Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 8 Solutions Please write neatly, and in complete sentences when possible.
|
|
- Daniela Simon
- 6 years ago
- Views:
Transcription
1 Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 8 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.3.5, 4.3.7, 4.3.8, 4.3.9, 4.4., 4.4.2, 4.4.9, Problem G. Use the Intermediate Value Theorem to show that every polynomial of odd degree has a root in R. (Recall that a polynomial p(x) = a n x n a 0, with a n 0, has degree n.) Problem H. () Give an example of a continuous function on (, ) that is not bounded. (2) Give an example of a uniformly continuous function on (, ) that is not bounded. (3) Suppose that f is uniformly continuous on (, ). Prove that f is bounded. Problem I. Let t(x) be defined by if x = p Q in lowest terms. q q t(x) = 0 if x R \ Q. () Show that t is not continuous at x Q. (2) Show that t is continuous at x R \ Q. Solution Suppose g is continuous at x = c and g(c) 0. Let ɛ = g(c). By continuity, there is a δ > 0 so that x c < δ implies that g(x) g(c) < ɛ. For such x, by the triangle inequality, g(x) g(c) g(c) g(x), so that g(c) g(x) < ɛ. Rearranging, we find g(x) > g(c) ɛ = 0, so that g(x) 0 for any x c < δ. This implies that f/g is defined on V δ (c), an open interval containing c. Solution Let c be a limit point of K, so that there exists a sequence {x n } K \ {c} converging to c. By continuity of h, we have h(c) = lim n h(x n ). Since x n K, we have h(x n ) = 0 for each n, so that h(c) = 0, and c K. Since K contains its limit points, K is closed.
2 2 Solution 4.3.8(a). Suppose f : R R is a continuous function with f(x) = 0 for all x Q. Given c R, since Q is dense in R, there exists a sequence {r n } Q converging to c. By continuity, f(c) = lim n f(r n ) = 0. Thus f is identically 0 on R. Solution 4.3.8(b). If f and g are not both assumed to be continuous, then f and g may be distinct functions of R, even if f and g are equal on Q. For example, we may have f(x) = 0 for all x, and g equal to Dirichlet s function (called g as well on p. 00). On the other hand, if f and g are both continuous, then so is f g. The latter is a continuous function that vanishes on Q by assumption, so that it is identically zero by part (a). Solution 4.3.9(a). Given ɛ > 0 and x R, let δ = ɛ/c. Then, for any x y < δ, we have f(x) f(y) c x y < c δ = ɛ. (Note that we ve actually proved uniform continuity of f on R). Solution 4.3.9(b). Let C = y 2 y. We claim first that y n+ y n c n C. This is easily seen by induction: y n+ y n = f(y n ) f(y n ) c y n y n, and the latter is less than or equal to c c n C = c n+ C by the inductive hypothesis. Notice now that this allows us to estimate y n y for any n: n n y n y = (y k+ y k ) y k+ y k k= k= n n c k C = C c n. k= The last sum can be computed explicitly using the formula for a geometric series, and we obtain k= y n y C cn c C c.
3 (Note that we have used 0 < c < in the last inequality, since we needed that c n is positive, i.e. c n <.) Finally, we observe that f k (y l ) = y k+l, for any k, l N. m > n, we have y m y n = f m (y ) f n (y ) = f n (y m n ) f n (y ) c n y m n y c n C c. 3 Thus, for Since 0 < c <, we have lim n c n = 0 (see example on p. 56). Thus, for any ɛ > 0, there is an N so that for all n N we have c n < ɛ( c)/c. For any pair m, n N with m > n N, we have y m y n c n C c < ɛ( c) C so that {y n } is a Cauchy sequence, as desired. C c = ɛ, Solution 4.3.9(c). Consider evaluating the function f along the sequence {y n } obtained above. Since f is continuous on R, we have ( ) f(y) = f lim y n = lim f(y n ) = lim y n+ = y, n n n so that y is a fixed point of f. This fixed point is unique: Suppose that y R were another fixed point of f, and note that f(y) f(y ) c y y by the contraction property of f. Since y and y are both fixed points of f, we have y y = f(y) f(y ) c y y. If y y, then y y 0, and we may cancel it from both sides above and obtain c, contradicting the assumption that c <. Thus y is the unique fixed point of f. Solution 4.3.9(d). The argument in part (b) did not rely on the value of y, so that we may perform the above argument with y = x, for any x R. We conclude that the sequence of iterates {x, f(x), f(f(x)),...} is Cauchy, and hence converges, say to y. The continuity argument in part (c) applies again, so that f(y ) = y, and y is a fixed point of f. By the unicity demonstrated in part (c), we conclude that y = y, so that the sequence of iterates {x, f(x), f(f(x)),...} approaches y, for any choice of x R.
