim x→∞ xr x→∞ rx r−1 1 = lim = 0. x→∞ rx r lim
≥
≤
35
4.4. L’HOPITAL’S RULE
Exercise 4.4.7. If we write x ln x = L’Hopital’s, the limit is equal to lim
ln x then the limit is of the form 1/x
1/x = lim x = 0. 1/x2 x→0
−
→0 −
x
−∞/∞.
Exercise 4.4.8. The limit is of the form 0/0 hence it is equal to
−
cos x 1 . x→0 3x2 lim
It is still of the form 0/0 hence equal to lim
x
→0
− sin x . 6x
Applying L’Hopital’s again, the limit equals lim
x
→0
− cos x = 0. 6
Exercise 4.4.9. Exercise 4.4.10. Let g(x) = log f (x) = x log x. Then lim x log x = lim log x/(1/x).
x
→0
x
→0
±∞, hence we may apply L’Hopital’s to get lim x log x = lim (1/x)/(−1/x ) = lim −x = 0. → → →
This has numerator and denominator
2
x
Exercise 4.4.11. Exercise 4.4.12. Exercise 4.4.13. Exercise 4.4.14. Exercise 4.4.15. Exercise 4.4.16.
0
x
0
x
0
By
36
CHAPTER 4. THE DERIVATIVE
Chapter 5 The Integral 5.1
Definition of the Integral
Exercise 5.1.1. We have U (f, P ) = 1
· 14 + 45 · 14 + 23 · 14 + 47 · 14 = 319/40.
Moreover,
4 1 2 1 4 1 1 1 + + + = 533/840. 5 4 3 4 7 4 2 4 Exercise 5.1.2. Since x is an increasing function, we know that M k = k/n and mk = (k 1)/n hence n 1 n(n + 1) k1 n + 1 = 2 = U (f, P n ) = 2 2n nn n k=1 L(f, P ) =
·
−
and
·
− n
L(f, P n ) =
·
k
k=1
·
−
−
11 1 n(n 1) n 1 = 2 = . 2 2n n n n
→ ∞
Since these have the same limit as n , we know that lim n→∞ U (f, P n ) as desired. Exercise 5.1.3. The base case is clearly true, so suppose that k
j 2 =
j=1
− L(f, P ) = 0,
k(k + 1)(2k + 1) . 6
Then adding (k + 1) 2 to both sides yields k+1
j 2 =
j=1
(k + 1)(k + 2)(2(k + 1) + 1) k(k + 1)(2k + 1) + (k + 1) 2 = , 6 6
as desired. 37
n
38
CHAPTER 5. THE INTEGRAL
Exercise 5.1.4. By using exercise 3 above, one finds that n
U (f, P n ) =
ka n
k=1
2
a a3 n(n + 1)(2n + 1) = , 6 n n2
(n 1)a a < < < a and M k = (ka/n)2 since f (x) = x 2 is increasing. n n Moreover, notice that the limit is a 3 /3. We now find that
{
−
···
where P n = 0 <
}
n
L(f, P n ) =
(k
− 1)a n
k=1
2
a n
which we may find evaluates to a3 n(n n2
− 1)(2n − 1) . 6
Notice that this also has limit a3 /3, hence the limit of the difference of the two is zero, and the integral exists. Moreover, it equals lim U (f, P n ) = a3 /3. Exercise 5.1.5. For any P , we know that U (f, P ) = 1 since U (f, P ) = M k (xk xk−1 ) =
k
k
xk
− x −
k 1
= 1
− 0 = 1 as the supremum of f on any interval is 1.
−
On the
other hand we know L(f, P ) = 0 since the infimum of f on any interval is 0. Hence for all partitions P , U (f, P ) L(f, P ) = 1, hence by theorem 5.1.7 the function is not Riemann integrable. Exercise 5.1.8. We know that
−
b
f (x) dx = sup L(f, Q) : Q
{
a
P {
∈ P},
where is the set of partitions of [a, b]. This is equal to partition a = x 0 < x1 < < xn = b of [a, b]. But we know
···
}
n
n k=1 mk (xk
− x − ) for some k 1
n
−x − )
mk (xk
k 1
k=1
≥
m(xk
k=1
− x − ) = m(b − a). k 1
This gives the first inequality in the chain of inequalities. The second inequality is obtained by the definition of supremum and infimum. The third inequality is found by noting that M k M . Exercise 5.1.9. For any partition,
≤
n
U (f, P ) =
n
M j (x j
− x − ) =
k=1
−
j 1
j=1
n
k(b
− a) =
j=1
mk (b
− a) = L(f, P ).
Hence U (f, P ) L(f, P ) = 0 < for any partition P , so by theorem 5.1.7 the integral exists. Applying definition 5.1.6 and the definition of upper and lower integral, we find that the n integral equals j=1 k(b a).
−
39
5.2. EXISTENCE AND PROPERTIES OF THE INTEGRAL
Exercise 5.1.10. See definition 5.1.1. If P partitions [a, b] into n equal subintervals, then xk xk−1 = (a b)/n hence the difference is
−
−
a
−b n
5.2
n
(M k
k=1
− m ). k
Existence and Properties of the Integral
Exercise 5.2.1. If f = g h and g, h are non-decreasing, then g , h 0. We will then have g h , g = h or g h . In the first and third cases, we know that either f 0 or f 0. In either case, f is monotone on a closed bounded interval [a, b], hence by theorem 5.2.1 it is integrable. On the other hand if g = h then f = 0, hence f is constant and it was proved that a constant function is integrable in the previous homework. Exercise 5.2.6. We know 1 + x2n is continuous and never zero, hence 1/(1 + x2n ) is also continuous on [ 1, 1], hence by theorem 5.2.2 it is integrable. Moreover, M = sup[−1,1] f = 1 and m = inf [−1,1] f = 1/2. By corollary 5.2.5, we know that the integral is therefore trapped between m(b a) = 12 2 = 1 and M (b a) = 1 2 = 2. Hence the desired inequality. Exercise 5.2.11. Let f (x) = 1 if x [0, 1] is rational, and 1 if x is irrational. Then for reasons similar to problem 5.1.5, this function is not integrable. However, f (x) = 1 for all x [0, 1], and hence it is integrable. Exercise 5.2.12. We know that f (x) is continuous, hence integrable on a closed bounded interval [a, b]. It is also therefore bounded, and so M = sup[a,b] f and m = inf [a,b] f are finite so f actually attains these values. Moreover, we know by the IVT that f attains every value between m and M (possibly these values). In particular, f attains the value
≥
≤
−
≥
≥
≤
−
−
·
− ∈
·
−
|
∈
1 b
−a
|
b
f (x) dx,
a
by corollary 5.2.5.
5.3
The Fundamental Theorems of Calculus
Exercise 5.3.1. We know that the antiderivative is f (x) = x 2 sin(1/x) and that it satisfies all the hypotheses of theorem 5.3.1 from example 5.3.2, and so the integral equals f (2/π) f (4/π) = 4/π2 8 2/π2 . Exercise 5.3.2. It is clear that the derivative is cos(1 /x). Exercise 5.3.4. The integral may be split up as
−
− √
x
0
t2
e− dt
−
1/x
2
e−t dt,
0
2
2
and hence the derivative of the first integral is e−x and of the second it is e−1/x /x2 by the chain rule and second fundamental theorem of calculus. Exercise 5.3.5. f (x) = 1/x is not continuous at x = 0 [ 1, 1], violating the hypotheses of theorem 5.3.1.
