Calculus-1 Lecture Notes PDF

Calculus-1 Lecture Notes M.C. October 17, 2024 Contents 1 Syllabus 3 2 Introduction 5 2.1 What is Calculus....................................... 5 2.2 A Brief History of Calculus................................ 7 2.3 Why Do Calculus?...................................... 7 2.3.1 What things could be considered the greatest achievements of the human mind? 7 2.3.2 Calculus in Today’s World.............................. 8 3 Preliminaries 12 3.1 What is a Set?........................................ 12 3.1.1 Sets of Numbers.................................... 12 3.1.1.1 Natural Numbers (N)........................... 13 3.1.1.2 Integers (Z)................................. 15 3.1.1.3 Rational Numbers (Q)........................... 15 3.1.1.4 Irrational Numbers (I or Q′ )........................ 15 3.1.1.5 Algebraic Numbers............................. 15 3.1.1.6 Transcendental Numbers.......................... 16 3.1.1.7 Real Numbers (R)............................. 16 3.1.1.8 Complex Numbers (C)........................... 16 3.1.2 Understanding the Necessity of Each Number Set................. 16 3.2 Types of Mathematical Proofs................................ 17 3.2.1 Direct Proof...................................... 18 3.2.1.1 Mathematical Induction.......................... 18 3.2.2 Indirect Proof..................................... 24 3.2.2.1 Proof by Contradiction (Çelişki Bulma)................. 24 3.2.2.2 Proof by Contrapositive (Olmayana Ergi)................ 25 3.3 Important Mathematical Symbols and Their Meanings.................. 26 3.3.1 What is if and only if?.............................. 27 4 Sequences 29 4.1 Recursive Sequences..................................... 31 4.2 Operations with sequences.................................. 32 4.3 Some Definitions and Theorems............................... 32 4.4 Convergence of Sequences (Limit).............................. 36 4.4.1 Geometric Interpretation of Limit.......................... 42 4.4.2 Some Definitions and Theorems for Limit of Sequences.............. 42 4.4.2.1 Limit Laws................................. 48 4.4.2.2 Proofs of Limit Laws (Not required for the exams)........... 50 1 4.5 Examples........................................... 53 5 Appendix 67 5.1 Absolute Value Laws..................................... 67 References 68 Index 69 2 1 Syllabus Introduction and General Informations – Week 1: Introduction and General Information on Analysis / Mathematical Induction Sequences – Week 2: Real Number Sequences / Increasing-Decreasing Sequences / Monotonic Sequences / Boundedness / Convergence / Algebraic Operations on Sequences – Week 3: Limits and Convergence of Sequences / Some Limit Theorems (Limits of the sum, difference, product, and quotient of sequences, concept of subsequence and its limit) – Week 4: Theorems on Bounded and Monotonic Sequences / Cauchy Sequence (general defi- nition) / Realization of Convergence and Limits of Sequences (Induction and Epsilon Tech- nique) Functions – Week 5: Real-Valued Functions / Classification of Functions / Definition Set and General Graphs of Functions (polynomial, rational, irrational, absolute value, sign, Heaviside, inte- ger value, exponential, logarithmic, trigonometric, inverse trigonometric, hyperbolic, inverse hyperbolic, composite) Limits – Week 6: Concept of Limit in Real-Valued Functions / Left-Right Limits, Relationship Be- tween Left and Right Limits / Technique for Limits / Limit Theorems, Squeeze Theorem / Infinite Limits, Limits at Infinity / Indeterminate Forms. Continuity – Week 6: Continuity in Functions / Concept of Continuity-Discontinuity / Continuity-Discontinuity at a Point / Left-Right Continuity / Relationship Between Limit and Continuity / Types of Discontinuity, Removable Discontinuity / Continuity on Closed-Ended Intervals / Uniform Continuity / Continuity Theorems (Intermediate Value Theorem) Derivative – Week 7: Derivatives and Rules of Differentiation / Definition of Derivative, Geometric and Physical Interpretation / Leibniz Notation / Tangent and Normal Lines and Slope / Deriva- tive of Functions Involving Sum, Difference, Product, and Quotient / Linearity of the Deriva- tive Operator (general expression) – Week 8: Midterm Exam 1 – Week 9: Derivative of Composite Functions and Chain Rule / Derivatives of Polynomial, Ra- tional, Irrational, Absolute Value, Sign, Heaviside, Step, Exponential, Logarithmic, Trigono- metric, Inverse Trigonometric, Hyperbolic, Inverse Hyperbolic Functions / Higher Order Derivatives / Derivatives of Parametric Functions (parametric derivative) / Derivatives of Implicit Functions / Derivative of Inverse Function 3 Applications of Derivative – Week 10: Mean Value Theorem / Generalized Mean Value Theorem / Rolle’s Theorem – Week 11: Applications of Derivatives and Curve Sketching / First Derivative, Relationship Between Increasing andDecreasing Functions / Critical, Singular, Endpoint / Extrema (Local Min., Local Max.) Points / Absolute Min., Absolute Max. Values / Second Derivative Test, Relationship Between Second Derivative and Concavity / Concept of Asymptote / Examination of Curve Changes and Sketching of Graphs – Week 12: Midterm Exam 2 Applications of Derivative – Week 13: Application of Derivative to Indeterminate Limit Forms (L’Hôpital’s Rule) / Differential / Geometric Interpretation of the Differential / Differential Geometry of Plane Curves (arc differential) / Arc Differential for Parametric and Polar Curves / Differential and Approximate Calculation / – Week 14: Linear Approximation Function / Maximum-Minimum Problems – Week 15: General Practice and Make-up Exams – Week 16: Final Exam Remark 1.1. This file can be updated. Therefore, check it from time to time and download the latest version. Remark 1.2. All notes in this folder may contain minor typographical and computational errors since it is still in the draft. It is highly recommended that each of you carefully review all materials at every stage of your studies. Ensuring accuracy and understanding of the content is your responsibility. Please approach these materials with a critical eye to verify correctness and clarity. 4 Introduction 2 Figure 1: Important Topics of Calculus 2.1 What is Calculus Calculus is the mathematical study of continuous change. Calculus is, in short, the science of change. In the same way that geometry is the study of shape, and algebra is the study of generalizations of arithmetic operations. Originally called infinitesimal calculus or ”the calculus of infinitesimals”, it has two major branches, differential calculus and integral calculus. The former concerns instantaneous rates of change, and the slopes of curves, while the latter concerns accumulation of quantities, and areas under or between curves. In addition to differential calculus and integral calculus, the term is also used for naming specific methods of computation or theories that imply some sort of computation. Examples of this usage include propositional calculus, Ricci calculus, calculus of variations, lambda calculus, sequent calculus, and process calculus. Furthermore, the term ”calculus” has variously been applied in ethics and philosophy, for such systems as Bentham’s felicific calculus, and the ethical calculus. 