4 4 Solution 4.4.(a). We ve seen in class that g(x) = x is continuous at each c R. By the Algebraic Continuity Theorem, lim f(x) = lim g(x) g(x) g(x) = g(c) g(c) g(c) = x c x c c3, so that f is continuous at c R, for each c R. Solution 4.4.(b). Let x n = n + and y n n = n for each n N, so that x n y n = /n, which converges to 0 as n. We have ( f(x n ) f(y n ) = n + 3 n n) 3 = 3n + 3 n + 3, n 3 for all n. By Theorem 4.4.6, f cannot be uniformly continuous on R. Solution 4.4.(c). Let A be a bounded subset of R, so that there is an R > 0 with x R for all x A. Note that we have x 3 y 3 = x y x 2 + xy + y 2 x y ( x 2 + x y + y 2). For x, y A, we thus have x 3 y 3 x y (R 2 + R 2 + R 2 ) = 3R 2 x y. Given ɛ > 0, let δ = ɛ. Then, for any x, y A with x y δ, we 3R 2 have f(x) f(y) = x 3 y 3 3R 2 x y < 3R 2 ɛ 3R = ɛ, 2 so that f is uniformly continuous on A. Solution Note that we have x 2 y 2 = y 2 x 2 x 2 y 2 For x, y [, ), we have x + y x 2 y 2 = x y x + y x 2 y 2. = x + y x 2 y 2 = xy 2 + x 2 y 2. Thus, for any ɛ > 0, we let δ = ɛ/2. Then for x y < δ we have x x + y 2 y 2 = x y 2 x y < 2δ = ɛ, x 2 y 2 so that /x 2 is uniformly continuous on [, ).
5 On the other hand, consider x n = /n, while y n = /(n + ). Then we have x n y n = n n + = n 2 + n, which goes to 0 as n. Meanwhile, we have x 2 n yn 2 = n2 (n + ) 2 = 2n +. By Theorem 4.4.6, /x 2 is not uniformly continuous on (0, ]. Solution 4.4.9(a). Given ɛ > 0, let δ = ɛ/m. Then, for any x y < δ, we have f(x) f(y) M x y < M δ = ɛ, so that f is uniformly continuous on A. Solution 4.4.9(b). Not every uniformly continuous function is Lipschitz: Consider f(x) = x on [0, ]. As explained in example 4.3.8, f is continuous on [0, ]. By Theorem 4.4.8, f is uniformly continuous on [0, ]. On the other hand, f is not Lipschitz. This can be seen graphically, since the slope of the tangent line to the graph of f is equal to f (x) = /(2 x), which approaches as x 0. More precisely, suppose that f were Lipschitz, with constant M > 0. Namely, for any x, y [0, ] we would have M x y x y x y = ( x + y)( x y) =. x + y By choosing x and y distinct and very small, we will find that M for all x, y [0, ] is impossible. Namely, consider the points x+ y x n = /n 2 and y n = /(4n 2 ) in [0, ], for each n N. Then we have M + = 2n/3, n 2n for all n. Since 2n/3 as n, this is impossible, and we conclude that f is not Lipscitz on [0, ], though it is uniformly continuous. Solution There is no continuous function f on R with range f(r) = Q, since the continuous image of a connected set is connected. More precisely, suppose that f was such a function. Since R satisfies the interval property, it is connected by Theorem By Theorem we must have that f(r) = Q is connected. On the other hand, 5
6 6 Q doesn t satisfy the interval property: The real number c = 2 is in [0, 0], and 0, 0 Q, but c / Q. By Theorem again, Q is not connected, a contradiction. Solution G. Let p(x) = a n x n + a n x n a x + a 0, with n odd and a n 0. Consider the sequence {p(k)/k n }. Since p(x) x n = a n + a n x + a n 2 x a x n + a 0 x n, it is evident that the sequence {p(k)/k n } approaches a n as k goes to. This means that, for k large enough, p(k)/k n a n < a n, so that p(k) a n k n < a n k n and p(k) and a n k n have the same sign. Similarly, {p( k)/( k) n } also approaches a n, so that, again for k large enough, p( k) and a n ( k) n have the same sign. Since n is odd, we have a n ( k) n = a n k n, and thus (for k large enough) p(k) and p( k) have opposite signs. In particular, 0 is between the values of p at k and k. By the Intermediate Value Theorem, there is a value c ( k, k) so that p(c) = 0, as desired. Solution H(). Let f(x) =. Then f((n )/n) = n, so that f is x not bounded on (, ). Solution H(2). Let f(x) = x. This function is evidently uniformly continuous: For ɛ > 0, let δ = ɛ. Then x y < δ implies that f(x) f(y) = x y < ɛ. On the other hand, f is also not bounded: For any M > 0, f( M ) = M, which is not bounded in absolute value by M. Solution H(3). Choose ɛ = > 0, so that by the uniform continuity of f, there is a δ > 0 so that x y < δ implies that f(x) f(y) <, for any x, y (, ). For each integer n Z, let x n = δn/2, so that x n x n = δ/2 < δ. Choose x > 0, and let N = 2x/δ. (Note that a indicates the greatest integer less than a). Then x N = δn/2 x, and x N + δ/2 = x N+ = δ(n + )/2 > x, so that x [x N, x N + δ/2). Note that this implies that x x N < δ/2. We may now use the uniform continuity
