SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then xy x + y. Proof. First we prove that if x is a real number, then x 0. The product of two positive numbers is always positive, i.e., if x 0 and y 0, then xy 0. In particular if x 0 then x = x x 0. If x is negative, then x is positive, hence x) 0. But we can conduct the following computation by the associativity and the commutativity of the product of real numbers: 0 x) = x) x) = )x) )x) = )x)) ))x = )x )))x = ) ))x)x = x)x = xx = x. The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x y which is defined to be x + y). Therefore we conclude that 0 x + y)) and compute: 0 x + y)) = x + y))x + y)) = xx + y)) + y)x + y)) = x + x y) + y)x + y) = x + y + xy) + xy) = x + y + xy); adding xy to the both sides, xy = 0 + xy x + y + xy)) + xy = x + y ) + xy) + xy) = x + y ) + 0 = x + y. Therefore, we conclude the inequality: xy x + y for every pair of real numbers x and y.
SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem. If a and b are real numbers with a < b, then there exists a pair of integers m and n such that a < m n < b, n 0. Proof. The assumption a < b is equivalent to the inequality 0 < b a. By the Archimedian property of the real number field, R, there exists a positive integer n such that nb a) >. Of course, n 0. Observe that this n can be if b a happen to be large enough, i.e., if b a >. The inequality nb a) > means that nb na >, i.e., we can conclude that na + < nb. Let m be the smallest integer such that na < m. Does there exists such an integer? To answer to the question, we consider the set A = {k Z : k > na} of integers. First A. Because if na 0 then A and if na > 0 then by the Archimedian property of R, there exists k Z such that k = k > na. Hence A. Choose l A and consider the following chain: l > l > l > > l k, k N. This sequence eventually goes down beyond na. So let k be the first natural number such that l k na, i.e., the natural number k such that l k na < l k +. Set m = l k + and observe that na < m = l k + na + < nb. Therefore, we come to the inequality na < m < nb. Since n is a positive integer, we devide the inequlity by n withoug changing the direction of the inequality: a = na n < m n < nb n = b. Page 4, Problem 6. Generate the graph of the following functions on R and use it to determine the range of the function and whether it is onto and one-to-one: a) fx) = x 3. b) fx) = sin x. c) fx) = e x. d) fx) = +x 4.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 3 Solution. a) The function f is bijection since fx) < fy) for any pair x, y R with the relation x < y and for every real number y R there exists a real numbe x R such that y = fx). b) The function f is neither injective nor surjective since fx + π) = fx) x + π x, x R, and if y > then there is no x R such that y = fx). c) The function f is injective because fx) < fy) if x < y, x, y R, but not surjective as a map from R to R, because there exists no x R such that fx) =.
4 SOLUTION SET FOR THE HOMEWORK PROBLEMS d) The function f is not injective as fx) = f x) and x x for x 0, nor surjective as there is no x R such that fx) =. Page 4, Problem 8. Let P be the set of polynomials of one real variable. If px) is such a polynomial, define Ip) to be the function whose value at x is Ip)x) x 0 pt)dt. Explain why I is a function from P to P and determine whether it is one-to-one and onto. Solution. Every element p P is of the form: px) = a 0 + a x + a x + + a n x n, x R, with a 0, a,, a n real numbers. Then we have Ip)x) = x 0 a 0 + a t + a t + + a n t n )dt = a 0 x + a x + a 3 x3 + + a n n xn. Thus Ip) is another polynomial, i.e., an element of P. Thus I is a function from P to P. We claim that I is injective: If have Ip)x) = Iq)x), x R,i.e., px) = a 0 + a x + a x + + a m x m ; qx) = b 0 + b x + b x + + b n x n a 0 x + a x + a 3 x3 + + a m m xm = b 0 x + b x + b 3 x3 + + b n n xn. Let Px) = Ip)x) and Qx) = Iq)x). Then the above equality for all x R allows us to differentiate the both sides to obtain P x) = Q x) for every x R,
SOLUTION SET FOR THE HOMEWORK PROBLEMS 5 in particular a 0 = P 0) = Q 0) = b 0. The second differentiation gives P x) = Q x) for every x R, in particular a = P 0) = Q 0) = b. Suppose that with k N we have P k) x) = Q k) x) for every x R. Then the differentiation of the both sides gives P k+) x) = Q k+) x) forevery x R, in particular a k+ = P k+) 0) = Q k+) 0) = b k+. Therefore the mathematical induction gives a 0 = b 0, a = b,, a m = b m and m = n, i.e., p = q. Hence the function I is injective. We claim that I is not surjective: As Ip)0) = 0, the constant polynomial qx) = cannot be of the form qx) = Ip)x) for any p P, i.e., there is no p P such that Ip)x) =. Hence the constant polynimial q is not in the image IP). Page 9, Problem 3. Prove that: a) The union of two finite sets is finite. b) The union of a finite sent and a countable set is countable. c) The union of two contable sets is countable. Proof. a) Let A and B be two finite sets. Set C = A B and D = A B. First, let a, b and c be the total number of elements of A, B and C respectively. As C A and C B, we know that c