0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2.

Size: px
Start display at page:

Download "0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2."

Transcription

1 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.

2 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.

3 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 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,

5 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 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

7 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 = ) and y = ) 0 0 0

8 8 SOLUTION SET FOR THE HOMEWORK PROBLEMS and compute ) ) 0 x y = and x y) 0 = ; 0 0 ) ) ) ) x 0 0 =, y =, xy =, yx = ; ) ) x xy + y 0 0 = = x y) 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 >.

9 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 ) } > 0 o 3 L) for all t R as 9 3 <. Page 5, Problem 8. Prove that for all positive integers n, 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 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.

10 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 >

11 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.

12 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

13 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.

14 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 = 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+)

15 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! q) 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.

16 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 = ) 4 = = 56 = = 03 56, + 5 = + + 5) = 0 7, ) ) = ) =. n 4 =.5. = )

17 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 k n k + + n n = + nn )! n + nn )n ) 3! n 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 ) = a 4 = , a 5 = , a 6 = a 7 = , a 8 = , a 9 = , a 0 = 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. )

18 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) = 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,

19 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 a n 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. = 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),

20 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

21 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 = ) = 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.

22 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 ) +

23 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.

24 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 = ). n b) c) d) a n = 3n + n +. a n = n + 6 3n. a n = 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 n ) by Theorem ) 5 by Theorem..4. n

25 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 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 = n. As /n 0, Theorem..5 yields that /n 0 and therefore + 6 n = 3 n d) As seen before, we have Thus we get by Theorem..4 and Theorem n n 0 and n + 5 3n = + 5 n 3 n 3. a n = 5 + ) 3 n + n n + = 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).

26 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 = 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

27 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.

28 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

29 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 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.

31 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 ) ε = ε ε

32 3 SOLUTION SET FOR THE HOMEWORK PROBLEMS Hence if we take 0 < δ ε/ ε ), 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.

1 if 1 x 0 1 if 0 x 1

1 if 1 x 0 1 if 0 x 1 Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

More information

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

More information

x if x 0, x if x < 0.

x if x 0, x if x < 0. Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

More information

INTRODUCTORY SET THEORY

INTRODUCTORY SET THEORY M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,

More information

Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set.

Discrete Mathematics: Solutions to Homework (12%) For each of the following sets, determine whether {2} is an element of that set. Discrete Mathematics: Solutions to Homework 2 1. (12%) For each of the following sets, determine whether {2} is an element of that set. (a) {x R x is an integer greater than 1} (b) {x R x is the square

More information

1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. 1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

More information

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5. PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

More information

Metric Spaces Joseph Muscat 2003 (Last revised May 2009)

Metric Spaces Joseph Muscat 2003 (Last revised May 2009) 1 Distance J Muscat 1 Metric Spaces Joseph Muscat 2003 (Last revised May 2009) (A revised and expanded version of these notes are now published by Springer.) 1 Distance A metric space can be thought of

More information

God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886)

God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work

More information

Lies My Calculator and Computer Told Me

Lies My Calculator and Computer Told Me Lies My Calculator and Computer Told Me 2 LIES MY CALCULATOR AND COMPUTER TOLD ME Lies My Calculator and Computer Told Me See Section.4 for a discussion of graphing calculators and computers with graphing

More information

Mathematics Review for MS Finance Students

Mathematics Review for MS Finance Students Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,

More information

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely

More information

PART I. THE REAL NUMBERS

PART I. THE REAL NUMBERS PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the representation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS

More information

We now explore a third method of proof: proof by contradiction.

We now explore a third method of proof: proof by contradiction. CHAPTER 6 Proof by Contradiction We now explore a third method of proof: proof by contradiction. This method is not limited to proving just conditional statements it can be used to prove any kind of statement

More information

MA651 Topology. Lecture 6. Separation Axioms.