7 of f to deduce f(0) f(x) = f(0) f(x ) + f(x ) f(x 2 ) f(x N ) f(x N ) + f(x N ) f(x) f(0) f(x ) + f(x ) f(x 2 ) +... < N +. + f(x N ) f(x N ) + f(x N ) f(x) Finally, since N < δx/2 < δ/2, we conclude that f(0) f(x) < δ 2 +. The triangle inequality now implies that f(x) δ f(0). Thus f(x) is bounded in absolute value by δ/2 + + f(0), for all x > 0. The case x < 0 is almost identical ( N should be replaced with the least integer greater than δx/2), and we conclude that f is bounded. 7 Solution I(). Since R \ Q is dense in R, for any c Q there is a sequence {x n } R \ Q converging to c. Since f(x n ) = 0 for all n N, the limit lim n f(x n ) = 0, but on the other hand f(c) 0. Thus lim f(x n ) f(c), and f is not continuous at c. Solution I(2). For a fixed ɛ > 0, consider the set F ɛ = {x : t(x) ɛ}. Evidently, F ɛ Q. Given x F ɛ written in lowest terms x = p/q, t(x) ɛ implies that q /ɛ, so that the elements of F ɛ all have denominators bounded above by /ɛ. We show first that, for any ɛ > 0 and any interval [a, b], the intersection F ɛ [a, b] is finite. Suppose x = p/q F ɛ [a, b], and note that the above argument implies that there are only finitely many possibilities for q (namely, integers less than /ɛ). Then a p q b so that aq p bq. For each q, there are finitely many integers p in the interval [aq, bq], so that there are only finitely many choices for p and q so that x = p/q F ɛ [a, b]. This implies that F ɛ has no limit points: If there were a limit point c of F ɛ, then the interval [c, c + ] must intersect infinitely many points of F ɛ, which we know is impossible by the above argument. We
8 conclude that F ɛ is closed. Finally, this implies that t is continuous at x 0 R \ Q: Since F ɛ is closed, R \ F ɛ is open, and x 0 R \ Q R \ F ɛ implies that there is some δ > 0 so that V δ (x 0 ) (R \ F ɛ ). For any x V δ (x 0 ), we have x / F ɛ, so that t(x) < ɛ. Since t(x) 0 for all x R, this means t(x) < ɛ, for any x such that x x 0 < δ, and t is continuous at any irrational point.
Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible.
Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 2.6.3, 2.7.4, 2.7.5, 2.7.2,
More informationMath 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible.
Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.2.1, 4.2.3, 4.2.6, 4.2.8,
More informationMath 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 2 Solutions Please write neatly, and in complete sentences when possible.
Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 2 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 1.4.2, 1.4.4, 1.4.9, 1.4.11,
More informationMath 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 9: OPTIONAL Please write neatly, and in complete sentences when possible.
Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 9: OPTIONAL Please write neatly, and in complete sentences when possible. Do the following problems from the book: 5.2., 5.2.2, 5.2.3, 5.2.5, 5.2.8,
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationSolutions Final Exam May. 14, 2014
Solutions Final Exam May. 14, 2014 1. (a) (10 points) State the formal definition of a Cauchy sequence of real numbers. A sequence, {a n } n N, of real numbers, is Cauchy if and only if for every ɛ > 0,
More informationDetermine whether the formula determines y as a function of x. If not, explain. Is there a way to look at a graph and determine if it's a function?