−
−
∈ −
40
CHAPTER 5. THE INTEGRAL
Exercise 5.3.6. Letting g(x) = f (x), applying (5.3.5) gives that the integral equals
−
b
f (b)
2
− f (a)
2
(f (x))2 dx.
a
1 n+1 and t n + 1 f (t) = 1/t. Both f and g are continuous on [0, ) and differentiable on (0, ). We know f g and gf are continuous, hence by (5.3.5) the integral equals
→ t and g(t) = t
Exercise 5.3.8. Let x
n
and f (t) = ln t. Then g(t) =
∞
−
1 n+1 ln t t n + 1
x 0
x
0
∞
1 n+1 1 t dt. n + 1 t
Simplifying this and making the necessary evaluations, the integral comes out as 1 xn+1 ln x n + 1
− (n +1 1)
2
xn+1 .
||
Exercise 5.3.10. Theorem 5.3.1 requires f (x) = x/ x be continuous at every point in [ 1, 1], but f (x) is not continuous at x = 0.
−
5.4
Logs, Exponentials, Improper Integrals
Chapter 6 Infinite Series 6.1
Convergence of Infinite Series
6.2
Tests for Convergence
6.3
Absolute and Conditional Convergence
6.4
Power Series
Exercise 6.4.2. Prove that f (x) =
∞
k=1
sin(kx) 2k
is continuous on the entire real line.
Proof. We use the Ratio test to show that the sum converges on the entire real line: sin((k + 1)x) 2k sin(kx + x) = 2k 2 sin(kx) 2sin(kx)
≤ 12 .
−k = 2, This proves convergence on all of R . We notice that sin(kx)2−k 2 −k . Since ∞ k=0 2 ∞ sin(kx) converges uniformly on all R. This we know by the Weierstrass M-test that k=1 2k proves that this sum is continuous on the entire real line.
≤
∞
1 k k=1 k3k x .
Exercise 6.4.4. Radius of convergence of
We have c k = (k3k )−1 . We proceed to take the k-th root: (k3k )−1/k =
1 1 = 3−1 . k 3 k
√
We then solve using the formula for the radius of convergence: R =
1 lim sup
√ c k
=
k
Therefore, the radius of convergence is 3. 41
1 = 3. limsup3−1
42
CHAPTER 6. INFINITE SERIES
Exercise 6.4.5. Radius of convergence of
( 1)k−1 k k=0 k+1 (x + 2) .
∞
−
1)k−1 We have c k = (−k+1 . We take the k-th root:
(ck )1/k = =
1 k
− − − ( 1)k−1 k + 1 ( 1)k/k
( 1)k k + 1 1
=
− −
=
(k + 1) 1/k
− −
(k + 1) 1/k
1 k
=
1 (k + 1) 1/k
.
We then use the formula for the radius of convergence: R =
1 lim sup
√ c k
= k
1 limsup(k + 1) −1/k
=
1 = 1. 1
And hence, the radius of convergence is 1.
Exercise 6.4.8. Radius of convergence of
∞ 2k x2k . k=0
We have c k = 2k . We then take the k-th root: 2k the the radius of convergence: R =
1 lim sup
√ c k
=
k
1 k
= 2k/k = 2. We use the formula for
1 1 = . limsup2 2
Thus we can see that the radius of convergence is 1 /2.
{ }
Exercise 6.4.10. Let ak be a non-increasing sequence of non-negative numbers which k+1 converges to 0. Use theorem 6.3.2 to show that the power series ∞ ak xk converges k=0 ( 1) uniformly on [0, 1], and hence converges to a continuous function on this interval.
−
k+1 Proof. According to theorem 6.3.2, the series ∞ ak converges provided that ak k=0 ( 1) is a non-increasing sequence of non-negative numbers. We notice that this power series is uniformly Cauchy, since the sequence of partial sums is uniformly Cauchy according to the theorem. Therefore, this series converges uniformly. Notice that for all x [0, 1], ( 1)k+1 ak xk ( 1)k+1 ak . Therefore, we see that the series ∞ ( 1)k+1 a xk is bounded by ∞ ( 1)k+1 a , which by the Weierstrass M-test implies k k k=0 k=0 ∞ k+1 k that k=0 ( 1) ak x is uniformly convergent on [0, 1].
−
∈
−
−
−
−
{ }
≤ −
Exercise 6.4.12. Prove that if f (x) is the sum of a power series centered at a and with radius of convergence R, then f is infinitely differentiable on (a R, a + R) - that is, its derivative of order m exists on this interval for all m N.
∈
−
∞ c (x a)k . Then, according to theorem 6.4.12, f (x) Proof. Suppose that f (x) = k=0 k is differentiable on (a R, a + R) and its derivative is f (x) = ∞ a)k−1 . This k=1 kc k (x means we are left with a k 1 degree differentiable polynomial. By induction we see that we can then integrate this power series again. Since we take k to infinity, this means we can take an infinite number of derivatives. Additionally, any polynomial has an infinite number of derivatives: the kth derivative is a constant, and all others are zero.
−
−
−
−
43
6.5. TAYLOR’S FORMULA
6.5
Taylor’s Formula n
Exercise 6.5.1. Prove that lim xn! = 0 for all x. Proof. Since we take the limit as n , n will take on every value of N . We assume that x R . Therefore, at some point we find have that x n; at the first point this occurs, we n−1 then consider the rational expression nx C where C is some constant of the form (nx −1)! . If we xx consier the next term in the sequence, n (n+1) C , we find that n (n + 1) x x, since x n. n Therefore, it is obvious that lim n→∞ xn! = 0 for all x.
→ ∞
∈
≤
≥
≤
Exercise 6.5.3. Use Taylor’s formula to estimate the error if cos(x) is approximated by 2 1 x2 on the interval [ 0.1, 0.1].
−
−
We estimate as follows: x4 4! Thus we can bound the error by 5
≤
0.14 4!
6
≤ 5 × 10−
6
× 10− .
√ t + x with a = 0. √ We begin by taking the first few derivatives of f (x) = t + x.
Exercise 6.5.5. Taylor’s formula for f (x) =
1 √ 2 t + x 1 f (x) = − f (x) =
f (3) (x) = f (4) (x) =
3
4 (t + x) 2 3 5
8 (t + x) 2 15
−
7
16 (t + x) 2
Then the Taylor formula for the above function for the first 5 terms, with remainder, at a = 0 will be as follows: f (x) =
∞
n=0
f (n) (a) (x n!
n
− a)
1 1 1 3 1 15 1 ≈ √ t + 2√ x− x + x − x t 4t 2 8t 3! 16t 4! √ 1 1 5 x = t + √ − x + x − x + 2
3
3 2
2
2 t
8t
3 2
7 2
3
16t
5 2
4
128t
1 5 x 5! 32 (c + t) 7 5 9 x . 256(c + t) 2 4
5 2
7 2
+
Exercise 6.5.7. Taylor’s formula for f (x) = ln(1 + x) with a = 0.
105
9 2
44
CHAPTER 6. INFINITE SERIES
We again begin by taking the first four derivatives of f . f (x) = ln(1 + x) 1 f (x) = x + 1 1 f (x) = (x + 1)2 2 f (3) (x) = (x + 1)3 6 f (4) (x) = (x + 1)4
− −
Then the Taylor formula for the above function for the first 5 terms, with remainder, at a = 0 will be as follows: f (x) =
∞
n=0
f (n) (a) (x n!