a and c b. We then see that the union: D = C A\C) B\C) is a disjoint union, i.e., the sets C, A\C and B\C are mutually disjoint. Thus the total number d of elements of D is precisely c + a c) + b c) = a + b c which is a finite number, i.e., D is a finite set with the total number d of elements. b) Let A be a finite set and B a countable set. Set C = A B and D = A B. Since C is a subset of the finite set A, C is finite. Let m be the total number of elements of C and {c, c,, c m } be the list of elemtns of C. Let n be the total number of elements of A and let {a, a,, a n m } be the leballing of the set A\C. Arrange an enumeration of the elements of B in the following fashion: Arranging the set A in the following way: B = {c, c,, c m, b m+, b m+, }. A = {a, a,, a n m, c, c,, c m },
6 SOLUTION SET FOR THE HOMEWORK PROBLEMS we enumerate the elements of D = A B in the following way: a i for i n m; d i = c i n m for n m < i n; b i n+m for i > n. This gives an enumeration of the set D. Hence D is countable. c) Let A and B be two countable sets. Let A = {a n : n N} and B = {b n : n N} be enumerations of A and B respectively. Define a map f from the set N of natural numbers in the following way: fn ) = a n, n N; fn) = b n, n N. Then f maps N onto D = A B. The surjectivity of the map f guarantees that f d) for every d D. For each d N, let gd) N be the first element of f d). Since f d) f d ) = for every distinct pair d, d D, gd) gd ) for every distinct pair d, d D. Hence the map g is injective. Now we enumerate the set D by making use of g. Let d D be the element of D such that gd ) is the least element of gd). After {d, d,, d n } were chosen, we choose d n+ D as the element such that gd n+ ) is the least element of gd\{d, d,, d n }). By induction, we choose a sequence {d n } of elements of D. Observe that gd ) < gd ) < < gd n ) < in N. Hence we have n gd n ). This means that every d N appears in the list {d, d, }. Hence D is countable. Page 4, Problem. Give five examples which show that P implies Q does not necessarily mean that Q implies P. Examples. ) P is the statement that x = and Q is the statement that x =. ) P is the statement that x and Q is the statement that x. 3) Let A be a subset of a set B with A B. P is the statement that x is an element of A and Q is the statement that x is an element of B. 4) P is the statement that x is a positive real number and Q is the statement that x is a positive real number. 5) P is the statement that x = 0 or x = and Q is the statement that xx )x ) = 0. Page 4, Problem 3. Suppose that a, b, c, and d are positive real numbers such that a/b < c/d. Prove that a b < a + c b + d < c d. Proof. The inequality a/b < c/d is equivalent to the inequality bc ad > 0. We compare two numbers by subtracting one from the other. So we compare the first two of the above
SOLUTION SET FOR THE HOMEWORK PROBLEMS 7 three fractions first and then the second pair of the fractions: a + c b + d a b c d a + c b + d = ba + c) ab + d) bb + d) = cb + d) a + c)d cb + d) = = bc ad bb + d) > 0; bc ad cb + d) > 0. Therefore the desired inequalities follows. Page 4, Problem 4. Suppose that 0 < a < b. Prove that a) a < ab < b. b) ab < a+b. Proof. a) We compute, based on the fact that the inequality a < b implies the inequality a < b, b) We simply compute: b = b b > a b = ab > a a = a. a + b ab = a) + b) a b = a b) > 0, where we used the fact that x + y) = x xy + y which follows from the distributive law and the commutativity law in the field of real numbers as seen below: x y) = x y)x x y)y by the distributive law = x yx xy + y = x xy + y by the commutativity law. Remark. In the last computation, is x y) = x xy + y obvious? If so, take x = ) 0 0 0 and y = ) 0 0 0
8 SOLUTION SET FOR THE HOMEWORK PROBLEMS and compute ) ) 0 x y = and x y) 0 = ; 0 0 ) ) ) ) x 0 0 =, y 0 0 0 0 0 =, xy =, yx = ; 0 0 0 0 0 0 0 ) ) x xy + y 0 0 = = x y). 0 0 0 Therefore, the formula x y) = x xy + y is not universally true. This is a consequence of the distributive law and the commutative law which governs the field R of real numbers as discussed in the very early class. Page 4, Problem 5. Suppose that x and y satisfy x + y 3 =. Prove that x + y >.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 9 Proof. The point x, y) lies on the line: x + y 3 = L) which cuts through x-axis at,0) and y-axis at 0,3). The line L is also discribed by parameter t,3 t)), t R. So we compute x + y = t) + 3 t)) = 4t + 9 t) = 4t + 9t 8t + 9 = 3t 8t + 9 = 3 t 8 3 t + 9 ) 3 { = 3 t 9 ) + 9 ) } 9 3 3 > 0 o 3 L) for all t R as 9 3 <. Page 5, Problem 8. Prove that for all positive integers n, 3 + 3 + + n 3 = + + + n). Proof. Suppose n =. Then the both sides of the above identity is one. So the formula hold for n =. Suppose that the formula hold for n, i.e., Adding n + ) 3 to the both sides, we get 3 + 3 + + n 3 = + + + n). + + + n 3 + n + ) 3 = + + + n) + n + ) 3 ) nn + ) = + n + ) 3 by Proposition.4.3 n = n + ) ) + 4n + 4 = n + ) n + ) 4 4 = + + + n + n + ) by Proposition.4.3. Thus the formula holds for n+. Therefore mathematical induction assures that the formula holds for every n N.