MA651 Topology. Lecture 6. Separation Axioms. MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples

More information

TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

More information

Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)

Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x) SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

More information

Mathematical Methods of Engineering Analysis

Mathematical Methods of Engineering Analysis Mathematical Methods of Engineering Analysis Erhan Çinlar Robert J. Vanderbei February 2, 2000 Contents Sets and Functions 1 1 Sets................................... 1 Subsets.............................

More information

4. An isosceles triangle has two sides of length 10 and one of length 12. What is its area?

4. An isosceles triangle has two sides of length 10 and one of length 12. What is its area? 1 1 2 + 1 3 + 1 5 = 2 The sum of three numbers is 17 The first is 2 times the second The third is 5 more than the second What is the value of the largest of the three numbers? 3 A chemist has 100 cc of

More information

Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook. Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

More information

Worksheet on induction Calculus I Fall 2006 First, let us explain the use of for summation. The notation

Worksheet on induction Calculus I Fall 2006 First, let us explain the use of for summation. The notation Worksheet on induction MA113 Calculus I Fall 2006 First, let us explain the use of for summation. The notation f(k) means to evaluate the function f(k) at k = 1, 2,..., n and add up the results. In other

More information

Mathematical Procedures

Mathematical Procedures CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics,

More information

Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 5 9/17/2008 RANDOM VARIABLES

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 5 9/17/2008 RANDOM VARIABLES MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 5 9/17/2008 RANDOM VARIABLES Contents 1. Random variables and measurable functions 2. Cumulative distribution functions 3. Discrete

More information

Taylor and Maclaurin Series

Taylor and Maclaurin Series Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions

More information

HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)! Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

More information

CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12 CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

More information

Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

More information

Practice with Proofs

Practice with Proofs Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using

More information

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,

More information

Geometric Transformations

Geometric Transformations Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted

More information

CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 Solutions CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

More information

FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES

FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES FUNCTIONAL ANALYSIS LECTURE NOTES: QUOTIENT SPACES CHRISTOPHER HEIL 1. Cosets and the Quotient Space Any vector space is an abelian group under the operation of vector addition. So, if you are have studied

More information

ADVANCED CALCULUS. Rudi Weikard

ADVANCED CALCULUS. Rudi Weikard ADVANCED CALCULUS Lecture notes for MA 440/540 & 441/541 2015/16 Rudi Weikard log x 1 1 2 3 4 5 x 1 2 Based on lecture notes by G. Stolz and G. Weinstein Version of September 3, 2016 1 Contents First things

More information

3. Mathematical Induction

3. Mathematical Induction 3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

More information

Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture. Brad Groff

Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture. Brad Groff Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture Brad Groff Contents 1 Congruent Numbers... 1.1 Basic Facts............................... and Elliptic Curves.1

More information

1 Review of complex numbers

1 Review of complex numbers 1 Review of complex numbers 1.1 Complex numbers: algebra The set C of complex numbers is formed by adding a square root i of 1 to the set of real numbers: i = 1. Every complex number can be written uniquely

More information

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

More information

Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2

Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 2 Proofs Intuitively, the concept of proof should already be familiar We all like to assert things, and few of us

More information

SECTION 10-2 Mathematical Induction

SECTION 10-2 Mathematical Induction 73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms

More information

x a x 2 (1 + x 2 ) n.

x a x 2 (1 + x 2 ) n. Limits and continuity Suppose that we have a function f : R R. Let a R. We say that f(x) tends to the limit l as x tends to a; lim f(x) = l ; x a if, given any real number ɛ > 0, there exists a real number

More information

4. How many integers between 2004 and 4002 are perfect squares?

4. How many integers between 2004 and 4002 are perfect squares? 5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started

More information

LIES MY CALCULATOR AND COMPUTER TOLD ME

LIES MY CALCULATOR AND COMPUTER TOLD ME LIES MY CALCULATOR AND COMPUTER TOLD ME See Section Appendix.4 G for a discussion of graphing calculators and computers with graphing software. A wide variety of pocket-size calculating devices are currently