≈ x − 12 x = x
−
− a)
2 + x3 6 1 2 1 3 x + x 2 3 2
n
− 246 x
24 5 5x 120(c + 1) 1 4 1 5 x + 5x . 4 5 (c + 1) 4
−
+
− ∈ −
Exercise 6.5.11. If f is an infinitely differentiable function on (a r, a + r) and there is some constant K such that f (n) (x) K n! for all n N and all x (a r, a + r), then the rn Taylor series for f at a converges to f on (a r, a + r).
≤
−
∈
This appears to be a backwards reasoning argument. We that if f is a convergent taylor series on (a R, a + R), then it is infitinely differentiable on that same interval. We can see that if there is some constant K such that f (n) (x) K n! for all n N and rn all x (a r, a + r), then f (n) (x) is always bounded.
−
∈ −
2
≤
∈
Exercise 6.5.13. If g(x) = e −1/x for x = 0 and g(0) = 0, show that g is infinitely differentiable on the entire real line but all of its derivatives at 0 are 0. Argue that this means that g cannot be analytic at 0. Hint: Use the previous exercise to help compute the derivatives of g at 0.
−1/x2 2 First note that g (x) = 2 e x3 . The Taylor expansion of e−1/x centered around a is k f (k) (a) f (k) (a) of the form ∞ (x . This is a power series where = and as such is a) c k k=0 k! k! differentiable according to theorem 6.4.12, and infinitely so (cf. exercise 6.4.12). Likewise, 2 e−1/x is infinitely differentiable, and all of its diefferentials are clearly defined everywhere except at x = 0. This is due to the fact that all n-th derivatives of g are sums of the form n −1/x2 x−(n+k+1) . Therefore, analytically g(x) is not defined at 0. a e k k=1
−
−1/x2
Remember however that lim x→0 e xn = 0. Therefore, if we define g(x) and and all of its derivatives to be 0 at 0, g(x) is infinitely differentiable.
Part II Multivariable - 3220
45
Chapter 7 Convergence in Euclidean Space 7.1
Euclidean Space
|| | − |y|| ≤ |x − y|. Proof. We note that x = x − y + y. This implies that |x| = |x − y + y |. We use the triangle Exercise 7.1.5. Prove that x inequality:
|x − y + y| ≤ |x − y| + |y| |x| ≤ |x − y| + |y| |x| − |y| ≤ |x − y|
which is equivalent to y
|−| |
This proves the result.
Exercise 7.1.6. Prove that equality holds in the Cauchy-Schwartz inequality if and only if one of the vectors u, v is a scalar multiple of the other. Proof. Let v be a scalar multiple of u, such that v = αu. Then u v = u αu = αu 21 + αu22 + . . . + αu2n
·
·
= α u21 + . . . + u2n 2
n
= α
= α u 2 .
u2i
i=1
||
Therefore, we find that u 2 v 2 = u α u 2 = α u 2 , and hence, equality holds. The proof for the case where u is a scalar multiple of v is analogous.
| | | | | | | |
||
7.2
Convergent Sequences of Vectors
7.3
Open and Closed Sets
{
Exercise 7.3.1. Prove that the set (x, y)
∈R
2
47
: y > 0 is an open subset of R 2 .
}
48
CHAPTER 7. CONVERGENCE IN EUCLIDEAN SPACE
Proof. According to theorem 7.3.10, a set A R d is closed if and only if every convergent sequence in A converges to a point x A. Consider the sequence n1 , n1 . This sequence is clearly contained in the set A = (x, y) R2 : y > 0 . However, n1 , n1 (0, 0), and (0, 0) / A. Therefore, the set A is not closed. Since A is not closed, A is an open set.
∈ ∈ { ∈
∈
{ } { } →
}
Exercise 7.3.4. Find the interior, closure, and boundary for the set
{
A = (x, y)
2
∈ R : |(x, y)| < 1} ∪ {(x, y) ∈ R
2
}
: y = 0, 2 < x < 2 .
The solution is as follows: 2
A◦ = (x, y)
2
{ ∈ R : |(x, y)| < 1} ∪ {(x, y) ∈ R : y = 0, 2 < x < 2} A = {(x, y) ∈ R : |(x, y)| ≤ 1 ∨ y = 0, |x| ≤ 2 } ∂A = {(x, y) ∈ R : |(x, y)| = 1 ∨ x = −2, y = 0 ∨ x = 2, y = 0} Exercise 7.3.6. Let A be an open set and B a closed set. If B ⊂ A, prove that A\B is open. If A ⊂ B, prove that B \A is closed. Proof. Case 1: B ⊂ A. Since B is a closed set, for any sequence {x } ∈ B , there is some L ∈ B such that {x } → L . Since B ⊂ A, we know that {x }, L ∈ A for all n, i ∈ N. If B is a proper subset of A, then for every L there exists some sequence {y } such that {y } ∈ A\B and {y } → L . Since L ∈ B , we find that L ∈/ A \B. But then not all limit points of A\B are actually in A \B, thus showing that A\B is an open set. Case 2: A ⊂ B. Since A is an open set, ∃L , {x } such that L ∈ B, { x } ∈ A and {x } → L . Consider now the set B \A. Since L ∈ B, but L ∈/ A, we find that L ∈ B \A ∀i. Therefore, B \A is a closed set. Exercise 7.3.9. If A and B are subsets of R show that A ∪ B = A ∪ B. Is the analogous statement true for A ∩ B? Justify your answer. 2 2
n
i
n
i
n
i
i
n
n
i
n
i
i
i
n
i
n
i
i
i
n
i
d
Existing solution utilizes a graphic. Try to find analytic solution instead.
Exercise 7.3.15. If E is a subset of R d , show that (E )c = (E c )◦ . As is evident from Figure 1, this identity holds.
7.4
Compact Sets
Exercise 7.4.1. If K is a compact subset of R d and U 1 U 2 . . . U k . . . is a nested upward sequence of open sets with K k U k , then prove that K is contained in one of the sets U k .
⊂ ⊂ ⊂ ⊂
⊂
⊂ ⊂
Proof. This seems obvious. Since U K is a nested upward sequence, we have that U i U i+1 for all i N . If K K we find that xi k U k , then xi k U k . Since K is a compact set, it has a finite subcoverage. This means, there is a finite n N such that n U n K . Therefore, there exists a finite m N such that K n+m U m+n .
∈
⊂
∀ ∈ ∈
⊂
∈
∈
49
7.4. COMPACT SETS
Exercise 7.4.4. Prove that if K is a compact subset of Rd , then K contains a point of maximal norm. That is, there is a point x 1 K such that
∈ |x| ≤ |x | for all x ∈ K. 1
{| | ∈ K } and consider the open balls B
Hint: set m = sup x : x
m 1/n (0).
−
Proof. A continuous image of a compact space is compact. Consider the space K Rd , and let f : K R such that f (x) = x . This is a continuous function since the space is compact. As a continuous, real valued function, it attains both a maximum and a minimum per 3.2.1. Therefore, there exists an xi K such that f (xi ) f (xk ) for all k N. This proves that x : x xi for all i N.
→
∃ | | ≥ | |
∈
∈
| | ∈
≥
∈
Exercise 7.4.6. Prove that the conclusion of the previous exercise also holds if we only assume that K is a closed subset of R d . Hint: replace K by its intersection with a suitably large closed ball centered at y. Proof. Let f (x) : K R such that f (x) = x y where y Rd . Since K is a compact set imaged into a compact set, we know that f is a continuous function. As such, this function attains both a minimum and a maximum per 3.2.1. Therefore, there exists a point xi such that f (xi ) f (xk ) for all k N, which proves that there exists a point x i such that xi y x y for all x.