1.2 Functions and Their Properties Name: Objectives: Students will be able to represent functions numerically, algebraically, and graphically, determine the domain and range for functions, and analyze
More informationPrinciple of Mathematical Induction
Advanced Calculus I. Math 451, Fall 2016, Prof. Vershynin Principle of Mathematical Induction 1. Prove that 1 + 2 + + n = 1 n(n + 1) for all n N. 2 2. Prove that 1 2 + 2 2 + + n 2 = 1 n(n + 1)(2n + 1)
More informationChapter 2. Limits and Continuity. 2.1 Rates of change and Tangents to Curves. The average Rate of change of y = f(x) with respect to x over the
Chapter 2 Limits and Continuity 2.1 Rates of change and Tangents to Curves Definition 2.1.1 : interval [x 1, x 2 ] is The average Rate of change of y = f(x) with respect to x over the y x = f(x 2) f(x
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationMATH 409 Advanced Calculus I Lecture 12: Uniform continuity. Exponential functions.
MATH 409 Advanced Calculus I Lecture 12: Uniform continuity. Exponential functions. Uniform continuity Definition. A function f : E R defined on a set E R is called uniformly continuous on E if for every
More information1.2 Functions and Their Properties Name:
1.2 Functions and Their Properties Name: Objectives: Students will be able to represent functions numerically, algebraically, and graphically, determine the domain and range for functions, and analyze
More informationSolutions Final Exam May. 14, 2014
Solutions Final Exam May. 14, 2014 1. Determine whether the following statements are true or false. Justify your answer (i.e., prove the claim, derive a contradiction or give a counter-example). (a) (10
More informationProblem Set 5: Solutions Math 201A: Fall 2016
Problem Set 5: s Math 21A: Fall 216 Problem 1. Define f : [1, ) [1, ) by f(x) = x + 1/x. Show that f(x) f(y) < x y for all x, y [1, ) with x y, but f has no fixed point. Why doesn t this example contradict
More informationInfinite Continued Fractions
Infinite Continued Fractions 8-5-200 The value of an infinite continued fraction [a 0 ; a, a 2, ] is lim c k, where c k is the k-th convergent k If [a 0 ; a, a 2, ] is an infinite continued fraction with
More informationIn N we can do addition, but in order to do subtraction we need to extend N to the integers
Chapter 1 The Real Numbers 1.1. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {1, 2, 3, }. In N we can do addition, but in order to do subtraction we need
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More informationThus f is continuous at x 0. Matthew Straughn Math 402 Homework 6
Matthew Straughn Math 402 Homework 6 Homework 6 (p. 452) 14.3.3, 14.3.4, 14.3.5, 14.3.8 (p. 455) 14.4.3* (p. 458) 14.5.3 (p. 460) 14.6.1 (p. 472) 14.7.2* Lemma 1. If (f (n) ) converges uniformly to some
More informationMT804 Analysis Homework II
MT804 Analysis Homework II Eudoxus October 6, 2008 p. 135 4.5.1, 4.5.2 p. 136 4.5.3 part a only) p. 140 4.6.1 Exercise 4.5.1 Use the Intermediate Value Theorem to prove that every polynomial of with real
More informationLines, parabolas, distances and inequalities an enrichment class
Lines, parabolas, distances and inequalities an enrichment class Finbarr Holland 1. Lines in the plane A line is a particular kind of subset of the plane R 2 = R R, and can be described as the set of ordered
More informationMath 421, Homework #9 Solutions
Math 41, Homework #9 Solutions (1) (a) A set E R n is said to be path connected if for any pair of points x E and y E there exists a continuous function γ : [0, 1] R n satisfying γ(0) = x, γ(1) = y, and
More informationMATH 2400: PRACTICE PROBLEMS FOR EXAM 1
MATH 2400: PRACTICE PROBLEMS FOR EXAM 1 PETE L. CLARK 1) Find all real numbers x such that x 3 = x. Prove your answer! Solution: If x 3 = x, then 0 = x 3 x = x(x + 1)(x 1). Earlier we showed using the
More information(a) For an accumulation point a of S, the number l is the limit of f(x) as x approaches a, or lim x a f(x) = l, iff
Chapter 4: Functional Limits and Continuity Definition. Let S R and f : S R. (a) For an accumulation point a of S, the number l is the limit of f(x) as x approaches a, or lim x a f(x) = l, iff ε > 0, δ
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 4 Solutions
Math 0: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 30 Homework 4 Solutions Please write neatly, and show all work. Caution: An answer with no work is wrong! Problem A. Use Weierstrass (ɛ,δ)-definition
More informationHomework 1 Solutions
MATH 171 Spring 2016 Problem 1 Homework 1 Solutions (If you find any errors, please send an e-mail to farana at stanford dot edu) Presenting your arguments in steps, using only axioms of an ordered field,
More informationMath 140A - Fall Final Exam
Math 140A - Fall 2014 - Final Exam Problem 1. Let {a n } n 1 be an increasing sequence of real numbers. (i) If {a n } has a bounded subsequence, show that {a n } is itself bounded. (ii) If {a n } has a
More informationHOMEWORK ASSIGNMENT 5
HOMEWORK ASSIGNMENT 5 DUE 1 MARCH, 2016 1) Let f(x) = 1 if x is rational and f(x) = 0 if x is irrational. Show that f is not continuous at any real number. Solution Fix any x R. We will show that f is
More informationMath 46 Final Exam Review Packet
Math 46 Final Exam Review Packet Question 1. Perform the indicated operation. Simplify if possible. 7 x x 2 2x + 3 2 x Question 2. The sum of a number and its square is 72. Find the number. Question 3.