5 Topic Description Calculus Calculus is the branch of mathematics that deals with the study of rates of change (differential calculus) and the accumulation of quantities (integral calculus). It is primarily concerned with the techniques for calculating derivatives and integrals, as well as solving problems related to them. It is often taught as an introductory course in mathematics and focuses on practical methods for solving problems involving rates of change and areas under curves. Calculus is more concerned with concrete examples and computations. It is applied in physics, engineering, economics, and various other fields to model real-world phenomena. Analysis Analysis is a broader and more rigorous field that encompasses calculus but extends far beyond it. It deals with the rigorous study of the concepts underlying calculus, such as limits, continuity, differentiation, integration, and infinite series. The primary focus of analysis is on providing a solid theoretical foundation for calculus, using precise definitions and proofs. It includes topics like real analysis, complex analysis, functional analysis, and measure theory. Analysis explores the properties of mathematical functions and sequences in a much more general and abstract context, often dealing with infinite-dimensional spaces and more advanced mathematical structures. In essence, while calculus provides the tools and techniques for computation, analysis is concerned with the rigorous justification of these tools. Summary Calculus is a practical field focused on techniques and applications, whereas analysis is more theoretical and abstract, dealing with the foundational aspects of calculus and beyond [4, 5, 6]. 6 2.2 A Brief History of Calculus The discovery of calculus is often attributed to two men, Isaac Newton and Gottfried Leibniz, who independently developed its foundations. Although they both were instrumental in its creation, they thought of the fundamental concepts in very different ways. While Newton considered variables chang- ing with time, Leibniz thought of the variables x and y as ranging over sequences of infinitely close values. He introduced dx and dy as differences between successive values of these sequences. Leibniz knew that dy/dx gives the tangent, but he did not use it as a defining property. On the other hand, Newton used quantities x′ and y ′ , which were finite velocities, to compute the tangent. Of course, neither Leibniz nor Newton thought in terms of functions, but both always thought in terms of graphs. For Newton, calculus was geometrical, while Leibniz took it towards analysis. It is interesting to note that Leibniz was very conscious of the importance of good notation and put a lot of thought into the symbols he used. Newton, on the other hand, wrote more for himself than anyone else. Consequently, he tended to use whatever notation he thought of on that day. This turned out to be important in later developments. Leibniz’s notation was better suited to generalizing calculus to multiple variables, and in addition, it highlighted the operator aspect of the derivative and integral. As a result, much of the notation that is used in calculus today is due to Leibniz. The development of calculus can roughly be described along a timeline which goes through three periods: Anticipation, Development, and Rigorization. In the Anticipation stage, techniques were being used by mathematicians that involved infinite processes to find areas under curves or maximize certain quantities. In the Development stage, Newton and Leibniz created the foundations of calculus and brought all of these techniques together under the umbrella of the derivative and integral. However, their methods were not always logically sound, and it took mathematicians a long time during the Rigorization stage to justify them and put calculus on a sound mathematical foundation. In their development of calculus, both Newton and Leibniz used ”infinitesimals,” quantities that are infinitely small and yet nonzero. Of course, such infinitesimals do not really exist, but Newton and Leibniz found it convenient to use these quantities in their computations and their derivations of results. Although one could not argue with the success of calculus, this concept of infinitesimals bothered mathematicians. Lord Bishop Berkeley made serious criticisms of the calculus, referring to infinitesimals as ”the ghosts of departed quantities.” 287 BC 1600 1665 1675 1820 1900 Ancient Roots: Newton and Fundamental Formalization Archimedes’ Development of Modern Leibniz develop Theorem of by Cauchy and work on areas Infinitesimals Developments calculus Calculus Weierstrass and volumes 2.3 Why Do Calculus? 2.3.1 What things could be considered the greatest achievements of the human mind? Let us imagine the greatest achievements of the human mind. 7 Consider that all these things emerged because of technological advances. Those advances relied on CALCULUS! Calculus has made it possible to: Build giant bridges Travel to the moon Predict patterns of population change etc. 2.3.2 Calculus in Today’s World Nowadays, calculus is used in very diverse fields all over the world. 8 Quantum mechanics uses the Schrödinger equation, which is defined using concepts that come from calculus. Financial experts use the famous Black-Scholes equation to find the price for options in a stock portfolio. It also uses advanced concepts from calculus. Albert Einstein developed his general theory of relativity using math based on calculus. Table 1: Calculus in Today’s World 9 Economists use calculus to analyze markets and see the current trends. In medicine, calculus is used to figure out the proper dosing strategy for a new drug that has been developed. Scientists at NASA use calculus to help direct their rockets during spaceflight. Table 2: Calculus in Today’s World Let us consider more example as follows: Satellite Calculus is used in satellites to determine the trajectories and orbits. The differential equations help model gravitational forces, velocity, and changes in position as they orbit Earth. Rocket Launch The path of rockets during launch is modeled using differential equations. These equations account for changing velocity, fuel consumption, and gravitational forces as the rocket ascends through the atmosphere. Artificial Intelligence Robot In AI and machine learning, calculus is vital for optimizing models through gradient descent algorithms, which minimize error functions using derivatives. Electrical Circuit In electrical circuits, calculus is used to model the flow of current and voltage over time, particularly in inductors and capacitors where the current or voltage changes at varying rates. GPS Systems Calculus is essential in GPS technology to calculate the changing positions of satellites and to synchronize time data, providing accurate location tracking. Airplane The flight dynamics of an airplane—such as lift, drag, and thrust—are described by complex differential equations. Calculus helps optimize these factors for safe and efficient flight. Bridge The design of bridges involves calculus to calculate the forces acting on different parts of the structure. Engineers use integration to determine load distribution and structural stability. 10 Skyscraper Skyscraper design relies on calculus to ensure that the forces due to wind, gravity, and earthquakes are properly distributed throughout the building, ensuring safety and stability. Race Car In race car design, calculus is used to optimize the aerodynamics, fuel efficiency, and traction. Engineers use derivatives to model acceleration and braking forces. Computer The design of computer chips and processors involves calculus to analyze electrical currents, signal processing, and the optimization of logic gates within the circuit architecture. Astronaut Astronauts rely on calculus during space missions to calculate trajectories, orbits, and the forces experienced during launch and re-entry, ensuring precise navigation and safety. 