→
| − |≤| − |
≤
| − |
∈
∈
Exercise 7.4.9. Show that it is true that the union of any finite collection of compact subsets of R d is compact, but it is not true that the union of an infinite collection of compact subsets is necessarily compact. Show the latter statement by finding an example of an infinite union of compact sets which is not compact. Any compact set has a finite subcoverage. Having a finite subcoverage - given an open cover - is sufficient for a set to be considered compact. The union of 2 compact sets then has two sets of compact subcoverages. The union of subcoverages is likewise finite. Assume we have n compact sets with finite subcoverage. The union of these n sets with an additional compact set results in the union of a set of finite subcoverages of the union of the n sets together with a finite subcoverage of the n + 1-th set. This is then evidently also finite. Hence, any finite union of compact sets is also compact. However, an infinite collection is not necessarily compact as this infinite union may lead to the existence of an infinite number of subcoverages, which would make this union no longer compact.
Exercise 7.4.10. Prove that if A and B are compact subsets of a Rd , then A B and A B are also compact.
∪
∩
Proof. Any compact set has a finite open coverage. Hence, there is a finite open cover U A of A and an open coverage U B of B. We now consider the union U A U B . Since both coverages are finite, the union is finite as well. This union now is a finite coverage of A B. Since A B has a finite coverage, it is also compact.
∪
∪
∪
50
7.5
CHAPTER 7. CONVERGENCE IN EUCLIDEAN SPACE
Connected Sets
Exercise 7.5.1. A = (x, y) R 2 : (x, y) < 1 (x, y) R 2 : 1 x 2, y = 0 . This set is connected, since it is the union of two connected sets whose intersect is non-empty.
{
∈
|
|
}∪{
∈
≤ ≤
}
Exercise 7.5.3. A = (x, y) R2 : 1 < (x, y) < 2 . This set is an open set in the form of a disk, and as such is connected.
{
∈
|
|
}
Exercise 7.5.5. What are the connected components of the complement of the set of integers in R ? The connected components of this set consists of all sets A = x R : m < x < m + 1 for all m Z.
{ ∈
∈
}
Exercise 7.5.7. Which subsets of R are both compact and connected? Justify. Any closed interval in R will be both compact and connected. That is any set A = x R : m x n for some m, n R. It is important that this interval not contain any holes, as it is possible to create a compact set with holes which would not be connected. For example, the Cantor set is considered to be both compact and totally disconnected.
≤ ≤ }
{ ∈
∈
Exercise 7.5.10. Prove that the closure of a connected set is connected. Proof. We need to consider two distinct cases: either A is an open set, or a closed set. If A is a closed set, then A = A and hence, the closure is connected. If however A is an open set, then there exists at least one sequence xn such that xn A for all n N , xn L, but L / A. Consider now the set A. Now, L A and A A. Assume now that A is not a connected set. Then there exists some xi such that xi xn but x i / A. But by assumption, x i xn and therefore x i A. Since A A, which shows that x i A. Contradiction. Hence, A must also be a connected set.
∈ { } → ∈ ∈
∈ ∈ { }
{ }
∈
∈ ⊂
∈ ⊂ ∈ { }
Chapter 8 Functions on Euclidean Space 8.1
Continuous Functions of Several Variables
Exercise 8.1.2. Give a simple reason why the function γ : R (t, sin(t), et , t2 ) is continuous in R .
→R
4
defined by γ (t) =
The function γ consists solely of continuous functions in R , and as such γ (t) is likewise continuous (theorem 8.1.5).
Exercise 8.1.4. Consider the function f : R2 f (x, y) =
→ R defined by
x y : x y > 0. 0 : x y 0.
≤
at which points of R 2 is this function continuous? This function is continuous whenever
• x,y > 0 • x,y < 0 • x < 0, y > 0 • or x > 0, y < 0. The function is also continuous at zero if we approach (0, 0) along the axis from the same quadrant.
Exercise 8.1.5. For the function f : R2
→ R defined by
x2 y f (x, y) = 4 x + y 2 51
52
CHAPTER 8. FUNCTIONS ON EUCLIDEAN SPACE
→ →
show that f has a limit 0 as (x, y) (0, 0) along any straight line through the origin but that it does not have a limit as ( x, y) (0, 0) in R 2 . Assume thatwe approach along the x = 0 axis. In that case, the function has the limit 0. Assume now we approach along some line, which has the form y = mx. Then we consider x2 (mx) x3 m mx lim = lim = lim = 0. x,y →(0,0) x4 + (mx2 )2 x,y→(0,0) x4 + m2 x4 x,y→(0,0) x2 + m2 The function does not however have a limit itself. For a function to have a limit, its limit must be path-independent. This is not the case for the function in question. Assume you approach via the path y = mx2 . The resulting limit x2 (mx2 ) x4 m m lim = lim = lim x,y→(0,0) x4 + (mx2 )2 x,y→(0,0) x4 + x4 m x,y→(0,0) 1 + m clearly depends on m, and as such is not path-independent.
Exercise 8.1.12. Let B1 (0) be the open unit ball in R2 . Is it true that every continuous function f : B1 (0) R takes Cauchy sequences to Cauchy sequences?
→
I can’t come up with a counter example; as such, I think this holds true given that we are considering an open ball which includes the point (0 , 0). If we however choose the go towards the functions via a path of the form y = mx 2 .
Exercise 8.1.14. Find a parameterized curve γ (t) in R2 , with parameter interval [0, ), that begins at (1, 0), spirals inward in the counterclockwise direction, and approaches (0 , 0) as t .
∞
→∞
We use the function f (r, θ) = (r cos(θ), r sin(θ)).
8.2
Properties of Continuous Functions
Exercise 8.2.3. If K is a compact, connected subset of R p and f : K function, what can you say about f (K )?
→ R is continuous
We know that a continuous function maps a compact set to a compact set, and that a continuous function also maps a connected set to a connected set. As such, I would argue that f (K ) is a compact, connected set.
Exercise 8.2.5. The image of a compact set under a continuous function is compact, hence closed by theorem 8.2.3. Is the image of a closed set under a continuous function necessarily closed? Prove that it is or give a counter example. A simple counter example would be the function f : R R+ where f (x) = ex . Then the preimage of f is closed, yet the image of f is not, as the limit point lim x→−∞ ex = 0 is not part of the image set.
→
53
8.3. SEQUENCES OF FUNCTIONS
{
Exercise 8.2.7. Is the sphere (x,y,z ) know?
∈R
3
: x 2 + y 2 + z 2 = 1 connected? How do you
}
The sphere described above is connected. We know that any path-connected set is a connected set. The surface of a sphere is path connected, as we can join any two points through a continuous path function. As such, I would argue that the sphere is simply connected, which implies that it is a connected set.
Exercise 8.2.10. Is the function f : R2
\{(2, 0} → R defined by
f (x, y) =
(x
−
1 2)2 + y 2
uniformly continuous on B1 (0, 0)? Is it uniformly continuous on B2 (0, 0)? Justify your answers. I think it is uniformly continuous on B1 (0, 0), as well as on B2 (0, 0). In both cases, we can bound the function appropriately such that it goes to 0 without depending on x and y.