0 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5, Problem 9. Let x > and n be a positive integer. Prove Bernoulli s inequality: + x) n + nx. Proof. If x 0, then the binary expansion theorem for real numbers gives ) ) ) + x) n n = + nx + x n + x 3 n + + x n + x n 3 n + nx as x, x 3,, x n and x n are all non-negative. If < x < 0, then the inequality is more delicate. But we can proceed in the following way: + x) n = x + x) n + + x) n + + + x) + ) ; n + x) n + + x) n + + + x) + as + x <. As x < 0, we get ) nx x + x) n + + x) n + + + x) +, consequently the desired inequality: ) + x) n = x + x) n + + x) n + + + x) + nx. Page 5, Problem. Suppose that c < d. a) Prove that there is a q Q so that q < d c. b) Prove that q is irrational. c) Prove that there is an irrational number between c and d. Proof. a) Choose a = and b = and observe that a = < < 4 = b consequently a < < b. Consider a + b )/ and square it to get a + b ) = 9 4 >
SOLUTION SET FOR THE HOMEWORK PROBLEMS and put a = a and b = a + b )/. Suppose that a, a,, a n and b, b,, b n were chosen in such a way that i) if ) ak + b k <, then a k = a k + b k )/ and b k = b k ; ii) if ) ak + b k >, then a k = a k and b k = a k + b k )/. Thus we obtain sequences {a n } and {b k } of rational numbers such that a a a 3 a n b n b n b b and b k a k = b k a k ; a n < < b n, n N, hence a n < < b n. As b a =, we have b n a n = / n. For a large enough n N we have / n < d c. Now we conclude 0 < a n b n a n = n < d c and a n Q. Thus q = a n has the required property. b) Set p = q. If p Q, then = p + q Q which is impossible. Therefore, p cannot be rational. c) From b), p = q is an irrational number and 0 < p < d c from a). Thus we get c < c + p < d. As seen in b), c + p cannot be rational. Because if c + p = a is rational, then p = a c has to be rational which was just proven not to be the case.
SOLUTION SET FOR THE HOMEWORK PROBLEMS Problems. ) Negate the following statement on a function f on an interval [a, b], a < b. The function f has the property: fx) 0 for every x [a, b]. ) Let fx) = x + bx + c, x R. What can you say about the relation on the constants b and c in each of the following cases? a) fx) 0 for every x R. 3) Let b) fx) 0 for som x R. fx) = x + 4x + 3, x R. Which of the following statements on f is true? State the proof of your conclusion. a) fx) < 0 for some x R; b) c) d) fx) > 0 for some x R; fx) 0 for every x R. fx) < 0 for every x R. Solution. ) There exists an x 0 [a, b] such that fx 0 ) < 0. -a) First, we look at the function f closely: fx) = x + bx + c = x + bx + b 4 + c b 4 = x + b ) 4c b 4c b + for every x R. 4 4
SOLUTION SET FOR THE HOMEWORK PROBLEMS 3 The function f assumes its smallest value f b ) = 4c b 4 at x = b. Thus fx) 0 for every x R if and only if f b ) = 4c b 4 0 if and only if 4c b. -b) If 4c b, then fx) 0 for every x R, in particular f0) 0. If 4c < b, then we have b + ) b f 4c = 0. Therefore the condition that fx) 0 for some x R holds regardless of the values of b and c. So we have no relation between b and c. 3) First, we factor the polynomial f and draw the graph: fx) = x + 4x + 3 = x + 3)x + ). We conclude that fx) 0 is equivalent to the condition that 3 x. Therefore we conclude that a) and b) are both true and that c) and d) are both false. Page 5, Problem. Prove that the constant e = k=0 k! is an irrational number. Proof. Set s k = k j=0 j!, k N.