11 Preliminaries 3 In this section, we will introduce fundamental concepts and definitions that will be necessary throughout the course, including basic mathematical sets, type of mathematical proofs etc. 3.1 What is a Set? A set is a collection of different things; these things are called elements or members of the set and are typically mathematical objects of any kind: numbers, symbols, points in space, lines, other geometrical shapes, variables, or even other sets. 3.1.1 Sets of Numbers Natural Numbers (N) Integers (Z) Rational Numbers (Q) Irrational Numbers (I or Q′ ) Real Numbers (R), that is, R = Q ∪ Q′ Complex Numbers (C) Figure 2: Number sets 12 Figure 3: Another figure 3.1.1.1 Natural Numbers (N) The natural numbers are the set of positive integers that start from 1 and go on indefinitely. N = {1, 2, 3, 4,... } Remark 3.1. Whether zero is considered a natural number differs across definitions. There is no general consensus on its inclusion or exclusion from the set. Discussion: Is zero a Natural number or not? The set of natural numbers is often defined to begin with 1, not including zero. This characteri- zation aligns with the Peano Axioms, one of the foundational structures for natural numbers, which commences with 1 as the first natural number and constructs subsequent members recur- sively (Peano, 1889 ). In the original formulation by Peano in 1889, the axioms began with 1 as the first natural number. However, contemporary versions often start with 0, especially in set theory and computer science, to simplify the definitions and theorems that follow. However, it is worth noting that some contexts do include zero in the set of natural numbers, denoting it as N = {0, 1, 2,... }. Consequently, whether zero is considered a natural number depends on the specific requirements of the field and the definitions employed. Nonetheless, in most traditional mathematical settings, zero is not considered a natural number. 13 Figure 4: original formulation of Peano’s Axioms Peano’s Axioms 1. 1 ∈ N, i.e., 1 is a natural number. To move on from 1, we need the successor function, S (2 = S(1), 3 = S(S(1)), etc.): 2. n ∈ N ⇒ S(n) ∈ N, i.e., if a number is a natural number, its successor is also a natural number. 3. ∀m, n ∈ N, m = n ⇐⇒ S(m) = S(n), i.e., S is an injection. 4. ∀n ∈ N, S(n) ̸= 1, i.e., there is no natural number whose successor is 1. And to tie it all up: 5. If K is a set such that: 1∈K ∀n ∈ N, if n ∈ K then S(n) ∈ K Then K contains all of the natural numbers. This is called the axiom of induction. 14 Remark 3.2. Other Peano Axioms are: ∀x, x = x, i.e., equality is reflexive. ∀x, y ∈ N, if x = y then y = x, i.e., equality is symmetric. ∀x, y, z ∈ N, if x = y and y = z then x = z, i.e., equality is transitive. ∀a, b ∈ N and a = b then b ∈ N, i.e., the naturals are closed under equality. 3.1.1.2 Integers (Z) Integers extend the concept of natural numbers to include their negatives and zero. The set of integers is represented as: Z = {... , −3, −2, −1, 0, 1, 2, 3,... } 3.1.1.3 Rational Numbers (Q) Rational numbers are numbers that can be expressed as the ratio of two integers, where the denominator is not zero. The set of rational numbers includes integers, fractions, and terminating or repeating decimals. na o Q= | a, b ∈ Z, b ̸= 0 b Example: The fraction 34 and the decimal 0.75 are rational numbers. 3.1.1.4 Irrational Numbers (I or Q′ ) Irrational numbers are numbers that cannot be expressed as the ratio of two integers. They have non-terminating, non-repeating decimal expansions. I=R\Q √ Example: The numbers 2 and π (pi) are irrational numbers because their decimal expansions go on forever without repeating. Remark 3.3. Irrational numbers fill the ”gaps” between rational numbers on the real number line. 3.1.1.5 Algebraic Numbers If r is a root of a nonzero polynomial equation an xn + an−1 xn−1 + · · · + a1 x + a0 = 0, where the ai s are integers (or equivalently, rational numbers) and r satisfies no similar equation of degree < n, then r is said to be an algebraic number of degree n. Remark 3.4. A number that is not algebraic is said to be transcendental. Example 3.1. Take this simple polynomial for example: P (x) = x2 + 2x − 15 The coefficients 1, 2, and -15 are rational numbers (integers are part of rational numbers). If this polynomial is set to zero and solved, the answers are called solutions or roots, and these are also called algebraic numbers: x2 + 2x − 15 = 0 (x + 5)(x − 3) = 0 x = {−5, 3} Therefore, simply put, the solution set of x is the set of algebraic numbers. 15 3.1.1.6 Transcendental Numbers Transcendental numbers are a subset of real numbers that are not the root of any non-zero polynomial equation with integer coefficients. In formal terms, a number x is transcendental if it is a real or complex number that is not an algebraic number. Examples: π, e In the above schema, transcendental numbers are a subset of the irrational numbers, which in turn are a subset of the real numbers. Thus, they can be denoted as follows: N⊆Z⊆Q⊆R⊆C Remark 3.5. Irrational Numbers I=Algebraic ∪Transcendental= √ R \ Q. It should be noted that not all irrational numbers are transcendental. For example, 2 is irrational but not transcendental, as it is a root of the polynomial equation x2 − 2 = 0 with integer coefficients. Therefore, the set of transcendental numbers is strictly contained within the set of irrational numbers. 3.1.1.7 Real Numbers (R) The real numbers include all the numbers that can be found on the number line. This set includes both rational and irrational numbers. R=Q∪I √ Example: The number line can represent both 32 (a rational number) and 2 (an irrational num- ber). Key Point: Real numbers can be either positive, negative, or zero, and they include all possible decimal expansions. 3.1.1.8 Complex Numbers (C) Complex numbers extend the real numbers by including the square root of negative one, denoted by i, where i2 = −1. A complex number has a real part and an imaginary part. √ C = {a + bi | a, b ∈ R, i = −1} Example: The number 3 + 4i is a complex number, where 3 is the real part and 4 is the imaginary part. Key Point: Complex numbers are essential in advanced mathematics, particularly in solving equa- tions that have no real solutions, such as x2 + 1 = 0. 3.1.2 Understanding the Necessity of Each Number Set In this section, we will explain the necessity of each number set. Natural Numbers (N): Initially, humans needed natural numbers for counting and basic cal- culations. They played a fundamental role in tasks such as object enumeration, trade, and daily activities. However, subtraction operations don’t always yield natural numbers. Integers (Z): The set of natural numbers does not include negative numbers. However, when subtraction yields a negative result (for example, 2−5 = −3), natural numbers become insufficient. Thus, the set of integers (Z) was introduced, which includes negative numbers, zero, and positive numbers (−3, −2, −1, 0, 1, 2, 3,...). Rational Numbers (Q): Integers alone are insufficient when division between two integers is considered. For instance, dividing 1 by 2 results in 0.5, which is not an integer. Hence, the set of rational numbers (Q) was defined. Rational numbers can be expressed as a ratio of two integers ( ab , where a and b are integers, and b ̸= 0). 16 Real Numbers √ (R): Some numbers cannot be expressed as a ratio of two integers. For example, numbers like 2 and π are irrational and cannot be written as a fraction. These numbers led to the creation of the real number set (R), which includes both rational and irrational numbers. Algebraic Numbers (AR ): Some real numbers are roots of polynomials with integer coefficients √ √ 1+ 5 (such as 2 or 2 ). These are known as real algebraic numbers, which explain the relationship between rational numbers and certain irrational numbers via algebraic equations. Transcendental Numbers: Numbers that cannot be defined by any algebraic equation, such as π and e, are called transcendental numbers. They form a subset of real numbers but cannot be derived from algebraic expressions. Complex Numbers: When we find the root of the equations such as x2 + 1 = 0 (that is, when taking the square root of negative numbers), real numbers become insufficient. To handle such cases, the set of complex numbers (C) was introduced. Complex numbers have both a real and an imaginary component, and this set includes all previous sets, offering more general solutions to equations. 