Exercise 8.2.11. If D R p , prove that if a function F : D R q is uniformly continuous on D, then F (xn ) is a Cauchy sequence in R q whenever xn is a Cauchy sequence in D.
{
}
⊂
→ { }
{ } | − | { } | − | | − |
Proof. Assume F is defined as above, with xn being a Cauchy sqeuence in D. Since F is a a uniformly continuous mapping, we know that for all > 0, there exists some δ > 0 such that f (xn ) f (xm ) < whenever xn xm < δ . Since xn is Cauchy, we know that there exists some N > 0 such that xn xm < δ whenever n, m > N . Due to uniform continuity, we know that δ is independent of x. Hence, if xn xm < δ , we have f (xn ) f (xm ) < for m, n > N . As such, the image of the Cauchy sequence is again Cauchy.
|
8.3
−
|
|
−
|
Sequences of Functions
Exercise 8.3.1. Show that the sequence γ n (t) , where
{
γ n (t) =
}
1 t , , 1 + nt n
does not converge uniformly on [0, 1]. For the function to converge uniformly, both parts of the function need to individually converge uniformly. While it is true that t/n converges uniformly on [0, 1] since it can be bound by 1/n, it is not true that 1/(1 + nt) converges uniformly. Specifically, we find that
1 1 + n
≤
1 1 + nt
≤
1.
As can be seen, this bound is dependent on t, since we need n large whenever t approaches 0 to be within > 0. Hence, this part of the function fails to converge uniformly, which results in our sequence γ n (t) failing to converge uniformly.
54
CHAPTER 8. FUNCTIONS ON EUCLIDEAN SPACE
Exercise 8.3.3. Does the sequence (k −1 sin(kx), k−1 cos(ky)) converge pointwise in R 2 ? Does it converge uniformly in R 2 ? Justify your answers.
{
}
1 , 1 k csc(kx) k sec(ky)
We can rewrite this sequence as . Clearly, the limit of this sequence is (0, 0) for all x. As both the secant and cosecant functions aren’t properly bounded, I would argue that this convergence cannot be uniform, and must be pointwise.
Exercise 8.3.8. Does the series
(x, y)
Justify your answer.
∞ xk yk converge uniformly on the square k=0 2
∈ R : −1 < x < 1, −1 < y < 1
?
This series converges uniformly. It is an adaptation of the geometric series in multiple variables. The geometric series exhibits uniform convergence per the Cauchy criterion. As such, I would argue that the extension of the geometric series to R2 would converge uniformly as well.
n Exercise 8.3.10. Does the series ∞ x)n ) converge pointwise on [0, 1]? Does it k=0 (x , (1 converge pointwise on (0, 1)? On which subsets of (0, 1) does it converge uniformly? Justify your answers.
−
n The series does not converge pointwise on [0, 1], since ∞ k=0 1 does not converge. It does n however converge pointwise on (0, 1), since ∞ k=0 k converges pointwise provided k < 1. I would argue that the series in question converges uniformly for all rightclosed subsets of n (0, 1), i.e. all sets of the form (0, a] where a < 1. It is evident then that ∞ k=0 x passes the Weierstrass M-test. We can see though that (1 x)n will also pass this test. Assume we pick a value x = δ > 0. Then we can bound the series (1 δ )n with some function kn such that k = 1 > 1 δ for some δ > > 0. We are guaranteed such a point exists, and hence this series passes the Weierstrass M-test on all sets of the form (0 , a] with a < 1.
−
−
−
| |
−
Exercise 8.3.12. Prove that if D is a subset of R p and F n is a sequence of functions from D to R q , then F n fails to converge uniformly to 0 if and only if there is a sequence xn in D such that the sequence of numbers F n (xn ) does not converge to 0.
{ }
{ }
{
{ }
}
{ } { } }→
Proof. Assume that F n does not converge to 0, and that there does not exist a sequence xn such that F n (xn ) fails to converge to 0. Then for all sequences xi D, we find that F n (xi ) 0. This means that for all x, F n (x) < for some > 0 with x D and n N for some N . But this is the definition of convergence for F n . Hence, it cannot be that F n fails to converge provided every sequence of xn results in F n (xn ) going to 0.
{ } {
{ } ∈ | | ∈ ≥ { } { } { } { } Now, assume that {F } converges to 0, but there exists a sequence {x } ∈ D such that F (x ) does not converge to 0. But then, it is not true that |F (x)| < whenever x ∈ D and n ≥ N for some N, > 0. As such, {F } would no longer converge. Therefore, {F } converges to 0 if and only if for all sequences {x } ∈ D, {F (x )} → n
n
n
n
n
n
n
0.
n
n
n
8.4. LINEAR FUNCTIONS, MATRICES
8.4
Linear Functions, Matrices
8.5
Dimension, Rank, Lines, and Planes
55
56
CHAPTER 8. FUNCTIONS ON EUCLIDEAN SPACE
Chapter 9 Differentiation in Several Variables 9.1
Partial Derivatives
9.2
The Differential
Exercise 9.2.1. If L : R p R p is a linear function, show that dL = L. In other words, if L has a matrix A, then A is the differential matrix of the linear function L(x) = Ax.
→
−A h = 0. We need to then show that Proof. It suffices to show that lim h→0 L(a+h)−|hL(a) | L(a + h) L(a) A h. But by linearity we can write this is L(a) + L(h) L(a) A h = 0, from which it is evident that L(a) = A h, and hence, that dL = L.
−
−
−
−
Exercise 9.2.5. Find the differential of the real-valued function f (x,y,z ) = xy2 cos(xz ). Then find the best affine approximation to f at the point (1, 1, π/2).
−
The differential is df = xy2 z sin (xz ) + y 2 cos(xz ) 2xy cos(xz ) We find the linear approximation of f at the given point as follows: f (x,y,z )
2 2
−x y
sin(xz ) .
≈ f (1, 1, π/2) + f (1, 1, π/2)(x − 1) + f (1, 1, π/2)(y − 1) + f (1, 1, π/2)(z − π/2) π π = 0+ (x − 1) + 0 + −z + 2 2 1 π π = − z − (x − 1) = − (πx + 2z − 2π) . 2 2 2 We can see that we can now approximate f by the linear function L(x,y,z ) = − (πx + 2z − 2π). x
y
z
Evaluating L at the point of interest, we see that L(1, 1, π/2) = 0.
1 2
Exercise 9.2.7. Prove that if f is a real-valued function defined on an open subinterval of R containing a and if S is an affine function such that f (a) = S (a), and lim
h
→0
f (a + h)
− S (a + h) = 0 h
then S (a + h) = f (a) + f (a)h. 57
58
CHAPTER 9. DIFFERENTIATION IN SEVERAL VARIABLES
Proof. Let S (x) = m x + b. Then
− S (a + h) = f (a + h) − m a − m h − b. Since we assume f (a) = S (a), we find then that b = f (a) − m a. Then, we rewrite as f (a + h) − S (a + h) = f (a + h) − f (a) − m h. We find f (a + h) − S (a + h) f (a + h) − f (a) − lim m h . 0 = lim = lim f (a + h)
h
h
→0
h
h
→0
This implies that m = f (a) and S (x) = f (a)x + f (a) S (a + h) = f (a)(a + h) + f (a)
→0 h
h
− m a. If we let x = a + h, we find
− a f (a) = f (a) + f (a)h.
This completes the proof.