4 SOLUTION SET FOR THE HOMEWORK PROBLEMS Clearly we have = s s s 3 s k, i.e., the sequence {s k } is an increasing sequence. If k 3, we have s k = + + + 3 + + 3 k + + + + + < + = 3. k = + k Thus the sequence {s k } is bounded. The Bounded Monotonce Convergence Axiom MC) guarantees the convergence of {s k }. Thus the limit e of {s k } exists and e 3. a) If n > k, then we have s n s k = + +! + 3! + + k! + k + )! + + n! + +! + 3! + + ) k! = k + )! + k + )! + + n! = k + )! + ) k + + k + )k + 3) + k + )k + 3)k + 4) + +. k + ) n Since k + < k + < k + 3 < < n, and k + > k + k + 3 > > n, we have and k + ) > k + )k + ), k + ) 3 > k + )k + )k + 3), k + ) n k > k + )k + ) n s n s k < < n k k + )! l=0 kk + )!. k + ) l = k+) n k k + )! k+)
SOLUTION SET FOR THE HOMEWORK PROBLEMS 5 Taking the limit of the left hand side as n, we get e s k kk + )! < k k!. b) Now suppose that e is rational, i.e., e = p/q for some p, q N. Then we must have p q s q q q! ) p and 0 < q! q s q < q. But now q!p/q) is an integer and p!s q is also because!, 3!, and q!, appearing in the denominators of the summation of s q, all divide q!, i.e., q!s q = q! + +! + 3! + + ) q! = q! + q! + 3 4 q) + 4 5 q) + + k k + ) q) + + q +. Thus the number q!p/q s q ) is a positive integer smaller than /q, which is not possible. This contradiction comes from the assumption that e = p/q, p, q N. Therefore we conclude that there is no pair of natural numbers p, q such that e = p/q, i.e., e is an irrational number. Page 33, Problem. Compute enough terms of the following sequences to guess what their limits are: a) a n = nsin n. b) a n = + n) n. c) a n+ = a n +, a =. d) a n+ = 5 a n a n ), a = 0.3.
6 SOLUTION SET FOR THE HOMEWORK PROBLEMS Answer. a) It is not easy to compute sin/,sin /3 and so on. So let us take a closer look at the function sin x near x = 0: Consider the circle of radius with center 0, i.e., 0A =, and draw a line 0B with angle A0B = x. Let C be the intersection of the line 0B and the circle. Draw a line CD through C and perpendicular to the line 0A with D the intersection of C B x 0 D A the new line and the line 0A. So obtain the figure on the right. Now the length CD = sin x. We have the arc length AC = x and the line length AB = tan x. To compare the sizes of x, sin x and tan x, we consider the areas of the triangle 0AC, the piza pie cut shape 0AC and 0AB which are respectively sin x)/, x/ and tan x)/. As these three figures are in the inclusion relations: 0AC 0AC 0AB, we have Consequently we conclude that Therefore we have cos sin x ) n sin n b) We simply compute a few terms: a = a 3 = a 4 = a 5 = x tan x = cos x sin x x. sin x cos x ) and lim n n sin n + =, a = + ) ) = + + + ) 3 = + + 3 3 + 4) 4 = + + 6 96 + 6 + = 56 = = 03 56, + 5 = + + 5) 9 + 7 = 0 7, 5 6 + 4 64 + 56 ) 5 + 0 5 + 0 5 + 5 5 + 35 ) 5 3 5 + = 56 35. ) =. n 4 =.5. = 6 6 + 4 4 + 56 ) 5 4 65 + 35
SOLUTION SET FOR THE HOMEWORK PROBLEMS 7 It is still hard to make the guess of the limit of + /n) n. So let us try something else. a n = + n) n ) ) ) n n n = + + n + 3 n 3 + + k n k + + n n = + nn )! n + nn )n ) 3! n 3 + + nn )n ) n k + ) k! n k nn )n ) = +! a n+ = ) + n 3! n + ) n+ = + n +! + + k! + + ) n! ) + n k! + + n! n < k! = s n. k=0 ) n + + n + )! n n n ) n ) k ) n ) ) n ) n n n + n + 3! ) n + n + ) ) n + n + k ) + n + ) ) n+ ) n n + As each term of a n+ is greater than the corresponding term of a n, we have a n a n+, n N, i.e., the sequence {a n } is increasing and bounded by e as a n s n e. Therefore we conclude that the sequence {a n } converges and the limit is less than or equal to e. c) Skip. d) Let us check a few terms: With a = 0.3, a = 5 a a ) = 0.55 a 3 = 5 a a ) = 0.634375 a 4 =0.58690795898, a 5 = 0.60675666, a 6 = 0.5968476864 a 7 = 0.6055364, a 8 = 0.59983534, a 9 = 0.60038930979, a 0 = 0.599804966 fx) = 5 x x) = 5 x x ) = 5 ) ) 4 x 5 8 < In fact the limit of {a n } is the natural logarithm number e, which will be shown later. )