3.2 Types of Mathematical Proofs In mathematics, when you make a claim, propose a statement, or suggest a theorem, it is crucial to prove its correctness. A statement/proposition can only be universally accepted when it is correctly proven. * To prove something is true, you must show that it holds for all cases, but to prove it false, finding just one counterexample is enough. Example 3.2. Disprove that (x + 1)2 = x2 + 1 for every real number x. To disprove this statement, we need to find a counterexample, that is, a real number x for which: (x + 1)2 ̸= x2 + 1. Let’s choose x = 1 and check both sides: (x + 1)2 = (1 + 1)2 = 22 = 4, x2 + 1 = 12 + 1 = 1 + 1 = 2. Clearly, 4 ̸= 2, so x = 1 is a counterexample that disproves the original statement. Definition 3.1 (Mathematical Proof). A mathematical proof is a verification for establishing the truth of a proposition by a chain of logical deductions from a set of axioms. Mathematical proofs are one of the fundamental elements of mathematics and enhance abstract thinking. In mathematics, proving statements/theorems is generally done in two ways: Direct Proof (Doğrudan İspat) – Induction (Tümevarım) Indirect Proof (Dolaylı İspat) – Contradiction (Çelişki Bulma) – Contrapositive (Olmayana Ergi) 17 3.2.1 Direct Proof In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions. In order to directly prove a conditional statement of the form ”If p, then q”, it suffices to consider the situations in which the statement p is true. Logical deduction is employed to reason from assumptions to conclusion. Example 3.3. Let’s prove the proposition: ”If an integer is odd, then its square is also odd,” using the direct proof method. Proof. Remember that An integer n is an even number if there is an integer k such that n = 2k. An integer n is an odd number if there is an integer k such that n = 2k + 1. If n is odd, then n = 2k + 1, where k ∈ Z. Then, we have: n2 = (2k + 1)2 = 4k 2 + 4k + 1 Thus, n2 = 2(2k 2 + 2k) + 1, which is of the form 2m + 1 where m = 2k 2 + 2k, and therefore odd. 3.2.1.1 Mathematical Induction Mathematical Induction is a technique of proving a statement, the- orem or formula which is thought to be true, for any natural number n. By generalizing this in form of a principle which we would use to prove any mathematical statement is ‘Principle of Mathematical Induction‘. In mathematics, we come across many statements that are generalized in the form of n. To check whether that statement is true for all natural numbers we use the concept of mathematical induction. Steps of Mathematical Induction Consider a statement P (n), where n is a natural number. Then to determine the validity of P (n) for every n ∈ N, use the following principle: Step 1: (Base Case) Check whether the given statement is true for n = 1, that is, check P (1) is true. Step 2: (Inductive Hypothesis) Assume that given statement P (n) is also true for n = k, where k is any positive integer. Step 3: Prove that the result is true for P (k + 1) for any positive integer k. If the all conditions above are satisfied, then it can be concluded that P (n) is true for all n natural numbers. n(n+1) Example 3.4. By mathematical induction, prove that 1 + 2 + 3 + · · · + n = 2. Proof. Step 1: (Base Case) Let n = 1. The sum of the first 1 natural number is 1. On the right-hand side of the formula, n(n+1) 2 = 1(1+1) 2 = 1. Thus, the formula holds true for n = 1. Step 2: (Inductive Hypothesis) Assume the formula is true for some arbitrary integer k, such that: k(k + 1) 1 + 2 + 3 + ··· + k = 2 This is our inductive hypothesis. Step 3: 18 We now aim to prove that the formula is also true for k + 1. Using our inductive hypothesis: k(k + 1) 1 + 2 + 3 + · · · + k + (k + 1) = + (k + 1) 2 Now, factor out k + 1: k = (k + 1) +1 2 k+2 = (k + 1) 2 (k + 1)(k + 2) = 2 This is precisely the formula with n replaced by k + 1. Conclusion: n(n+1) Thus, by the principle of mathematical induction, the formula 1 + 2 + 3 + · · · + n = 2 holds true for all natural numbers n. Remark 3.6. In university entrance exams, speed was key. But now, it’s not about how fast you solve, it’s about delivering mathematically elegant and complete solutions with no ambiguity. Example 3.5. Show that 1 + 3 + 5 + · · · + (2n − 1) = n2. Proof. Step 1: (Base Case) Result is true for n = 1 That is 1 = (1)2 (True) Step 2: Assume that the result is true for n = k 1 + 3 + 5 + · · · + (2k − 1) = k 2 Step 3: Check for n = k + 1 i.e. 1 + 3 + 5 + · · · + (2(k + 1) − 1) = (k + 1)2 We can write the above equation as, 1 + 3 + 5 + · · · + (2k − 1) + (2(k + 1) − 1) = (k + 1)2 Using step 2 result, we get k 2 + (2(k + 1) − 1) = (k + 1)2 k 2 + 2k + 2 − 1 = (k + 1)2 k 2 + 2k + 1 = (k + 1)2 (k + 1)2 = (k + 1)2 L.H.S. and R.H.S. are the same. So the result is true for n = k + 1. Conclusion: By mathematical induction, the statement is true. 2 Example 3.6. Prove that 13 + 23 + 33 + · · · + n3 = n(n+1) 2. 19 Proof. Step 1: (Base Case) Now with the help of the principle of induction in Maths, let us check the validity of the given statement P (n) for n = 1. 2 2 1(1 + 1) 2 P (1) = = = 12 = 1 2 2 This is true. Step 2: Now as the given statement is true for n = 1, we shall move forward and try proving this for n = k, i.e., Step 3: Let us now try to establish that P (k + 1) is also true. 2 3 3 3 3 3 k(k + 1) 1 + 2 + 3 + · · · + k + (k + 1) = + (k + 1)(k + 1)2 2 2 2 k = (k + 1) + (k + 1) 4 2 2 k + 4k + 4 = (k + 1) 4 (k + 2)2 2 = (k + 1) 4 2 (k + 1)(k + 2) = 2 Conclusion: This proves that the statement is true for n = k + 1. Example 3.7. Let r ̸= 1. We want to prove that: 1 − rn 1 + r + r2 + · · · + rn−1 = (n ∈ N) 1−r Proof. Step 1: (Base Case) For n = 1: 1 − r1 1−r 1= = = 1 (True) 1−r 1−r Step 2: (Inductive Step) Assume that the statement is true for some n = k, i.e., 1 − rk 1 + r + r2 + · · · + rk−1 = 1−r Step 3: We need to show that the statement is true for n = k + 1, i.e., 1 − rk+1 1 + r + r2 + · · · + rk−1 + rk = 1−r Starting with the left-hand side (LHS) for n = k + 1, we have: 1 + r + r2 + · · · + rk−1 + rk Using the inductive hypothesis for n = k: 1 − rk = + rk 1−r Now, factor rk out: 20 1 − rk rk (1 − r) = + 1−r 1−r Simplify the expression: 1 − rk + rk − rk+1 = 1−r 1 − rk+1 = 1−r Thus, the formula is valid for n = k + 1. Conclusion: Since both the base case and the inductive step have been proven, by the principle of mathematical induction, the formula: 1 − rn 1 + r + r2 + · · · + rn−1 = 1−r is true for all natural numbers n. Example 3.8. (1 + a)n ≥ 1 + n · a for a ≥ −1 (a real number) and n ∈ N. This inequality is known as the Bernoulli’s Inequality. Prove that the inequatility using mathematical induction. Proof. Step 1: Base Case First, we test the base case, n = 1. (1 + a)1 ≥ 1 + 1 · a This simplifies to: 1+a≥1+a Clearly, 1 + a = 1 + a, so the inequality holds true for n = 1. Step 2: Inductive Hypothesis Next, we assume that the inequality holds for some arbitrary positive integer n = k, i.e., (1 + a)k ≥ 1 + k · a We will refer to this assumption as our inductive hypothesis. Step 3: We now need to prove that the inequality holds for n = k + 1, i.e., we need to show that (1 + a)k+1 ≥ 1 + (k + 1) · a Starting with the left-hand side of the inequality for n = k + 1: (1 + a)k+1 = (1 + a)k · (1 + a) Using our inductive hypothesis (1 + a)k ≥ 1 + k · a, we can substitute into the equation: (1 + a)k+1 ≥ (1 + k · a) · (1 + a) Expanding the right-hand side: (1 + k · a) · (1 + a) = 1 + k · a + a + k · a2 21 = 1 + (k + 1) · a + k · a2 Since k · a2 is non-negative for a ≥ −1, we have: 1 + (k + 1) · a + k · a2 ≥ 1 + (k + 1) · a Thus, (1 + a)k+1 ≥ 1 + (k + 1) · a Conclusion: By the principle of mathematical induction, the inequality: (1 + a)n ≥ 1 + n · a is true for all n ∈ N and a ≥ −1. This completes the proof of the Bernoulli inequality. Example 3.9. Prove that the equation 1 · 1! + 2 · 2! + 3 · 3! +... + n · n! = (n + 1)! − 1 holds for all natural numbers n. Proof. We will use the principle of mathematical induction to prove this equation. Step 1: (Base Case) For n = 1: 1 · 1! = (1 + 1)! − 1 1 = 2! − 1 = 2 − 1 = 1 The base case holds true. Step 2: (Inductive Step) Assume the formula holds for some arbitrary positive integer k. That is, assume: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! = (k + 1)! − 1 Step 3: We need to show that: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! + (k + 1) · (k + 1)! = ((k + 1) + 1)! − 1 = (k + 2)! − 1 Starting from the inductive hypothesis: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! = (k + 1)! − 1 Adding (k + 1) · (k + 1)! to both sides: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! + (k + 1) · (k + 1)! = (k + 1)! − 1 + (k + 1) · (k + 1)! Factor out (k + 1)! on the right-hand side: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! + (k + 1) · (k + 1)! = (k + 1)! · (1 + (k + 1)) − 1 Simplify the expression: 1 · 1! + 2 · 2! + 3 · 3! +... + k · k! + (k + 1) · (k + 1)! = (k + 1)! · (k + 2) − 1 = (k + 2)! − 1 Thus, the equation holds for n = k + 1, completing the inductive step. Conclusion: By the principle of mathematical induction, the formula 1 · 1! + 2 · 2! + 3 · 3! +... + n · n! = (n + 1)! − 1 holds for all natural numbers n. 22 Example 3.10. Prove by mathematical induction that for n ≥ 5, the inequality 3n−1 < n! holds. Proof. Step 1: Base Case Let us check the inequality for n = 5: 35−1 = 34 = 81 and 5! = 5 × 4 × 3 × 2 × 1 = 120 Clearly, 81 < 120, so the base case holds for n = 5. Step 2: Inductive Step Assume that the statement holds for some n = k ≥ 5, i.e., 3k−1 < k!. Step 3: We need to show that the inequality holds for n = k + 1, i.e., we need to prove: 3(k+1)−1 = 3k < (k + 1)!. From the inductive hypothesis, we know that: 3k−1 < k!. Now, we need to check whether multiplying both sides by 3 preserves the inequality: 3k = 3 · 3k−1 < 3 · k!. Thus, we need to prove that 3 · k! < (k + 1)!, which simplifies to: 3 · k! < (k + 1) · k!. This inequality holds if and only if 3 < k + 1, which is true for all k ≥ 3. Since k ≥ 5 by assumption, this is certainly true. Therefore, the inequality 3k < (k + 1)! holds, completing the inductive step. Conclusion: By the principle of mathematical induction, the inequality 3n−1 < n! holds for all n ≥ 5. Example 3.11. Prove by mathematical induction that for n ≥ 4, the inequality 2n ≤ n! holds. Proof. We will use the method of mathematical induction to prove this statement. Step 1: Base Case Let us check the inequality for n = 4: 24 = 16 and 4! = 4 × 3 × 2 × 1 = 24 Clearly, 16 ≤ 24, so the base case holds for n = 4. Step 2: Inductive Step Assume that the statement holds for some n = k ≥ 4, i.e., 2k ≤ k!. Step 3: We need to show that the inequality holds for n = k + 1, i.e., we need to prove: 2k+1 ≤ (k + 1)!. From the inductive hypothesis, we know that: 2k ≤ k!. Now, we multiply both sides by 2: 2k+1 = 2 · 2k ≤ 2 · k!. Thus, we need to prove that 2 · k! ≤ (k + 1)!, which simplifies to: 2 · k! ≤ (k + 1) · k!. This inequality holds if and only if 2 ≤ k + 1, which is true for all k ≥ 1. Since k ≥ 4 by assumption, this is certainly true. Therefore, the inequality 2k+1 ≤ (k + 1)! holds, completing the inductive step. Conclusion: By the principle of mathematical induction, the inequality 2n ≤ n! holds for all n ≥ 4. 23 3.2.2 Indirect Proof In contrast, an indirect proof may begin with certain hypothetical scenarios and then proceed to eliminate the uncertainties in each of these scenarios until an inescapable conclusion is forced. For example, instead of showing directly p ⇒ q, one proves its contrapositive q ′ ⇒ p′ (one assumes q ′ and shows that it leads to p′ ). Since p ⇒ q and q ′ ⇒ p′ are equivalent by the principle of transposition (see law of excluded middle), p ⇒ q is indirectly proved. Indirect proof methods include proof by contradiction, including proof by infinite descent. Contradiction (Çelişki Bulma) Contrapositive (Olmayana Ergi) Remark 3.7 ( The difference between the Contrapositive method and the Contradiction method). The difference between the Contrapositive method and the contradiction method is subtle. Let’s examine how the two methods work when trying to prove ”If P, then Q”. Method of Contradiction: Assume P and Not Q and prove some sort of contradiction. (In the method of Contradiction, your goal is to prove a contradiction, but it is not always clear what the contradiction is going to be at the start.) Method of Contrapositive: Assume Not Q and prove Not P. (The method of Contrapositive has the advantage that your goal is clear: Prove Not P.) Remark 3.8. As we add more proof techniques, it is important to realize that you are not expected to know which technique to use when you start a proof. Proof-writing often takes some trial and error. First try a direct proof, if you get stuck, you may think about whether breaking your set into cases will help, or whether negating a statement will make it easier to use. It is also quite possible that different methods can be used to prove the same statement. 3.2.2.1 Proof by Contradiction (Çelişki Bulma) The steps of contrapositive proof method are: Assume the statement to be proved is false. Show that this assumption leads to a logical contradiction. Conclude that the original statement must be true. √ Example 3.12. Prove that 2 is an irrational number. (In other words, there does not exist a rational number r such that r2 = 2.) √ Solution. √ Let’s assume, for the sake of contradiction, that 2 is a rational number. a Then 2 can be written as an irreducible fraction b such that a and b are coprime integers (having no common factor) which additionally means that at least one of a or b must be odd. Given this assumption, we can square both sides of the equation to obtain: a2 2= b2 Multiplying both sides by b2 gives: a2 = 2b2 This equation implies that a2 is an even number because it is equal to 2b2 , which is clearly even. Since a2 is even, it follows that a must also be even (the square of an odd number is always odd). Therefore, we can express a as a = 2k for some integer k. Substituting a = 2k into the equation a2 = 2b2 , we get: (2k)2 = 2b2 24 4k 2 = 2b2 Dividing both sides by 2 gives: 2k 2 = b2 This equation shows that b2 is also even, which means that b must be even as well. At this point, we have established that both a and b are even, which contradicts our initial assump- tion that ab is in its simplest form (since both a and b cannot have a common factor other than 1 if they are both even). √ √ This contradiction implies that our initial assumption that 2 is rational-is false. Therefore, 2 cannot be a rational number and must be irrational. Example 3.13. Let’s prove the proposition, ”If 6 divides a, then 2 divides a,” using the contradiction method. Assume that P ⇒ Q holds: 6 | a and 2 | a. If a is odd, then 6 | a implies a = 6k = 2(3k) for some integer k. However, this contradicts the assumption that a is odd. 3.2.2.2 Proof by Contrapositive (Olmayana Ergi) We know that P ⇒ Q ≡ Q′ ⇒ P ′. where P ′ and Q′ denote the negations of P and Q, respectively. By using this equivalence, instead of proving the proposition P ⇒ Q, we can prove its contrapositive, which leads to the same result. This method is called proof by contradiction. The steps of contrapositive proof method: Write the statement to be proved in the form: ∀x ∈ D, if P (x) then