Exercise 9.2.11. If f : R p R p is differentiable at a R p , then show that, for each h R p the function g : R R defined by g(t) = f (a + th) has a derivative at t = 0. Can you compute it in terms of df (a) and h?
→
→
∈
∈
Proof. MISSING A SOLUTION
Rq is affine if and only if it is differentiable Exercise 9.2.12. Prove that a function F : R p everywhere and its differential matrix is constant.
→
Proof. We start by proving that if a function F : R p R q is affine, then it is differentiable everywhere with a constant differential matrix. This is evident when we consider that an affine function F (x) is of the form F (x) = b + L(x), with L : R p R q , with an associated matrix p q -matrix A. Also, b, x R p . By exercise 9.2.1, we see that for a linear function L(x) with associated matrix A, dL = A. From this we see that F is differentiable everywhere with a constant differential matrix. We now proceed to consider a function which is differentiable everywhere, and has a constant differential matrix. The function F = (F 1 (x), F 2 (x), . . . , F p (x)). Since the differential is constant, we can write F i = a1,i x1 + a2,i x2 + . . . + a p,i x p + bi for some constants a1,i , a2,i ,...,bi . This implies that F (x) = b + Ax where A = a j,i and b = [b1 , b2 , . . . , bq ]T . Hence, F (x) is affine.
→
×
9.3
∈
→
The Chain Rule
Exercise 9.3.1. If F is a function from an open subset U of R p to Rq which is differentiable at a and if B is an r q matrix, then show that d(BF )(a) = B dF (a). Here, BF (x) is the matrix B applied to the vector F (x) and BdF (a) is the product of the matrix B and the matrix dF (a).
×
59
9.3. THE CHAIN RULE
Proof. F is differentiable at a if and only if there exists some function q (h) for h near 0 such that F (a + h) F (a) = Q(h)h, and Q(0) = dF (a). Let Q be a function satisfying these conditions. Now let B be some p q -matrix. Then the following holds:
−
×
− −
B (F (a + h) F (a)) = B Q(h) h B F (a + h) B F (a) = B Q(h) h Note that BQ(h) is continuous at 0 and that B F is differentiable at a. We then have d(B F )a = BQ(0) = B dF (a). ∂g Exercise 9.3.4. If f is a differentiable function on R and g(x, y) = f (xy), show that x ∂x ∂g = 0. y ∂y
−
Proof. MISSING A SOLUTION
Exercise 9.3.6. If u is a variable which is a differentiable function of ( x, y) in an open set R2, and U R2 , if x and y are differentiable function of (s, t) V for an open set V if (x, y) U whenever (s, t) V , then use the chain rule to obtain an expression for ∂u ∂s and ∂u on V in terms of the partial derivatives of u with respect to x and y and the partial ∂t derivatives of x and y with respect to s and t.
⊂
∈
∈
∈
⊂
∂d ∂d ∂x ∂ d ∂y = + ∂s ∂x ∂s ∂y ∂s ∂d ∂d ∂x ∂ d ∂y = + ∂t ∂x ∂t ∂y ∂t
Exercise 9.3.8. If F (x, y) = (f 1 (x, y), f 2 (x, y)) is a differentiable function from R2 to R2 and if we define G : R2 R2 by G(s, t) = F (st,s + t), find an expression for the differential matrix of G in terms of the partial derivatives of f 1 , f 2 .
→
Proof. MISSING A SOLUTION
Exercise 9.3.9. If (x,y,z ) are the Cartesian coordinates of a point in R 3 and the spherical coordinates of the same point are r, θ,φ, then x = r cos(θ) sin(φ)
y = r sin(θ) sin(φ)
z = r cos(φ)
Let u be a variable which is a differentiable function of ( x,y,z ) on R3 . Find a formula for the partial derivatives of u with respect to r, θ,φ in terms of its partial derivatives with respect to x, y,z .
60
CHAPTER 9. DIFFERENTIATION IN SEVERAL VARIABLES
∂u ∂d ∂x ∂ d ∂y ∂ d ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r ∂d ∂d ∂d = cos(θ)sin(φ) + sin(θ) sin(φ) + cos(φ) ∂x ∂y ∂z ∂u ∂d ∂x ∂ d ∂y ∂ d ∂z = + + ∂θ ∂x ∂θ ∂y ∂θ ∂z ∂θ ∂d ∂d = r sin(φ)sin(θ) + r sin(φ)cos(θ) ∂x ∂y ∂u ∂d ∂x ∂ d ∂y ∂ d ∂z = + + ∂φ ∂x ∂φ ∂y ∂φ ∂z ∂φ ∂d ∂d ∂d = r cos(φ)cos(θ) + r sin(θ)cos(φ) r sin(φ) ∂x ∂y ∂z
−
−
9.4
Applications of the Chain Rule
Exercise 9.4.3. Find the parametric equation for the tangent line to the curve γ (t) = (t3 , 1/t,e2t−2 ) at the point where t = 1.
− − − − −− −−
We note that γ (1) = (1, 1, 1), and dγ (1) = 3 be written as
2 τ (t) = 1 + (t 1
− 1)
1 2 . Then the tangent line τ (t) can
3 1 3t 1 = 1 + 1 2 1 2t
Exercise 9.4.6. Show that the gradient at x 2x.
∈R
p
3 3t t = 2 2 2t
2 t 1
of the function g(x) = x x is the vector
·
We note that we are considering the standard dot product as inner product, such that g(x) = x, x = x 22 + . . . x2n = ni=1 x2i . We notice that deg(x) = (2x1 , 2x2 , . . . , 2xn ) = 2x .
Exercise 9.4.11. Find the equation of the tangent plane to the cone z = x2 + y 2 at the point (1, 2, 5). We see that
∂z ∂x
= 2x and z
∂z ∂y
= 2y. Then:
∂z ∂ z − z = ∂x (x − x ) + (y − y ) ∂y ∂z ∂ z (x − x ) + (y − y ) + z z = ∂x ∂y z = 2x(x − x ) + 2y(y − y ) + z z = 2x(x − 1) + 2y(y − 2) + 5. 0
0
0
0
0
0
0
0
0
61
9.5. TAYLOR’S FORMULA
Exercise 9.4.12. Show that for each point (a,b,c) on the surface x2 + y 2 + z 2 = 1, there is a neighborhood of (a,b,c) in which the surface may be represented as a smoothly parameterized 2-surface. Hence, there is a tangent plane to this surface at every point. Let S denote the surface given by x2 + y 2 + z 2 = 1, which is a sphere. Consider the subsets of S given the graphs of the functions:
− −
y2
2
(y 2 + z 2 < 1)
2
2
(y 2 + z 2 < 1)
2
2
(x2 + z 2 < 1)
2
2
(x2 + z 2 < 1)
2
2
(x2 + y 2 < 1)
2
2
(x2 + y 2 < 1)
− z x− (y, z ) = 1 y − z √ y (x, z ) = 1 − x − z √ y− (x, z ) = 1 − x − z z (x, y) = 1 − x − y z − (x, y) = 1 − x − y x+ (y, z ) =
+
1
+
∈
Each of these functions is injective and differentiable. For any a S , we have that a is in the image of one of these functions, so we can construct a tangent plane to thee surface at any possible point.
Exercise 9.4.13. Find an equation for the tangent plane to the surface x 2 + y 2 each point on the surface.