More information

5.3 Improper Integrals Involving Rational and Exponential Functions

5.3 Improper Integrals Involving Rational and Exponential Functions Section 5.3 Improper Integrals Involving Rational and Exponential Functions 99.. 3. 4. dθ +a cos θ =, < a

More information

2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

More information

BANACH AND HILBERT SPACE REVIEW

BANACH AND HILBERT SPACE REVIEW BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but

More information

Solutions to Homework 10

Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

More information

Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

More information

BX in ( u, v) basis in two ways. On the one hand, AN = u+

BX in ( u, v) basis in two ways. On the one hand, AN = u+ 1. Let f(x) = 1 x +1. Find f (6) () (the value of the sixth derivative of the function f(x) at zero). Answer: 7. We expand the given function into a Taylor series at the point x = : f(x) = 1 x + x 4 x

More information

Understanding Basic Calculus

Understanding Basic Calculus Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other

More information

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +...

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +... 6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence

More information

Math 3000 Section 003 Intro to Abstract Math Homework 2

Math 3000 Section 003 Intro to Abstract Math Homework 2 Math 3000 Section 003 Intro to Abstract Math Homework 2 Department of Mathematical and Statistical Sciences University of Colorado Denver, Spring 2012 Solutions (February 13, 2012) Please note that these

More information

Section 6-2 Mathematical Induction

Section 6-2 Mathematical Induction 6- Mathematical Induction 457 In calculus, it can be shown that e x k0 x k k! x x x3!! 3!... xn n! where the larger n is, the better the approximation. Problems 6 and 6 refer to this series. Note that

More information

Appendix F: Mathematical Induction

Appendix F: Mathematical Induction Appendix F: Mathematical Induction Introduction In this appendix, you will study a form of mathematical proof called mathematical induction. To see the logical need for mathematical induction, take another

More information

TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

More information

13 Infinite Sets. 13.1 Injections, Surjections, and Bijections. mcs-ftl 2010/9/8 0:40 page 379 #385

13 Infinite Sets. 13.1 Injections, Surjections, and Bijections. mcs-ftl 2010/9/8 0:40 page 379 #385 mcs-ftl 2010/9/8 0:40 page 379 #385 13 Infinite Sets So you might be wondering how much is there to say about an infinite set other than, well, it has an infinite number of elements. Of course, an infinite

More information

POWER SETS AND RELATIONS

POWER SETS AND RELATIONS POWER SETS AND RELATIONS L. MARIZZA A. BAILEY 1. The Power Set Now that we have defined sets as best we can, we can consider a sets of sets. If we were to assume nothing, except the existence of the empty

More information

Limits and Continuity

Limits and Continuity Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function

More information

1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]

1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ] 1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not

More information

MATH PROBLEMS, WITH SOLUTIONS

MATH PROBLEMS, WITH SOLUTIONS MATH PROBLEMS, WITH SOLUTIONS OVIDIU MUNTEANU These are free online notes that I wrote to assist students that wish to test their math skills with some problems that go beyond the usual curriculum. These

More information

Review for Calculus Rational Functions, Logarithms & Exponentials

Review for Calculus Rational Functions, Logarithms & Exponentials Definition and Domain of Rational Functions A rational function is defined as the quotient of two polynomial functions. F(x) = P(x) / Q(x) The domain of F is the set of all real numbers except those for

More information

APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

More information

k, then n = p2α 1 1 pα k

k, then n = p2α 1 1 pα k Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

More information

Answer Key for California State Standards: Algebra I

Answer Key for California State Standards: Algebra I Algebra I: Symbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences.