More informationMaths 212: Homework Solutions
Maths 212: Homework Solutions 1. The definition of A ensures that x π for all x A, so π is an upper bound of A. To show it is the least upper bound, suppose x < π and consider two cases. If x < 1, then
More informationIn N we can do addition, but in order to do subtraction we need to extend N to the integers
Chapter The Real Numbers.. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {, 2, 3, }. In N we can do addition, but in order to do subtraction we need to extend
More informationTHE REAL NUMBERS Chapter #4
FOUNDATIONS OF ANALYSIS FALL 2008 TRUE/FALSE QUESTIONS THE REAL NUMBERS Chapter #4 (1) Every element in a field has a multiplicative inverse. (2) In a field the additive inverse of 1 is 0. (3) In a field
More informationPart 2 Continuous functions and their properties
Part 2 Continuous functions and their properties 2.1 Definition Definition A function f is continuous at a R if, and only if, that is lim f (x) = f (a), x a ε > 0, δ > 0, x, x a < δ f (x) f (a) < ε. Notice
More informationMATH 409 Advanced Calculus I Lecture 10: Continuity. Properties of continuous functions.
MATH 409 Advanced Calculus I Lecture 10: Continuity. Properties of continuous functions. Continuity Definition. Given a set E R, a function f : E R, and a point c E, the function f is continuous at c if
More informationDefinitions & Theorems
Definitions & Theorems Math 147, Fall 2009 December 19, 2010 Contents 1 Logic 2 1.1 Sets.................................................. 2 1.2 The Peano axioms..........................................
More informationMATH 131A: REAL ANALYSIS (BIG IDEAS)
MATH 131A: REAL ANALYSIS (BIG IDEAS) Theorem 1 (The Triangle Inequality). For all x, y R we have x + y x + y. Proposition 2 (The Archimedean property). For each x R there exists an n N such that n > x.
More informationMathematics 242 Principles of Analysis Solutions for Problem Set 5 Due: March 15, 2013
Mathematics Principles of Analysis Solutions for Problem Set 5 Due: March 15, 013 A Section 1. For each of the following sequences, determine three different subsequences, each converging to a different
More informationLimits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes
Limits at Infinity If a function f has a domain that is unbounded, that is, one of the endpoints of its domain is ±, we can determine the long term behavior of the function using a it at infinity. Definition
More informationWe are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero
Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
2 Limits 2.1 The Tangent Problems The word tangent is derived from the Latin word tangens, which means touching. A tangent line to a curve is a line that touches the curve and a secant line is a line that
More informationMath 5210, Definitions and Theorems on Metric Spaces
Math 5210, Definitions and Theorems on Metric Spaces Let (X, d) be a metric space. We will use the following definitions (see Rudin, chap 2, particularly 2.18) 1. Let p X and r R, r > 0, The ball of radius
More informationTo get horizontal and slant asymptotes algebraically we need to know about end behaviour for rational functions.