Q(x). Write the contrapositive of the statement: ∀x ∈ D, if Q(x)′ then P (x)′. Prove the contrapositive directly: Assume Q(x)′ , prove P (x)′. Contrapositive Notation Reminder: The contrapositive of P ⇒ Q is Q′ ⇒ P ′. In some cases, it is also represented as: ¬Q ⇒ ¬P which reads as ”if not Q, then not P ”. Here ¬P and ¬Q denote the negations of P and Q, respectively. Example 3.14. Let’s prove by contradiction the statement: ”If an integer n is odd, then n + 2 is also odd.” Proof: Its contrapositive is, ”If n + 2 is an even integer, then n is even,” which we can prove. Example 3.15. Prove the following statement using contrapositive reasoning: If n2 is even, then n is even. 25 Proof. We will prove the contrapositive of the given statement. The contrapositive is: If n is odd, then n2 is odd. Step 1: Assume n is odd. This means that n can be written as n = 2k + 1 for some integer k. Step 2: Now, we compute n2 : n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1. Since n2 = 2m + 1 for some integer m, we see that n2 is odd. Step 3: Therefore, if n is odd, n2 is odd. Thus, the contrapositive is true, and hence the original statement is also true. Conclusion: By proving the contrapositive, we have shown that if n2 is even, then n is even. 3.3 Important Mathematical Symbols and Their Meanings Meaning Symbol for all ∀ There exists a (at least) ∃ such that ∋ implies ⇒ if and only if ⇐⇒ therefore ∴ derivative (first) (prime) f ′ (x) derivative (second) (double prime) f ′′ (x) ∂ partial derivative R∂x integral RR double integral indicating that the proof is complete □ or Table 3: Important Mathematical Symbols and Their Meanings Figure 5: Symbols 26 3.3.1 What is if and only if ? In logic and mathematics, if and only if (sometimes abbreviated as iff ) is a logical operator denoting a logical biconditional (often symbolized by ⇔ or ↔). It is often used to conjoin two statements which are logically equivalent. In general, given two statements P and Q, the statement ”P if and only if Q” is true precisely when both P and Q are true or both P and Q are false. In which case, P can be thought of as the logical substitute of Q (and vice versa). We know that this statement asserts that both of the following conditional statements are true: If P, then Q. If Q, then P. So to prove “P if and only if Q,” we must prove two conditional statements. Recall that Q ⇒ P is called the converse of P ⇒ Q. Thus we need to prove both P ⇒ Q and its converse. These are both conditional statements, so we may prove them with either direct, contrapositive, or contradiction proof. Here is an outline: Outline for If-and-Only-If Proof Proposition: P if and only if Q. Proof: 1. ⇒ Prove P ⇒ Q (sufficient condition: if P is true, then Q must also be true). 2. ⇐ Prove Q ⇒ P (necessary condition: Q is required for P to be true). An ”if and only if” statement is also called a necessary and sufficient condition. P Q P⇔Q 0 0 1 0 1 0 1 0 0 1 1 1 Table 4: Truth table for biconditional (if and only if) operator Let’s start with a very simple example. Example 3.16. Water turns into ice if and only if it is below the freezing point. Explanation: If the water is frozen and has turned into ice, it indicates that the ambient tempera- ture is below the freezing point. Water cannot freeze above the freezing point. If the water has turned into ice, it means the temperature must be below freezing. Example 3.17. The integer n is odd if and only if n2 is odd. Proof. (⇒:) First, we show that n being odd implies that n2 is odd. Suppose n is odd. Then, by definition of an odd number, n = 2a + 1 for some integer a. Thus, n2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 + 2a) + 1. This expresses n2 as twice an integer, plus 1, so n2 is odd. (⇐:) Conversely, we need to prove that n2 being odd implies that n is odd. We use contrapositive proof. Suppose n is not odd. Then n is even, so n = 2a for some integer a (by definition of an even number). Thus, n2 = (2a)2 = 2(2a2 ), 27 so n2 is even because it’s twice an integer. Thus n2 is not odd. We’ve now proved that if n is not odd, then n2 is not odd, and this is a contrapositive proof that if n2 is odd then n is odd. Figure 6: What is wrong with this proof? What is wrong with this proof ? A proof of 1 = 2 1. a = b (Premise) 2. a2 = a × b (Multiply both sides of (1) by a) 3. a2 − b2 = a × b − b2 (Subtract b2 from both sides of (2)) 4. (a − b)(a + b) = b(a − b) (Algebra on (3)) 5. a + b = b (Divide both sides by a − b) 6. 2b = b (Replace a by b in (5) because a = b) 7. 2 = 1 (Divide both sides of (6) by b) Solution: Step 5 is incorrect because a − b = 0 based on the premise a = b, and division by zero is undefined. 28 Sequences 4 In this chapter, we will study the concept of sequences, focusing on various types of sequences, operations we can perform on them, and theorems related to their convergence. We will also delve into important definitions, limits, and geometric interpretations. Finally, practical examples will be presented to solidify understanding. Before giving the definition of the sequence, let us remember the definition of the function. Definition 4.1 (Function). A function f from a set D to a set Y is a rule that assigns a unique (single) element f (x) ∈ Y to each element x ∈ D. The set D of all possible input values is called the domain of the function. The set of all values of f (x) as x varies throughout D is called the range of the function. The domain and range of a function can be any sets of objects. The range may not include every element in the set Y. Example 4.1. For example; let us take the function f (x) = 2x. Definition 4.2 (Sequence). A sequence of real numbers (or a sequence in R) is a function defined on the set N = {1, 2,...} of natural numbers whose range is contained in the set R of real numbers. In other words, a sequence in R assigns to each natural number n = 1, 2,... a uniquely determined real number. If X : N → R is a sequence, we will usually denote the value of X at n by the symbol xn , rather than using the function notation X(n). The values xn are also called the terms or the elements of the sequence. We will denote this sequence by the notations X, (xn ), (xn : n ∈ N). 29 Figure 7: Terms of sequence Result 4.1. Basically and shortly; a sequence of real numbers is a function defined on natural numbers whose range is real numbers. That is, a sequence of real numbers is a function xn : N → R. Result 4.2. Similarly; a sequence of complex numbers is a function xn : N → C. Example 4.2. Sequences can be described by writing rules that specify their terms, such as √ 1 n−1 an = n, bn = (−1)n+1 , cn = , dn = (−1)n+1 , n n or by listing terms: n√ √ √ √ o {an } = 1, 2, 3,... , n,... 1 1 1 n+1 1 {bn } = 1, − , , − ,... , (−1) ,... 2 3 4 n 1 2 3 n−1 {cn } = 0, , , ,... , ,... 2 3 4 n {dn } = 1, −1, 1, −1, 1, −1,... , (−1)n+1 ,.... We also sometimes write a sequence using its rule, as with √ ∞ {an } = n n=1. Sequences can be represented as points on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis an is its value. Table 5 shows two ways to represent sequences graphically. The first marks the first few points from a1 , a2 , a3 ,... , an ,... on the real axis. The second method shows the graph of the function defining the sequence. The function is defined only on integer inputs, and the graph consists of some points in the xy-plane located at (1, a1 ), (2, a2 ),... , (n, an ),.... 1 Sequence an = n an 1.0 0.8 0.6 0.4 a5 a4 a3 a2 a1 0.2 0 1 - 1 - 1 - 1 - 1 1 … 0 n 5 4 3 2 1 2 3 4 5 6 7 8 9 10 … Table 5: Two ways to represent sequences graphically. 