8 SOLUTION SET FOR THE HOMEWORK PROBLEMS we have 0 fx) 5 8 = 0.65 for all x [0,], and consequently 0 a n+ = fa n ) 5 8 = 0.65, n 3. To compare a n and a n+ = fa n ), we consider x fx) = x 5 x x) = x 5x x ) = 5x 3x x5x 3) = { 0 for all x [0,0.6] 0 for all x / [0,0.6]. This means that a n+ a n if 0 a n 0.6 and a n+ a n if a n < 0 or a n > 0.6. But the case a n < 0 has been excluded by the above arguments. From the computation of the first three terms we observe that the sequence {a n } seems to oscillate. At any rate, if the sequence {a n } converges, then we must have a = lim n a n = lim n a n+, i.e., we must have a = fa), which narrows the candidate of the limit down to either 0 or 3/5 = 0.6. Let us examine the candidate 3/5 first. So we compute the error 3 5 fx) = 3 5 5 x x) = 6 5x x) 0 = 5x 5x + 6 5x )5x 3) = 0 0 = 5x x 3 5 = x 5 3 ) + 5 x 3 5 Therefore, if x 3/5 = δ, then 3 5 fx) 5δ + )δ. Thus if δ < /5, then with r = 5δ + )/ < we have 3 5 fx) rδ,
SOLUTION SET FOR THE HOMEWORK PROBLEMS 9 in other words r k 3 5 a n 3 rk 5 a n+ 3 rk 5 a n+ r k 3 3 5 a n+3 3 5 a n+k. Therefore, if we get 3/5 a n < /5 for some n N, then we have lim a k = 3 k 5. But we know 3 5 a 3 = 0.634375 0.6 = 0.034375 < 0. = 5. Therefore the limit of the sequence {a n } is 0.6 as seen in the first computation. Page 33, Problem. Prove directly that each of the following sequences converges by letting ε > 0 be given and finding Nε) so that a a n < ε for every n Nε). ) a) b) c) d) a n = + 0 n. a n = + 3 n. a n = 3 + n. a n = n n +. Solution. a) Obviously our guess on the limit a is a =. So let us try with a = to find Nε) which satisfy the condition ): a n = 0 n < ε for every n Nε),
0 SOLUTION SET FOR THE HOMEWORK PROBLEMS which is equivalent to the inequality: n > 0 ε n > 00 ε for every n Nε). Thus if we choose Nε) to be [ ] 00 Nε) = +, where [x], x R, means the largest integer which is less than or equal to x, i.e., the integer m such that m x < m +, then for every n Nε), we have ε 00 ε < Nε) n, hence ε > 00 n and ε > 0 n = a n. This shows that lim + 0 ) n =. n b) It is also easy to guess that the limit a of {a n } is. So let ε > 0 and try to find Nε) which satisfy the condition ) above which is: 3 n = a n < ε for every n Nε). So we look for the smallest integer N which satisfy 3 n < ε equivalently ε < 3 n, which is also equivalent to n > ε 3. So with Nε) = [ /ε 3] +, if n Nε), then ε 3 < Nε) n consequently 3 < ε. n d) First, we make a small change in the form of a n : n a n = n + = +, n
SOLUTION SET FOR THE HOMEWORK PROBLEMS and guess that the limit a of {a n } would be. So we compute: n n + n n + = = n + n) n + + n) n + n + n + + n) n + n = n + + n + )n n. Hence if n [/ε] +, then /n /[/ε] + ) < //ε) = ε, i.e., n n + < ε for every n Nε). Page 33, Problem 3. Prove directly that each of the following sequences converges by letting ε > 0 be given and finding Nε) so that a a n < ε for every n Nε). ) a) b) c) d) a n = 5 ln n a n = n + 6 n a n = 3n + n +. a n = n n!. for n. for n. Solution. a) From the form of the sequence, we guess that the limit a would be 5. So we try 5 as a: 5 a n = 5 5 + ) = ln n ln n, which we want to make smaller than a given ε > 0. So we want find how large n ought to be in order to satisfy the inequlity: ε > ln.