Concepts: Horizontal Asymptotes, Vertical Asymptotes, Slant (Oblique) Asymptotes, Transforming Reciprocal Function, Sketching Rational Functions, Solving Inequalities using Sign Charts. Rational Function
More information6.1 Polynomial Functions
6.1 Polynomial Functions Definition. A polynomial function is any function p(x) of the form p(x) = p n x n + p n 1 x n 1 + + p 2 x 2 + p 1 x + p 0 where all of the exponents are non-negative integers and
More informationMATH 301 INTRO TO ANALYSIS FALL 2016
MATH 301 INTRO TO ANALYSIS FALL 016 Homework 04 Professional Problem Consider the recursive sequence defined by x 1 = 3 and +1 = 1 4 for n 1. (a) Prove that ( ) converges. (Hint: show that ( ) is decreasing
More information56 CHAPTER 3. POLYNOMIAL FUNCTIONS
56 CHAPTER 3. POLYNOMIAL FUNCTIONS Chapter 4 Rational functions and inequalities 4.1 Rational functions Textbook section 4.7 4.1.1 Basic rational functions and asymptotes As a first step towards understanding
More informationKing Fahd University of Petroleum and Minerals Prep-Year Math Program Math Term 161 Recitation (R1, R2)
Math 001 - Term 161 Recitation (R1, R) Question 1: How many rational and irrational numbers are possible between 0 and 1? (a) 1 (b) Finite (c) 0 (d) Infinite (e) Question : A will contain how many elements
More informationChapter 8. Exploring Polynomial Functions. Jennifer Huss
Chapter 8 Exploring Polynomial Functions Jennifer Huss 8-1 Polynomial Functions The degree of a polynomial is determined by the greatest exponent when there is only one variable (x) in the polynomial Polynomial
More informationHence, (f(x) f(x 0 )) 2 + (g(x) g(x 0 )) 2 < ɛ
Matthew Straughn Math 402 Homework 5 Homework 5 (p. 429) 13.3.5, 13.3.6 (p. 432) 13.4.1, 13.4.2, 13.4.7*, 13.4.9 (p. 448-449) 14.2.1, 14.2.2 Exercise 13.3.5. Let (X, d X ) be a metric space, and let f
More informationReal Analysis - Notes and After Notes Fall 2008
Real Analysis - Notes and After Notes Fall 2008 October 29, 2008 1 Introduction into proof August 20, 2008 First we will go through some simple proofs to learn how one writes a rigorous proof. Let start
More information7. Let X be a (general, abstract) metric space which is sequentially compact. Prove X must be complete.
Math 411 problems The following are some practice problems for Math 411. Many are meant to challenge rather that be solved right away. Some could be discussed in class, and some are similar to hard exam
More informationHomework 3 Solutions, Math 55
Homework 3 Solutions, Math 55 1.8.4. There are three cases: that a is minimal, that b is minimal, and that c is minimal. If a is minimal, then a b and a c, so a min{b, c}, so then Also a b, so min{a, b}
More informationSummer Jump-Start Program for Analysis, 2012 Song-Ying Li
Summer Jump-Start Program for Analysis, 01 Song-Ying Li 1 Lecture 6: Uniformly continuity and sequence of functions 1.1 Uniform Continuity Definition 1.1 Let (X, d 1 ) and (Y, d ) are metric spaces and
More informationUndergraduate 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 informationmeans is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S.
1 Notation For those unfamiliar, we have := means equal by definition, N := {0, 1,... } or {1, 2,... } depending on context. (i.e. N is the set or collection of counting numbers.) In addition, means for
More informationconverges as well if x < 1. 1 x n x n 1 1 = 2 a nx n
Solve the following 6 problems. 1. Prove that if series n=1 a nx n converges for all x such that x < 1, then the series n=1 a n xn 1 x converges as well if x < 1. n For x < 1, x n 0 as n, so there exists
More informationChapter 2: Functions, Limits and Continuity
Chapter 2: Functions, Limits and Continuity Functions Limits Continuity Chapter 2: Functions, Limits and Continuity 1 Functions Functions are the major tools for describing the real world in mathematical
More informationSequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.
Chapter 3 Sequences Both the main elements of calculus (differentiation and integration) require the notion of a limit. Sequences will play a central role when we work with limits. Definition 3.. A Sequence
More informationCHAPTER 8: EXPLORING R
CHAPTER 8: EXPLORING R LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN In the previous chapter we discussed the need for a complete ordered field. The field Q is not complete, so we constructed
More informationSolution. 1 Solution of Homework 7. Sangchul Lee. March 22, Problem 1.1
Solution Sangchul Lee March, 018 1 Solution of Homework 7 Problem 1.1 For a given k N, Consider two sequences (a n ) and (b n,k ) in R. Suppose that a n b n,k for all n,k N Show that limsup a n B k :=
More informationContinuity and One-Sided Limits. By Tuesday J. Johnson
Continuity and One-Sided Limits By Tuesday J. Johnson Suggested Review Topics Algebra skills reviews suggested: Evaluating functions Rationalizing numerators and/or denominators Trigonometric skills reviews