More information

Mathematics for Computer Science

Mathematics for Computer Science Mathematics for Computer Science Lecture 2: Functions and equinumerous sets Areces, Blackburn and Figueira TALARIS team INRIA Nancy Grand Est Contact: patrick.blackburn@loria.fr Course website: http://www.loria.fr/~blackbur/courses/math

More information

Notes on metric spaces

Notes on metric spaces Notes on metric spaces 1 Introduction The purpose of these notes is to quickly review some of the basic concepts from Real Analysis, Metric Spaces and some related results that will be used in this course.

More information

36 CHAPTER 1. LIMITS AND CONTINUITY. Figure 1.17: At which points is f not continuous?

36 CHAPTER 1. LIMITS AND CONTINUITY. Figure 1.17: At which points is f not continuous? 36 CHAPTER 1. LIMITS AND CONTINUITY 1.3 Continuity Before Calculus became clearly de ned, continuity meant that one could draw the graph of a function without having to lift the pen and pencil. While this

More information

MATH 2300 review problems for Exam 3 ANSWERS

MATH 2300 review problems for Exam 3 ANSWERS MATH 300 review problems for Exam 3 ANSWERS. Check whether the following series converge or diverge. In each case, justify your answer by either computing the sum or by by showing which convergence test

More information

Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 126 Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

More information

6.8 Taylor and Maclaurin s Series

6.8 Taylor and Maclaurin s Series 6.8. TAYLOR AND MACLAURIN S SERIES 357 6.8 Taylor and Maclaurin s Series 6.8.1 Introduction The previous section showed us how to find the series representation of some functions by using the series representation

More information

2.1 Functions. 2.1 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair

2.1 Functions. 2.1 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 2.1 J.A.Beachy 1 2.1 Functions from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 21. The Vertical Line Test from calculus says that a curve in the xy-plane

More information

Cartesian Products and Relations

Cartesian Products and Relations Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special

More information

The Rational Numbers

The Rational Numbers Math 3040: Spring 2011 The Rational Numbers Contents 1. The Set Q 1 2. Addition and multiplication of rational numbers 3 2.1. Definitions and properties. 3 2.2. Comments 4 2.3. Connections with Z. 6 2.4.

More information

An Interesting Way to Combine Numbers

An Interesting Way to Combine Numbers An Interesting Way to Combine Numbers Joshua Zucker and Tom Davis November 28, 2007 Abstract This exercise can be used for middle school students and older. The original problem seems almost impossibly

More information

SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties

SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 3 Fall 2008 III. Spaces with special properties III.1 : Compact spaces I Problems from Munkres, 26, pp. 170 172 3. Show that a finite union of compact subspaces

More information

THE BANACH CONTRACTION PRINCIPLE. Contents

THE BANACH CONTRACTION PRINCIPLE. Contents THE BANACH CONTRACTION PRINCIPLE ALEX PONIECKI Abstract. This paper will study contractions of metric spaces. To do this, we will mainly use tools from topology. We will give some examples of contractions,

More information

Inner Product Spaces

Inner Product Spaces Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and

More information

FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper. FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is

More information

WASSCE / WAEC ELECTIVE / FURTHER MATHEMATICS SYLLABUS

WASSCE / WAEC ELECTIVE / FURTHER MATHEMATICS SYLLABUS Visit this link to read the introductory text for this syllabus. 1. Circular Measure Lengths of Arcs of circles and Radians Perimeters of Sectors and Segments measure in radians 2. Trigonometry (i) Sine,

More information

Math212a1010 Lebesgue measure.

Math212a1010 Lebesgue measure. Math212a1010 Lebesgue measure. October 19, 2010 Today s lecture will be devoted to Lebesgue measure, a creation of Henri Lebesgue, in his thesis, one of the most famous theses in the history of mathematics.

More information

So let us begin our quest to find the holy grail of real analysis.

So let us begin our quest to find the holy grail of real analysis. 1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers

More information

Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012

Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012 Some series converge, some diverge. Series Convergence Tests Math 22 Calculus III D Joyce, Fall 202 Geometric series. We ve already looked at these. We know when a geometric series converges and what it

More information

MATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers.

MATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers. MATHEMATICAL INDUCTION MIGUEL A LERMA (Last updated: February 8, 003) Mathematical Induction This is a powerful method to prove properties of positive integers Principle of Mathematical Induction Let P

More information

P (A) = lim P (A) = N(A)/N,

P (A) = lim P (A) = N(A)/N, 1.1 Probability, Relative Frequency and Classical Definition. Probability is the study of random or non-deterministic experiments. Suppose an experiment can be repeated any number of times, so that we

More information

Cardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.

Cardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set. Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection

More information

GRE Prep: Precalculus

GRE Prep: Precalculus GRE Prep: Precalculus Franklin H.J. Kenter 1 Introduction These are the notes for the Precalculus section for the GRE Prep session held at UCSD in August 2011. These notes are in no way intended to teach

More information

Graph Theory Problems and Solutions

Graph Theory Problems and Solutions raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

More information

sin(x) < x sin(x) x < tan(x) sin(x) x cos(x) 1 < sin(x) sin(x) 1 < 1 cos(x) 1 cos(x) = 1 cos2 (x) 1 + cos(x) = sin2 (x) 1 < x 2

sin(x) < x sin(x) x < tan(x) sin(x) x cos(x) 1 < sin(x) sin(x) 1 < 1 cos(x) 1 cos(x) = 1 cos2 (x) 1 + cos(x) = sin2 (x) 1 < x 2 . Problem Show that using an ɛ δ proof. sin() lim = 0 Solution: One can see that the following inequalities are true for values close to zero, both positive and negative. This in turn implies that On the

More information

Foundations of Mathematics I Set Theory (only a draft)

Foundations of Mathematics I Set Theory (only a draft) Foundations of Mathematics I Set Theory (only a draft) Ali Nesin Mathematics Department Istanbul Bilgi University Kuştepe Şişli Istanbul Turkey anesin@bilgi.edu.tr February 12, 2004 2 Contents I Naive

More information

Introduction to Topology

Introduction to Topology Introduction to Topology Tomoo Matsumura November 30, 2010 Contents 1 Topological spaces 3 1.1 Basis of a Topology......................................... 3 1.2 Comparing Topologies.......................................

More information

E3: PROBABILITY AND STATISTICS lecture notes

E3: PROBABILITY AND STATISTICS lecture notes E3: PROBABILITY AND STATISTICS lecture notes 2 Contents 1 PROBABILITY THEORY 7 1.1 Experiments and random events............................ 7 1.2 Certain event. Impossible event............................

More information

3. INNER PRODUCT SPACES

3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

More information

MATHEMATICS (CLASSES XI XII)

MATHEMATICS (CLASSES XI XII) MATHEMATICS (CLASSES XI XII) General Guidelines (i) All concepts/identities must be illustrated by situational examples. (ii) The language of word problems must be clear, simple and unambiguous. (iii)

More information

Probability Generating Functions

Probability Generating Functions page 39 Chapter 3 Probability Generating Functions 3 Preamble: Generating Functions Generating functions are widely used in mathematics, and play an important role in probability theory Consider a sequence

More information

2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.

2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. 2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then

More information

6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives 6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

More information

(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9 Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

More information

Mathematical induction. Niloufar Shafiei

Mathematical induction. Niloufar Shafiei Mathematical induction Niloufar Shafiei Mathematical induction Mathematical induction is an extremely important proof technique. Mathematical induction can be used to prove results about complexity of

More information

Lecture 16 : Relations and Functions DRAFT

Lecture 16 : Relations and Functions DRAFT CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence

More information

3. Equivalence Relations. Discussion

3. Equivalence Relations. Discussion 3. EQUIVALENCE RELATIONS 33 3. Equivalence Relations 3.1. Definition of an Equivalence Relations. Definition 3.1.1. A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric,

More information