SOLUTION SET FOR THE HOMEWORK PROBLEMS This inequality is equivalent to ln n > /ε. Taking the exponential of the both sides, we must have n > exp/ε). So if we take [ )] Nε) = exp +, ε then for every n Nε) the inequality ) holds. b) First we change the form of each term slightly: a n = 3n + n + = 3 + n +, n to make a guess on a. This indicates that the limit a would be 3. So we try to fulfil the requirement of ) with a = 3: 3 3n + n + = 3n + ) 3n + ) n + = 5 n + < 5 n. So if the inequality 5/n < ε holds, then 3 a n < ε holds. Thus Nε) = [5/ε]+ gives that 5 a n < ε for every n Nε). c) We alter the form of the sequence slightly: n a n = n + 6 n = + 6 in order to make a good guess on the limit a, which looks like /. Let us try with this a: n + 6 n = n ) n + 6) n = 7 n, for n. If n, then n ) n ) = 4n > 0, so that n ) < n and therefore n + 6 n < 7 n ). [ ] 7/ε) Thus if Nε) = +, then for every n Nε) we have n, n + 6 n < 7 n ) 7 Nε) ) = 7 [ ] < 7 7 ε = ε. 7 ε ) +
SOLUTION SET FOR THE HOMEWORK PROBLEMS 3 Therefore d) With we look at the ratio a n /a n+ : lim a n = n. a n = n /n! a n = n a n+ n! Therefore, we have for every k n + )! n+ = n + for n 3. a 3 k a 3+k equivalently a k+3 a 3 k = 8 6 k = 3 k < k < k. So for any ε > 0 if n Nε) = [ ε] + 5, then we have 0 < an < ε. Page 4, Problem 6. Suppose that a n a and let b be any number strictly less than a. Prove that a n > b for all but finitely many n. Proof. The assumption a > b yields b a > 0 so that there exists N N such that a a n < b a for every n N, equivalently b a < a a n < a b for every n N, hence b < a n for every n N. Thus the total number of n with b a n is at most N which is of course finite. Thus a n > b for all but finitely many n. Page 34, Problem 9. a) Find a sequence {a n } and a real number a so that a n+ a < a n a for each n, but {a n } does not converge to a. b) Find a sequence {a n } and a real number a so that a n a but so that the above inequality is violated for infinitely many n. Answer. a) Take a n = /n and a =. Then a n+ a = n + + < n + = a n a but a n a. b) Set a n = ) + )n n and a = 0.
4 SOLUTION SET FOR THE HOMEWORK PROBLEMS Then we have If n is odd, then a n = { for odd n; n for even n 3 n a n+ = and a n 0. 3 n + ) > n = a n. This occurs infinitely many times, i.e., at every odd n. Page 39, Problem. Prove that each of the following limits exists: a) a n = 5 + 3 ). n b) c) d) a n = 3n + n +. a n = n + 6 3n. a n = 5 + 3 n ) + n+5 3n Proof. a) First lim n / 3 n = 0 because for any given ε > 0 if n Nε) = [/ε 3 ]+ then Hence we get 3 n = Nε) 3 3 [ ] < ε + 3 lim 5 + ) n 3 = 5 n 3 = ε 3 ε) = ε. by the combination of Theorem..3, Theorem..4 and Theorem..5 as seen below: + 3 n by Theorem..3 + 3 n ) by Theorem..5 5 + 3 ) 5 by Theorem..4. n
SOLUTION SET FOR THE HOMEWORK PROBLEMS 5 b) We change the form of each term a n slightly: a n = a n = 3n + n + = 3 + n +. n We know that /n 0 and /n 0 as n. Thus we get the following chain of deduction: 3 + n 3 and + n by Theorem..3 3 + n + 3 by Theorem.6. n c) We change the form of each term a n in the following way: n a n = n + 6 3n = + 6 3 n. As /n 0, Theorem..5 yields that /n 0 and therefore + 6 n 3 + 6 0 3 0 = 3 n d) As seen before, we have Thus we get by Theorem..4 and Theorem..6. 3 n n 0 and n + 5 3n = + 5 n 3 n 3. a n = 5 + ) 3 n + n+5 5 + 4 0 0 3n + = 5 8 3 by a combination of Theorem..3, Theorem..4 and Theorem..6. Page 39, Problem 6. Let px) be any polynomial and suppose that a n a. Prove that lim pa n) = pa). n Proof. Suppose that the polynomial px) has the form: px) = p k x k + p k x k + + p x + p 0. We claim that a l n al for each l N. If l =, then certainly we have the convergence: a n = a n a = a. Suppose a l n a l. Then by Theorem..5 we have a l n = al n a n a l a = a l. By mathematical induction we have a l n al for each l N. Therefore, each term p l a l n converges to p la l for l =,,, k. A repeated use of Theorem..3 yields that pa n ) = p k a k n + p k a k n + + p a n + p 0 p k a k + p k a k + + p a + p 0 = pa).