2
− z = 1 at
Let w = x2 + y 2 + z 2 . Then w = (2x, 2y, 2z ), which means at point A = (a,b,c) we have w(A) = (2a, 2b, 2c). Then we get the following equation:
∇
∇
− 9.5
−
−
−
2a(x a) + 2b(y b) + 2c(z c) = 0 2a2 + 2ax 2b2 + 2by 2c2 + 2cz = 0 2ax + 2by + 2cz = 2a2 + 2b2 + 2c2 ax + by + cz = a2 + b2 + c2 .
−
−
Taylor’s Formula
Exercise 9.5.1. Find the degree n = 2 Taylor formula for f (x, y) = x2 + xy at the point a = (1, 2). We begin by finding the necessary partial derivatives. ∂f = 2x + y, ∂x ∂f = x, ∂y ∂ f ∂ 2 f = = 1. ∂x∂y ∂y∂x
∂ 2 f =2 ∂x 2 ∂ 2 f =1 ∂y 2
62
CHAPTER 9. DIFFERENTIATION IN SEVERAL VARIABLES
We now use the known equation with these derivatives at the point a = (1, 2).
1 ∂f (a, b) ∂ f (a, b) + f (a + x, b + y) f (a, b) y + 2 ∂x ∂y 1 = 3 + (4x + y) + 2x2 + 2xy + y 2 1 = 2x2 + 2xy + 8x + y 2 + 2y + 6 . 2
≈
∂ 2 f (a, b) 2 ∂ 2 f (a, b) ∂ 2 f (a, b) 2 x +2 xy + y ∂x 2 ∂x∂y ∂y 2
Exercise 9.5.4. Suppose U is an open convex set and f is a differentiable real-valued function on U . If there is a number M > 0 such that df (x) M for all x U , then
| | ≤ |f (x) − f (y) ≤ M |x − y|
∈
∈ U . Proof. Let x, y ∈ U . Then by Taylor’s theorem we have f (x) = f (y) + df (c)(x − y) for some c ∈ [y, x]. Then by algebra: |f (x) − f (y)| = |df (c)(x − y)| ≤ |df (c)||x − y| ≤ m|x − y|. for all x, y
This completes the proof.
Exercise 9.5.6. Show that the following form of the Mean Value Theorem is not true: if R2 is a differentiable function and a, b R2 , then there is a c on the line segment F : R2 joining a to b such that F (b) F (a) = dF (c)(b a). The problem here is that F is vectorvalued, not real valued.
→
∈ −
−
If the function F is vector-valued as above, we cannot make sense of the equation. Specifically, we find that dF (c)(b a) is defined as a scalar multiplication. In this case, the result is a scalar, whereas F (b) F (a) would be a case of vector addition. A scalar cannot equal a vector. Alternatively, we could attempt to solve this issue by redefining d F (c)(b a) as matrix multiplication. However, then we could only have equality in the case of d F begin a square matrix; otherwise, the dimensions wouldn’t match up. As such, it appears as though we can’t make sense of the equation as is for vector valued functions.
−
−
−
Exercise 9.5.8. Find all points of relative maximum and relative minimum and all saddle points for f (x, y) = 1 2x2 2xy y2 .
−
−
−
∂f We begin by finding the differential f (x, y) = ∂f = ( 4x 2y, 2x ∂x , ∂y is only zero at the point (0, 0). We calculate the Discriminant at this point:
∇
− − − (−2)
D = ( 4)( 2) Since D > 0,
∂ 2 f ∂x 2
2
=8
− − − − 2y). This
− 4 = 4.
< 0 we find that we are dealing with a point of local maximum.
63
9.6. THE INVERSE FUNCTION THEOREM
Exercise 9.5.9. Find all points of relative maximum and relative minimum and all saddle points for f (x, y) = y 3 + y 2 + x2 2xy 3y.
−
−
We begin by taking the various partial derivatives: ∂ 2 f =2 ∂x 2 ∂ 2 f = 6y + 2 ∂y 2
∂f = 2x 2y, ∂x ∂f = 3y2 + 2y 2x ∂y ∂ 2 f ∂ 2 f = = 2 ∂x∂y ∂y∂x
−
− − 3, −
We then find the points where f = 0. As f = (2x 2y, 3y2 + 2y 2x 3), we find that we have critical points at ( 1, 1) and (1, 1). At the point ( 1, 1) we have D = 2(6( 1) + 2) ( 2)2 = 12, and therefore we are dealing with a saddle point here. At the point (1, 1) we have D = 2(6 + 2) ( 2)2 = 4, and hence we have a point of local minimum here.
∇ − −
− −
9.6
∇ − −−
− − −
− −
−
The Inverse Function Theorem
R2 defined by F (x, y) = (x2 + y 2, xy) Exercise 9.6.2. Show that the function F : R2 has a smooth local inverse near points (x, y) where x = y. Find the inverse function F −1 on the set (x, y) : x < y < x and identify its domain. Calculate the differential of this inverse function (1) directly and (2) by using the Inverse Function Theorem. Verify that the two methods give the same answer.
→
{
−
}
±
Let u = x 2 + y2 and v = xy. Then we find that u + 2v = (x + y)2 and u 2v = (x y)2 . On the set (x, y) : x < y < x we find that x + y > 0 and x y > 0. This means that J F (x, y) = 2(x2 y2 ) = 0 and we have an inverse. We find then that
{
− −
x + y =
}
−
−
√ u + 2v
x
− y = √ u − 2v
From this we get that 1 x = 2 1 y = 2 We calculate the differential:
∂x ∂u ∂y ∂u
∂x ∂v ∂y ∂v
√ √
√ u + 2v + u − 2v √ u + 2v − u − 2v
√ 1 + 4√ u1−2v 4 u+2v = √ 1 − 4√ u1−2v 4 u+2v
√ 1 − 2√ u1−2v 2 u+2v √ 1 + 2√ u1−2v 2 u+2v
−
64
CHAPTER 9. DIFFERENTIATION IN SEVERAL VARIABLES
Now according to the inverse function theorem: dF −1 (b) = (dF (a))−1 = = =
− − ∂ (x2 +y 2 ) ∂x ∂ (xy) ∂x
2x 2y y x
x 2x2 2y 2 y 2x2 2y 2
− −
∂ (x2 +y 2 ∂y ∂ (xy) ∂y 1
−1
−
y x2 y 2 x x2 y 2
− −
√ 1 + 4√ u1−2v 4 u+2v = √ 1 − 4√ u1−2v 4 u+2v
and substituting:
√ 1 − 2√ u1−2v 2 u+2v √ 1 + 2√ u1−2v 2 u+2v
Exercise 9.6.5. Find a smooth local inverse function near (1 , π/2) for the function F of Example 9.6.6.
−
cos(θ) r sin(θ) We find that dF (r, θ) = . The determinant is non-zero whenever sin(θ) r cos(theta) r = 0. At the point F (a) = (0, 1). We know by the inverse function theorem that dF −1 (b) = (dF (a))−1 . We now use this information and find the inverse differential:
dF −1 (b) = (dF (a))−1 =
cos(θ) sin(θ) 0 1 = . 1 1 − − 1 0 r sin(θ) r cos(θ)
−
We find that the inverse function is F −1 (x, y) =
x2 + y 2 , tan−1 (y/x) .
Exercise 9.6.8. Show by example that the result of the previous problem is not true if U is only assumed to be connected, rather than convex. Hint: Try the function F (x, y) = (x2 y2 , 2xy) on R 2 0 .