More informationFunctional Limits and Continuity
Chapter 4 Functional Limits and Continuity 4.1 Discussion: Examples of Dirichlet and Thomae Although it is common practice in calculus courses to discuss continuity before differentiation, historically
More informationMath 328 Course Notes
Math 328 Course Notes Ian Robertson March 3, 2006 3 Properties of C[0, 1]: Sup-norm and Completeness In this chapter we are going to examine the vector space of all continuous functions defined on the
More informationMath 209B Homework 2
Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact
More informationStructure of R. Chapter Algebraic and Order Properties of R
Chapter Structure of R We will re-assemble calculus by first making assumptions about the real numbers. All subsequent results will be rigorously derived from these assumptions. Most of the assumptions
More informationMath 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1
Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x,
More informationChapter 3: Inequalities, Lines and Circles, Introduction to Functions
QUIZ AND TEST INFORMATION: The material in this chapter is on Quiz 3 and Exam 2. You should complete at least one attempt of Quiz 3 before taking Exam 2. This material is also on the final exam and used
More informationSection Properties of Rational Expressions
88 Section. - Properties of Rational Expressions Recall that a rational number is any number that can be written as the ratio of two integers where the integer in the denominator cannot be. Rational Numbers:
More informationMATH98 Intermediate Algebra Practice Test Form A
MATH98 Intermediate Algebra Practice Test Form A MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the equation. 1) (y - 4) - (y + ) = 3y 1) A)
More informationMcGill University Math 354: Honors Analysis 3
Practice problems McGill University Math 354: Honors Analysis 3 not for credit Problem 1. Determine whether the family of F = {f n } functions f n (x) = x n is uniformly equicontinuous. 1st Solution: The
More informationSupplementary Notes for W. Rudin: Principles of Mathematical Analysis
Supplementary Notes for W. Rudin: Principles of Mathematical Analysis SIGURDUR HELGASON In 8.00B it is customary to cover Chapters 7 in Rudin s book. Experience shows that this requires careful planning
More informationMAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9
MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended
More informationMath 341 Summer 2016 Midterm Exam 2 Solutions. 1. Complete the definitions of the following words or phrases:
Math 34 Summer 06 Midterm Exam Solutions. Complete the definitions of the following words or phrases: (a) A sequence (a n ) is called a Cauchy sequence if and only if for every ɛ > 0, there exists and
More informationMath 115 Spring 11 Written Homework 10 Solutions
Math 5 Spring Written Homework 0 Solutions. For following its, state what indeterminate form the its are in and evaluate the its. (a) 3x 4x 4 x x 8 Solution: This is in indeterminate form 0. Algebraically,
More informationImmerse Metric Space Homework
Immerse Metric Space Homework (Exercises -2). In R n, define d(x, y) = x y +... + x n y n. Show that d is a metric that induces the usual topology. Sketch the basis elements when n = 2. Solution: Steps
More informationMATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS
MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS TOMASZ PRZEBINDA. Final project, due 0:00 am, /0/208 via e-mail.. State the Fundamental Theorem of Algebra. Recall that a subset K
More informationALGEBRAIC GEOMETRY HOMEWORK 3
ALGEBRAIC GEOMETRY HOMEWORK 3 (1) Consider the curve Y 2 = X 2 (X + 1). (a) Sketch the curve. (b) Determine the singular point P on C. (c) For all lines through P, determine the intersection multiplicity
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include
PUTNAM TRAINING POLYNOMIALS (Last updated: December 11, 2017) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationAdvanced Calculus I Chapter 2 & 3 Homework Solutions October 30, Prove that f has a limit at 2 and x + 2 find it. f(x) = 2x2 + 3x 2 x + 2
Advanced Calculus I Chapter 2 & 3 Homework Solutions October 30, 2009 2. Define f : ( 2, 0) R by f(x) = 2x2 + 3x 2. Prove that f has a limit at 2 and x + 2 find it. Note that when x 2 we have f(x) = 2x2
More informationd(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N
Problem 1. Let f : A R R have the property that for every x A, there exists ɛ > 0 such that f(t) > ɛ if t (x ɛ, x + ɛ) A. If the set A is compact, prove there exists c > 0 such that f(x) > c for all x
More informationPUTNAM TRAINING PROBLEMS
PUTNAM TRAINING PROBLEMS (Last updated: December 3, 2003) Remark This is a list of Math problems for the NU Putnam team to be discussed during the training sessions Miguel A Lerma 1 Bag of candies In a
More informationSolutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010.
Solutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010. (24) The p-adic numbers. Let p {2, 3, 5, 7, 11,... } be a prime number. (a) For x Q, define { 0 for x = 0, x p = p n for x = p n (a/b),
More informationCaculus 221. Possible questions for Exam II. March 19, 2002
Caculus 221 Possible questions for Exam II March 19, 2002 These notes cover the recent material in a style more like the lecture than the book. The proofs in the book are in section 1-11. At the end there
More informationMATH 2400 (Azoff) Fall Notes to Homework Assignments. Problems from Chapter 1 of Spivak (Assignment 1)
MATH 2400 (Azoff Fall 2010 Notes to Homework Assignments Problems from Chapter 1 of Spivak (Assignment 1 1(ii. The fact that y( x (yx is not an axiom. Spivak proves it near the bottom of Page 7. 2. The
More informationCorrelation: California State Curriculum Standards of Mathematics for Grade 6 SUCCESS IN MATH: BASIC ALGEBRA
Correlation: California State Curriculum Standards of Mathematics for Grade 6 To SUCCESS IN MATH: BASIC ALGEBRA 1 ALGEBRA AND FUNCTIONS 1.0 Students write verbal expressions and sentences as algebraic
More informationProblem List MATH 5143 Fall, 2013
Problem List MATH 5143 Fall, 2013 On any problem you may use the result of any previous problem (even if you were not able to do it) and any information given in class up to the moment the problem was
More informationMATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM
MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is
More informationMATH 117 LECTURE NOTES
MATH 117 LECTURE NOTES XIN ZHOU Abstract. This is the set of lecture notes for Math 117 during Fall quarter of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1]. Contents 1. The set
More informationMATH 140B - HW 5 SOLUTIONS
MATH 140B - HW 5 SOLUTIONS Problem 1 (WR Ch 7 #8). If I (x) = { 0 (x 0), 1 (x > 0), if {x n } is a sequence of distinct points of (a,b), and if c n converges, prove that the series f (x) = c n I (x x n
More informationEconomics 204 Fall 2011 Problem Set 2 Suggested Solutions
Economics 24 Fall 211 Problem Set 2 Suggested Solutions 1. Determine whether the following sets are open, closed, both or neither under the topology induced by the usual metric. (Hint: think about limit
More informationMAT 544 Problem Set 2 Solutions
MAT 544 Problem Set 2 Solutions Problems. Problem 1 A metric space is separable if it contains a dense subset which is finite or countably infinite. Prove that every totally bounded metric space X is separable.
More informationFunctions. Chapter Continuous Functions
Chapter 3 Functions 3.1 Continuous Functions A function f is determined by the domain of f: dom(f) R, the set on which f is defined, and the rule specifying the value f(x) of f at each x dom(f). If f is
More informationSolutions for Homework Assignment 2
Solutions for Homework Assignment 2 Problem 1. If a,b R, then a+b a + b. This fact is called the Triangle Inequality. By using the Triangle Inequality, prove that a b a b for all a,b R. Solution. To prove
More informationREVIEW FOR THIRD 3200 MIDTERM
REVIEW FOR THIRD 3200 MIDTERM PETE L. CLARK 1) Show that for all integers n 2 we have 1 3 +... + (n 1) 3 < 1 n < 1 3 +... + n 3. Solution: We go by induction on n. Base Case (n = 2): We have (2 1) 3 =
More informationMath 104: Homework 1 solutions
Math 10: Homework 1 solutions 1. The basis for induction, P 1, is true, since 1 3 = 1. Now consider the induction step, assuming P n is true and examining P n+1. By making use of the result (1 + +... +
More informationIowa State University. Instructor: Alex Roitershtein Summer Homework #5. Solutions
Math 50 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 205 Homework #5 Solutions. Let α and c be real numbers, c > 0, and f is defined
More informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationResearch Methods in Mathematics Homework 4 solutions
Research Methods in Mathematics Homework 4 solutions T. PERUTZ (1) Solution. (a) Since x 2 = 2, we have (p/q) 2 = 2, so p 2 = 2q 2. By definition, an integer is even if it is twice another integer. Since
More informationMATH 409 Advanced Calculus I Lecture 11: More on continuous functions.
MATH 409 Advanced Calculus I Lecture 11: More on continuous functions. Continuity Definition. Given a set E R, a function f : E R, and a point c E, the function f is continuous at c if for any ε > 0 there
More informationMAT115A-21 COMPLETE LECTURE NOTES
MAT115A-21 COMPLETE LECTURE NOTES NATHANIEL GALLUP 1. Introduction Number theory begins as the study of the natural numbers the integers N = {1, 2, 3,...}, Z = { 3, 2, 1, 0, 1, 2, 3,...}, and sometimes
More informationPreCalculus Notes. MAT 129 Chapter 5: Polynomial and Rational Functions. David J. Gisch. Department of Mathematics Des Moines Area Community College
PreCalculus Notes MAT 129 Chapter 5: Polynomial and Rational Functions David J. Gisch Department of Mathematics Des Moines Area Community College September 2, 2011 1 Chapter 5 Section 5.1: Polynomial Functions
More informationEconomics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011
Economics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011 Section 2.6 (cont.) Properties of Real Functions Here we first study properties of functions from R to R, making use of the additional structure
More information