6 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 39, Problem 7. Let {a n } and {b n } be sequences and suppose that a n b n for all n and that a n. Prove that b n. Proof. The divergence a n means that for every M there exists N N such that a n > M for every n N. The assumption that a n b n gives M < a n b n for every n N. Hence b n. Page 39, Problem 9. a) Let {a n } be the sequence given by Prove that a n 4. b) Consider the sequence defined by a n+ = a n +, a = 0.5 a n+ = αa n +. Show that if α <, then the sequence has a limit independent of a. Proof. a) Based on the hint, we compute Hence we get a n+ 4 = a n + 4 = a n = a n 4). a n 4 = a n 4 = a n 4 = = n a 4 = 3.5 0. n b) We just compute α = αa α) n + α = αa n α α = α a n ) ; α ) = α a n a n+ α = αa n + a n α = α a n α = α n a α ) = α ) 0 as α <. Hence {a n } converges and which is independent of a. lim a n = n α
SOLUTION SET FOR THE HOMEWORK PROBLEMS 7 Page 40, Problem 0. For a pair x, y) of real numbers, define x, y) = x + y. a) Let x, x, y, y be real numbers. Prove that x x + y y x + x y + y. b) Prove that for any two dimensional vectors x, y ), x, y ) R x, y ) + x, y ) x, y ) + x, y ). c) Let p n = x n, y n ) be a sequence of points in the plane R and let p = x, y). We say that p n p if p n p 0. Prove that p n p if and only if x n x and y n y. Proof. a) Let us compute: x + x )y + y ) x y + x y ) b) We also compute directly: = x y + x y + x y + x y x y + x y x y + x y ) = x y + x y x y x y = x y x y ) 0. x, y ) + x, y ) = x + x, y + y ) = x + x ) + y + y ) = x + x x + x + y + y y + y = x + x + x x + y y ) + y + y x + y x + + y x + y + x + y = x + x + y ) + y = x, y ) + x, y ) ). This shows the inequality: c) Since we have the inequalities: x, y ) + x, y ) x, y ) + x, y ). max{ x n x, y n y } x n x) + y n y) max{ x n x, y n y }, show that p n p 0 if and only if max{ x n x, y n y } 0 if and only if x n x 0 and y n y 0.
8 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 50, Problem. Prove directly that a n = + n is a Cauchy sequence. Proof. We just compute for m < n: a m a n = + ) + ) = m n m n m + n m since m < n. So if ε > 0 is given, then we take N to be [/ε) ]+ so that for every n > m N we have a m a n < m [ ) ] < ) = ε. + Page 50, Problem. Prove that the rational numbers are dense in the real numbers. Proof. We have to prove that for every ε > 0 a ε, a + ε) Q. Choose m = [/ε] + so that /m < ε. If a ε > 0, then the Archimedian property of R yields the existence of k N such that k m = k m > a ε. Let n be the first such a number. Then we have n m a ε < n m = n m + m a ε + m < a ε + ε = a. Therefore, we conclude that a ε < n m < a therefore n a ε, a + ε) Q. m If a ε < 0, then we apply the Archimedian property of R to the pair /m and ε a > 0 to find a natural number k N such that ε a < k m = k m. Let n N be the smallest natural number such that n/m ε a, so that n m < ε a n m, equivalently n m a ε < n m = n m + m. As we have chosen m N so large that /m < ε, the above inequality yields n m = m n m < ε n n ε + a ε) = a, i.e., a ε < m m < a. Therefore we have n a ε, a + ε) Q consequently a ε, a + ε) Q m for arbitrary ε > 0. Hence a is a limit point of Q. ε ε
SOLUTION SET FOR THE HOMEWORK PROBLEMS 9 Page 59, Problem 3. Suppose that the sequence {a n } converges to a and d is a limit point of the sequence {b n }. Prove that ad is a limit point of the sequence {a n b n }. Proof. By the assumption on the sequence {b n }, there exists a subsequence {b nk } of the sequence {b n } such that lim k b n k = d. The subsequence {a nk b nk } of the sequence {a n b n } converges to ad because the subsequence {a nk } of {a n } converges to the same limit a. Hence ad is a limit point of {a n b n }. Page 59, Problem 6. Consider the following sequence: a = ; the next three terms are 4,, 3 4 ; the next seven terms are 8, 4, 3 8,, 5 8, 3 4, 7 ; and so forth. What are the limit 8 points. Answer. The sequence {a n } consists of the numbers {k/ n : k =,,, n, n N}. Fix x [0,]. We are going to construct a subsequence {b n } of the sequence {a n } by induction. For n =, choose k = { 0 if x ; if < x. For each n >, let k n be the natural number such that k n n x < k n + n. Then the ratio b n = k n / n is in the sequence {a n } and b n x < 0 as n. n Hence lim n b n = x. Therefore, every x [0,] is a limit point of {a n }. Thus the sequence {a n } is dense in the closed unit interval [0,]. Page 59, Problem 8. Let {I k : k N} be a nested family of closed, finite intervals; that is, I I. Prove that there is a point p contained in all the intervals, that is p k= I k. Proof. The assumption means that if I k = [a k, b k ], k N, then a a a k b k b k b b. The sequence {a k } is increasing and bounded by any of {b l }. Fix k N. Then we have a = lim n a n b k for k N,