−
\{ }
−
2x 2y Consider the function f (x, y) = (x . Then the y , 2xy). Then df = 2y 2x determinant of df is greater than 0, namely it is 4(x2 + y 2 ). Consider now U = R2 (0, 0) , which is a connected but not a convex set. Here, df is non-singular but f is not injective since f (1, 1) = f ( 1, 1). 2
−
−
2
\{
}
−
Exercise 9.6.10. Show that the condition that dF (a) be non-singular is necessary in the Inverse Function Theorem by showing that if a function F from a neighborhood of a in R P to R p is differentiable at a and has an inverse function at a which is differentiable at F (a), then dF (a) is non-singular. We are trying to show that if F is locally invertible, its differential matrix is also invertible, i.e. non-singular.
65
9.7. THE IMPLICIT FUNCTION THEOREM
Assume the inverse of F exists, and refer to it as F −1 , such that F F −1 = I . If we differentiate, we find that we are looking for a function such that
◦
I = dF dF −1 (F ) . By linear algebra, this implies that (dF (a))−1 is the inverse differential of dF (a), which implies that dF (a) is non-singular.
Exercise 9.6.12. If F : R p R p is a C 1 function, what can you say about F at a point of R p where F has a local minimum? How about a point where F has a local maximum?
→
| |
| |
The norm of a function F is defined as F = (F 1 )2 + . . . + (F p )2 for a function with p-components. Differentiating both sides with respect to some variable x yields the equation
| |
dF | | d|xF | = 2F dF + . . . + 2F dx dx
2 F
1
1
p
p
We note that since we are dealing with local minima/maxima, that d|xF | = 0. As such, we get the equation p dF i 0= F i . dx i=1
This is an interesting equation.
9.7
The Implicit Function Theorem
Exercise 9.7.1. Are there any points on the graph of the equation x3 + 3xy + 2y3 = 1 where it may not be possible to solve for y as a smooth function of x in some neighborhood of the point? Consider the function f (x, y) = x 3 + 3xy2 + 2y3 1. By the implicit function theorem, we can solve this function for y as a smooth function of x whenever ∂f = 0. We find that ∂y ∂f = 6xy + 6y2 . Solving this for zero is equivalent to solving y(x + y) = 0. Therefore, we ∂y can solve for y as a smooth function of x at all points where y = 0 and y = x.
−
Exercise 9.7.3. Find
∂ (f 1 ,f 2 ) ∂ (u,v)
−
if
f 1 (x,y,u,v) = u 2 + v 2 + x2 + y 2 , f 2 (x,y,u,v) = xu + yv + x y.
−
At which points (x,y,u,v) is this matrix non-singular? We solve for the differential: ∂ (f 1 , f 2 ) = ∂ (u, v)
∂f 1 ∂u ∂f 2 ∂u
∂f 1 ∂v ∂f 2 ∂v
=
2u 2v x y
We find that this matrix is singular whenever 2 uv = 2vx. Hence, this matrix is non-singular for all points (x,y,u,v) S = (x,y,u,v) : 2uv = 2vx .
∈
{
}
66
CHAPTER 9. DIFFERENTIATION IN SEVERAL VARIABLES
Exercise 9.7.4. Show that the system of equations u2 + v 2 + 2u xy + z = 0 u3 + sin(v) xu + yv + z 2 = 0
−
−
has a solution for (u, v) as a smooth function of (x,y,z ), in some neighborhood of (0, 0, 0), with the property that (u, v) = (0, 0) when (x,y,z ) = (0, 0, 0). Let f 1 = u2 + v 2 + 2u xy + z and f 2 = u3 + sin(v) differential at (0, 0, 0, 0, 0):
−
2
− xu + yv + z .
−
∂f 2u + 2 2v = 2 3u x cos(v) + y ∂ (u, v) ∂f 2 0 (0, 0, 0, 0, 0) = 0 1 ∂ (u, v)
We solve for the
As the last differential is if non-singular, this is possible.
Exercise 9.7.6. For the equation xy + yz + xz = 1, at which points on the solution set S is there a neighborhood in which S is a smooth 2-surface? At each such point ( a,b,c), find an equation of the tangent plane. This function describes a two sheeted hyperbola. Let F (x,y,z ) : xy + yz + xz = 1. THen F = (y + z, x + z, y + x). We find that F = (0, 0, 0) only at the points (0, 0, 0) and (x, x, 0). Of these, the only points on the graph are (1 , 1, 0) and ( 1, 1, 0). We find that F (1, 1, 0) = F ( 1, 1, 0). This means that there exists a neighbord V of (1, 1, 0) and ( 1, 1, 0) with V R 3 there exists a level set S = u V : F (u) = 0 which is a smooth p-surface. The tangent space at (1, 1, 0) and ( 1, 1, 0) consists of the set of the solutions u to F (1, 1, 0)(u (1, 1, 0)) = 0 and F ( 1, 1, 0)(u ( 1, 1, 0)) = 0.
∇ −
−
∇
∇
−
⊂
−
−
− −
−
− ∇ −
−
−
{ ∈
}
−
−−
Exercise 9.7.8. For the system of equations x2 + y 2 + u2 2x + xy y + 3u2
−
− 3v = 1, − 9v = 0,
find all points on the solution set S for which there is a neighborhood in which S is a smooth 2-surface. Let f 1 = x 2 + y2 + u2
2
− 3v − 1 and f = 2x + xy − y + 3u − 9v. We find the differential 2
∂ (f 1 , f 2 ) = ∂ (x,y,u,v)
∂f 1 ∂x ∂f 2 ∂x
∂f 1 ∂y ∂f 2 ∂y
∂f 1 ∂u ∂f 2 ∂u
∂f 1 ∂v ∂f 2 ∂v
2x 2y 2u = 2 + y x 1 6u
−
−3 −9
From this we can see that we can express the system of equations in terms of x, y as functions of u, v provided that x = (y + 1) 2 .
±
Chapter 10 Integration in Several Variables 10.1
Integration over a Rectangle
10.2
Jordan Regions
10.3
The Integral over a Jordan Region
10.4
Iterated Integrals
10.5
The Change of Variables Formula
67
68
CHAPTER 10. INTEGRATION IN SEVERAL VARIABLES
Chapter 11 Vector Calculus 11.1
1-forms and Path Integrals
Exercise 11.1.1. Find a smooth curve in R2 which traces the straight line from (1 , 2) to (3, 0). The solution is the function γ (t) = u + t(v We rewrite: γ (t) = =
− u) where u = (1, 2), v = (3, 0), and t ∈ [0, 1].
− 1 + t 2 1 + t 2
3 0
1 2
2 1 + 2t = 2 2 2t
−
−
Exercise 11.1.2. We find the derivative γ (t) = (cos(t) proceed to find the norm
− t sin(t), sin(t) + t cos(t)).
We
√ − t sin(t)) + (sin(t) + t cos(t)) = 1 + t . √ 1 + t dt. We utilize a trig substitution, lettting To find the length, we solve the integral |γ (t)| =
2
(cos(t)
2
4π 0
t = tan(θ).
√ 4π
2
tan−1 (4π)
1 + t2 dt =
0
2
sec(θ)dθ
0
= ln (sec(θ) + tan(θ)) tan 0
|
−1 (4π
)
√ 2 1 − = ln sec tan (4π) + 4π = ln 4π + 16π + 1
The last step is justified by letting sec (tan −1 (4π)) = sec(θ), and solving tan2 (θ)+1 = sec2 (θ). 69