30 Remark 4.1. Let us compare sequence and function graphically. an : N → R f : A = (0, 5] ⊆ R → B ⊆ R 1 1 Sequence an = Function f (n)= n n an f(n) 2.5 1.0 2.0 0.8 1.5 0.6 0.4 1.0 0.2 0.5 … 0 n 1 2 3 4 5 6 7 8 9 10 … 1 2 3 4 5 n 1 1 Table 6: Graphs of an = n as n → ∞ (left) and function f (n) = n (right) , respectively. Example 4.3. List the first 5 terms of the following sequence: ∞ 4n n2 − 7 n=0 Solution: All we need to do is plug in the first five values of n into the formula for the sequence terms. Doing that gives, 4(0) n=0: =0 (0)2 − 7 4(1) 4 4 2 n=1: 2 = = =− (1) − 7 1−7 −6 3 4(2) 8 8 8 n=2: 2 = = =− (2) − 7 4−7 −3 3 4(3) 12 12 n=3: 2 = = =6 (3) − 7 9−7 2 4(4) 16 16 n=4: 2 = = (4) − 7 16 − 7 9 So, the first five terms of the sequence are: 2 8 16 0, − , − , 6, ,... 3 3 9 Note that the answer is given inside braces to indicate that we are dealing with a sequence, and the “...” at the end indicates that there are more terms in the sequence beyond those we listed. 4.1 Recursive Sequences So far, we have calculated each an directly from the value of n. But sequences are often defined recursively by giving: 1. The value(s) of the initial term or terms, and 2. A rule called a recursion formula for calculating any later term from terms that precede it. 31 Example 4.4. (a) The statements a1 = 1 and an = an−1 + 1 for n > 1 define the sequence 1, 2, 3,... , n,... of natural numbers. With a1 = 1, we have a2 = a1 + 1 = 2, a3 = a2 + 1 = 3, and so on. (b) The statements a1 = 1 and an = n · an−1 for n > 1 define the sequence 1, 2, 6, 24,... , n!,... of factorials. With a1 = 1, we have a2 = 2 · a1 = 2, a3 = 3 · a2 = 6, a4 = 4 · a3 = 24, and so on. (c) The statements a1 = 1, a2 = 1, and an+1 = an + an−1 for n ≥ 2 define the sequence 1, 1, 2, 3, 5,... of Fibonacci numbers. With a1 = 1 and a2 = 1, we have a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, a5 = 3 + 2 = 5, and so on. 4.2 Operations with sequences In this section, we will discuss different operations that can be performed on sequences. This includes summing, multiplying, and scaling sequences. Real numbers allow us to define the operations of sum, subtraction, multiplication and division. These operations can be naturally extended to the set of sequences. The way of extending the operations is performed term-wise. Let a = {an }n∈N and b = {bn }n∈N be two sequences. We define the following sequences: Summation: a + b = {an + bn }n∈N Subtraction: a − b = {an − bn }n∈N Multiplication: a · b = {an · bn }n∈N Division: a an = b bn n∈N But in this case, it is necessary to demand bn ̸= 0 so that the division is definite. Example 4.5. Let us have two sequences a = {n}n∈N and b = n1 n∈N. Summation: 2 1 n +1 a+b= n+ = n n∈N n n∈N Subtraction: 2 1 n −1 a−b= n− = n n∈N n n∈N Multiplication: 1 a·b= n· = {1}n∈N n n∈N Division: a n = n2 = 1 n∈N b n n∈N 4.3 Some Definitions and Theorems We will introduce essential definitions related to sequences in this section. These definitions form the foundation for understanding advanced concepts in sequence analysis. Definition 4.3 (Increasing, Decrasing, etc.). A sequence {an } is said to be increasing if the relation an < an+1 (∀n ∈ N) holds between its consecutive terms. 32 A sequence {an } is said to be decreasing if the relation an > an+1 (∀n ∈ N) holds between its consecutive terms. A sequence {an } is said to be non-decreasing if the relation an ≤ an+1 (∀n ∈ N) holds between its consecutive terms. A sequence {an } is said to be non-increasing if the relation an ≥ an+1 (∀n ∈ N) holds between its consecutive terms. Example 4.6. Let us consider an = 1/n n Example 4.7. Let us consider an = n+1 Definition 4.4 (Monotonic Sequence). A sequence that is increasing, decreasing, non-decreasing, or non-increasing is termed a monotonic sequence. Remark 4.2. In various textbooks, the term “non-decreasing” may be substituted for “increasing,” and “non-increasing” may be substituted for “decreasing.” To emphasize the meaning, some sources use the operators < and > instead of ≤ and ≥, and employ the qualifiers “strictly increasing” or “strictly decreasing.” In this text, definition 4.3 will be considered. Example 4.8. (a) The sequence 1, 2, 3,... , n,... is nondecreasing. (b) The sequence 21 , 23 , 43 ,... , n+1 n ,... is nondecreasing. 1 1 1 1 (c) The sequence 1, 2 , 4 , 8 ,... , 2n ,... is nonincreasing. (d) The constant sequence 3, 3, 3,... , 3,... is both nondecreasing and nonincreasing. (e) The sequence 1, −1, 1, −1, 1, −1,... is not monotonic. 33 Example 4.9. Determine whether the sequence (n + 2)! an = 4n is monotonic increasing or monotonic decreasing. Solution: We will use two methods to analyze the monotonicity of the sequence. Method 1: For n ∈ N, if an+1 − an ≥ 0, the sequence is monotonic increasing, and if an+1 − an ≤ 0, the sequence is monotonic decreasing. We calculate: (n + 3)! (n + 2)! (n + 3)! 4 · (n + 2)! an+1 − an = n+1 − n = −. 4 4 4n+1 4n+1 This simplifies to (n + 2)! · (n − 1) an+1 − an = 4n+1 Here, for n ∈ N, n + 2 ≥ 3 then (n + 2)! ≥ 3! = 6. And for n ∈ N, n − 1 ≥ 0 and 4n+1 ≥ 16. So, we get: (n + 2)! · (n − 1) ≥0 4n+1 Consequently, (n + 2)! · (n − 1) an+1 − an = ≥0 4n+1 Therefore, the sequence an is nondecreasing. Method 2: We consider the ratio an+1 an instead, which can sometimes provide an easier way to analyze mono- tonicity. We compute: (n+3)! an+1 4n+1 (n + 3) · (n + 2)! 4n n+3 = (n+2)! = n+1 · =. an n 4 (n + 2)! 4 4 For all n ∈ N, this ratio is always greater than 1. Thus, for all n ∈ N, we get: an+1 > 1 that is an+1 > an , an which shows that the sequence an is monotonic increasing. Example 4.10. Determine whether the following sequence is monotone: 2 n +1 (an ) = n 2 Solution. Given the sequence (an ) = n n+1 , let’s check if it is increasing. (n + 1)2 + 1 n2 + 1 an+1 − an = − n+1 n Expanding and simplifying: (n2 + 2n + 1 + 1) n2 + 1 n2 + 2n + 2 n2 + 1 an+1 − an = − = − n+1 n n+1 n Next, simplify the fractions: n3 + 2n2 + 2n − (n3 + n2 + n2 + 1) an+1 − an = n(n + 1) 34 n2 + n − 1 = n(n + 1) 1 Notice that an+1 − an = 1 − n(n+1) > 0, which implies that the sequence is increasing. Thus, the sequence is increasing. Example 4.11. Determine whether the following sequence is monotone: (an ) = (n + (−1)n ) Solution. Given the sequence (an ) = (n + (−1)n ), let’s analyze its behavior. Calculate the difference an+1 − an : an+1 − an = n + 1 + (−1)n+1 − (n + (−1)n ) Simplify the expression: an+1 − an = 1 + 2(−1)n+1 Now, analyze the expression based on the parity of n: For n = 2k + 1, an+1 − an is positive. For n = 2k, an+1 − an is negative. Thus, the sequence (an ) is not monotone. Definition 4.5 (Boundedness). A sequence X = (xn ) of real numbers is said to be bounded if and only if there exists a real number M > 0 such that |xn | ≤ M for all n ∈ N. Let us rewrite the definition using mathematical notations: A sequence (xn ) is bounded if ∃M > 0 such that ∀n ∈ N, |xn | ≤ M. (−1)n Example 4.12. Consider the sequence an =. Is this sequence bounded? 3n + 2 Solution: If the sequence (an ) is bounded, there exists an M such that |an | ≤ M for all n ∈ N. First, find the absolute value of an : (−1)n 1 |an | = =. 3n + 2 3n + 2 1 We know that for all n ∈ N, 3n+2 ≤ 1, because as n increases, the denominator 3n+2 grows, making the fraction smaller. Thus, |an | ≤ 1 for all n ∈ N, meaning the sequence (an ) is bounded by 1. 2n+1 Example 4.13. Now, consider the sequence an = n+1. Is this sequence bounded? Solution. We can express the absolute value of an as: 2n + 1 1 |an | = = 2−. n+1 n+1 1 Since for all n ∈ N, we have 2 − n+1 ≤ 2, the sequence |an | ≤ 2 is bounded by 2. Definition 4.6 (Bounded from Above, Upper Bound Etc.). A sequence {an } is bounded from above if there exists a number M such that an ≤ M for all n. The number M is an upper bound for {an }. If M is an upper bound for {an } but no number less than M is an upper bound for {an }, then M is the least upper bound (lub(an ) or LU B(an ) or supremum) for {an }. 