30 SOLUTION SET FOR THE HOMEWORK PROBLEMS where the convergence of {a n } is guaranteed by the boundedness of the sequence. Now look at the sequence {b k } whis is decreasing and bounded below by a. Hence it converges to b R and a b. Thus the situation is like the following: a a a k a b b k b k b b. Hence the interval [a, b] is contained in the intersection k= I k. Any point p in the interval [a, b] is a point of k= I k; in fact [a, b] = k= I k. Page 59, Problem 9. Suppose that {x n } is a monotone increasing sequence of points in R and suppose that a subsequence of {x n } converges to a finite limit. Prove that {x n } converges to a finite limit. Proof. Let {x nk } be the subsequence converging to the finite limit x 0. As n < n < < n k <, we have k n k for every k N. If ε > 0 is given, then choose K so large that x nk x 0 < ε for every k K, i.e., x 0 ε < x nk x 0 for every k K. Set N = n K. Then if m N, then we have x 0 ε < x nk = x N x m x nm x 0. Hence we have 0 x 0 x m < ε for every m N. Hence {x n } converges to the same limit x 0. Page 79, Problem 3. Let fx) be a continuous function. Prove that fx) is a continuous function. Proof. Let x [a, b] be a point in the domain [a, b] of the function f. If ε > 0 is given, then choose a δ > 0 so small that fx) fy) < ε whenever x y < δ. If x y < δ, then Hence f is continuous at x. fx) fy) fx) fy) < ε. Page 79, Problem 5. Suppose that f is a continuous function on R such that fq) = 0 for every q Q. Prove that fx) = 0 for every x R. Proof. Choose x R and ε > 0. Then there exists δ > 0 such that fx) fy) < ε whenever x y < δ. Take q Q x δ, x + δ), then fx) = fx) fq) < ε. Thus fx) is less than any ε > 0 which is possible only when fx) = 0.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 3 Page 79, Problem 7. Let fx) = 3x and let ε > 0 be given. How small δ be chosen so that x ε implies fx) < ε? Answer. To determine the magnitude of δ, assume that x < δ and see how the error becomes: fx) = 3x = 3x 3 = 3 x < 3δ. Thus if 3δ ε, i.e., if δ ε/3, then x < δ implies fx) < ε. Page 79, Problem 8. Let fx) = x and let ε > 0 be given. a) Find a δ so that x δ implies fx) ε. b) Find a δ so that x δ implies fx) ε. c) If n > and you had to find a δ so that x n δ implies fx) n ε, would the δ be larger or smaller than the δ for parts a) and b)? Why? Answer. a) Choose δ > 0 and see how the error grows from x δ: fx) = x = x + )x ) = x + x x + δ = x + + δ x + )δ δ + )δ. So we want to make δ + )δ ε. Let us solve this inequality: 0 δ + δ ε = δ + ) ε ε + δ + ) ε + δ ε +. But we know that δ must be positive. Hence 0 < δ ε + = ε/ ε + + ). If δ is chosen in the interval 0, ε + ), then the above calculation shows that x δ fx) ε. b) Now we continue to examine the case x δ: fx) 4 = x 4 = x + )x ) = x + x δ x + = δ x + 4 δ x + 4) δδ + 4). So we want to make δ + 4)δ ε, equivalently: ε δδ + 4) = δ + 4δ δ + 4δ ε 0 ε + 4 δ ε + 4 = ε + 4 ) ε + 4 + ) ε + 4 + = ε ε + 4 +.
3 SOLUTION SET FOR THE HOMEWORK PROBLEMS Hence if we take 0 < δ ε/ ε + 4 + ), then x < δ fx) 4 < ε. c) Similarly, we examine the case x n δ: fx) n = x n = x + n x n δ x + n So we want to make δ + n)δ ε, equivalently: δ x n + n δ x n + n) δδ + n). ε δδ + 4) = δ + nδ δ + nδ ε 0 ε + n n δ ε + n n = ε + n n) ε + n + n) ε + n + n Hence if we take δ > 0 so small that 0 < δ ε/ ε + n + n), then x n δ fx) n ε. = ε ε + n + n. The largest possible δ = ε/ ε + n + n) is squeezed to zero when n glows indefinitely. Page 79, Problem. Let fx) = x with domain {x : x 0}. a) Let ε > 0 be given. For each c > 0, show how to choose δ > 0 so that x c δ implies x c ε. b) Give a separate argument to show that f is continuous at zero. Solution. a) Once again we examine the growth of error by letting x c δ and compute: x c = x c + c c = δ c + c δ c. x c x c + c + c under the assumption δ c ) δ c δ + c Thus if 0 < δ min{c/, ε c/}, then Hence f is continuous at c > 0. b) If 0 x δ, then Hence if 0 < δ ε, then Therefore f is continuous at 0. x c δ x c ε. x 0 = x δ. 0 x δ 0 x ε.