35 A sequence {an } is bounded from below if there exists a number m such that an ≥ m for all n. The number m is a lower bound for {an }. If m is a lower bound for {an } but no number greater than m is a lower bound for {an }, then m is the greatest lower bound (glb(an ) or GLB(an ) or infimum) for {an }. Definition 4.7 (Bounded Sequence (Other Definition)). If {an } is bounded from above and below, then {an } is bounded. If {an } is not bounded, then we say that {an } is an unbounded sequence. Example 4.14. Consider the following sequences. For each sequence, determine whether it is mono- tonic and find its greatest lower bound (GLB) and least upper bound (LUB). (a) The sequence 1, 2, 3,... , n,.... (b) The sequence 21 , 32 , 34 ,... , n+1 n ,.... (c) The constant sequence 3, 3, 3,... , 3,.... Solution: (a) The sequence 1, 2, 3,... , n,... This sequence is increasing, as each term is greater than the previous term. The sequence is unbounded above, meaning there is no least upper bound (LUB). The greatest lower bound (GLB) is the first term: inf{an : n ∈ N} = 1, n (b) The sequence n+1 : This sequence is increasing, as each term satisfies: n n+1 <. n+1 n+2 The limit of the sequence as n → ∞ is. Thus, the LUB is 1. The first term is 21 , so the GLB is: 1 inf{an : n ∈ N} = , sup{an : n ∈ N} = 1. 2 (c) The constant sequence 3, 3, 3,...: This sequence is both nondecreasing and nonincreasing (constant sequence). The GLB and LUB are both equal to 3: inf{an : n ∈ N} = 3, sup{an : n ∈ N} = 3. 4.4 Convergence of Sequences (Limit) Sometimes the numbers in a sequence approach a single value as the index n increases. For example; 1 1 1 1 1, , , ,... , ,... , 2 3 4 n whose terms approach 0 as n gets large, and in the sequence 1 2 3 4 1 0, , , , ,... , 1 − ,... , 2 3 4 5 n whose terms approach 1. 36 On the other hand, sequences like n√ √ √ √ o 1, 2, 3,... , n,... have terms that get larger than any number as n increases, and sequences like 1, −1, 1, −1, 1, −1,... , (−1)n+1 ,... change between 1 and −1 again and again, never converging to a single value. Definition 4.8 (Epsilon). In mathematics, epsilon (ε) represents a small positive infinitesimal quan- tity. (Or in other words; epsilon is a positive real number that can be chosen as small as desired.) Definition 4.9 (Distance between two real numbers). Given two real numbers x and y, we define their distance to be |x − y|. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we specify any number ε > 0, then by going far enough out in the sequence, taking the index n to be larger than some value N , the difference between an and the limit of the sequence becomes less than ε. Definition 4.10 (Epsilon Definition of a Convergent Sequence). A sequence X = (xn ) in R is said to converge to x ∈ R, or x is said to be a limit of (xn ), if for every ε > 0 there exists a natural number K(ε) such that for all n ≥ K(ε), the terms xn satisfy |xn − x| < ε. If the sequence (xn ) converges to x ∈ R, it can be represented by lim xn = x or lim xn = x n→∞ or xn → x (n → ∞) Remark 4.3. limn→∞ xn = x is read as: “The limit of xn as n approaches infinity equals x” or “The limit as n approaches infinity of xn equals x.” Remark 4.4. Shortly, in sequences, the limit represents the fixed real value that the terms of the sequence approach as n tends to infinity. Remark 4.5. If a sequence X = (xn ) is not convergent, we say it is a divergent sequence and we say the limit does not exist (DNE). Definition 4.11 (The Short Definition of Limit:). lim xn = x ⇔ ∀ε > 0, ∃K(ε) ∈ N ∋ ∀n ≥ K(ε) ⇒ |xn − x| < ε n→∞ Remark 4.6. Another way to express convergence is as follows: If “almost all terms” of the sequence are within an ε > 0 neighborhood of x, then the sequence (xn ) is said to converge to x ∈ R. 37 Definition 4.12 (What is epsilon neighborhood of x?). For y ∈ R and ε > 0, recall that the ε- neighborhood of x is the set Vε (x) = {y ∈ R : |y − x| < ε} = {y ∈ R : x − ε < y < x + ε} y y x−ε x x+ε Here, |y − x| shows the distance between the points y and x. Theorem 4.1. Let X = (xn ) be a sequence of real numbers, and let x ∈ R. The following statements are equivalent. 1. X converges to x. 2. For every ε > 0, there exists a natural number K such that for all n ≥ K, the terms xn satisfy |xn − x| ≤ ε. 3. For every ε > 0, there exists a natural number K such that for all n ≥ K, the terms xn satisfy x − ε < xn < x + ε. 4. For every ε-neighborhood Vε (x) of x, there exists a natural number K such that for all n ≥ K, the terms xn belong to Vε (x). Definition 4.13 (Archimedean property). Archimedean property states that given two positive real numbers x and y, there is an integer n such that nx > y. It also means that the set of natural numbers is not bounded above. Symbolically; ∀x, y ∈ R+ ∃n ∈ N ∋ n · x > y. 1 Example 4.15. i) Show that limn→∞ n = 0 using the definition of the limit. 1 ii) Determine how many terms of the sequence are outside ε = 10 neighborhood of the limit L = 0. Solution. i) Method 1: First, let us remember the definition of the limit: ∀ε > 0, ∃K(ε) ∋ ∀n ≥ K(ε) ⇒ |xn − x| < ε The sequence n1 converges to 0, if for every ε > 0, there exists a natural number K = K(ε) such that for all n ≥ K, we have 1 1 − 0 = < ε. n n Step 1: Let ε > 0 be given. According to the definition of the limit, we need to find a natural number K such that for all n ≥ K, the inequality n1 < ε holds. Step 2: To determine K, recall the Archimedean Property, which states that for any two positive real numbers x and y, there exists a natural number n such that nx > y. This implies that the set of natural numbers is not bounded above. Step 3: Let’s choose x = ε and y = 1. By the Archimedean Property, there exists a natural number K such that 1 Kε > 1 or equivalently < ε. K 38 1 1 Step 4: Hence, for all n ≥ K, we have n ≤ K. Thus, we get: 1 1 ≤ < ε. n K Therefore, by the definition of the limit, the sequence n1 converges to 0. Conclusion: We have shown that for every ε > 0, there exists a natural number K such that for all n ≥ K, n1 < ε. This completes the proof that limn→∞ n1 = 0. Method 2: Before starting to the method, let us consider some definitions below: Definition 4.14 (Ceiling Function (Tavan Fonksiyonu)). The ceiling function, denoted by ⌈x⌉, is defined as the smallest integer greater than or equal to x. That is: ⌈x⌉ = min{n ∈ Z : n ≥ x}. For example, ⌈2.3⌉ = 3 and ⌈−1.5⌉ = −1. Definition 4.15 (Floor Function (Taban Fonksiyonu (Tam Değer))). The floor function, denoted by ⌊x⌋, is defined as the largest integer less than or equal to x. That is: ⌊x⌋ = max{n ∈ Z : n ≤ x}. For example, ⌊2.7⌋ = 2 and ⌊−1.5⌋ = −2. The sequence n1 converges to 0, if for every ε > 0, there exists a natural number K = K(ε) such that for all n ≥ K, we have 1 1 −0 =. n n We start with the same inequality: 1 n> , ε or 1 n≥ + 1, ε and define K as: 1 K(ε) := + 1. ε As you see, K depends on ε like ε definition. Here, 1ε gives the largest integer less than or equal to 1 ε , and adding 1 ensures that K is strictly greater than 1ε. For example; If ε = 0.1, then: 1 = 10, ε so ⌊10⌋ = 10, and we have: K = 10 + 1 = 11. Thus, for all n ≥ 11, the inequality n1 < 0.1 holds. For n = 9, let us check whether the inequality holds or not; 1 1 1 |an − L| = −0 = − 0 = ≈ 0.111111 > ε = 0.1. n 9 9 We obtain |an − L| > ε. So, the inequality does not hold. For n = 10, let us check whether the inequality holds or not; 1 1 1 |an − L| = −0 = −0 = = 0.1 = ε = 0.1. n 10 10 We obtain |an − L| = ε. So, the inequality does not hold. 39 For n = 11, let us check whether the inequality holds or not; 1 1 1 −0 = |an − L| = −0 = ≈ 0.0909091 < ε = 0.1. n 11 11 We obtain |an − L| < ε. So, the inequality holds. Now, let us show it graphically. 1 1+ n 2 n4 1 Sequence an Sin[n] an n n 4+ n 4 n ϵ 0.1 1.0 Max n Value |a1 - L| > ϵ Show an |a2 - L| > ϵ 0.8 |a3 - L| > ϵ Show L (Limit) |a4 - L| > ϵ |a5 - L| > ϵ Show L + ϵ 0.6 an |a6 - L| > ϵ |a1 - L| Show L - ϵ Limit Value (L) = 0. |a7 - L| > ϵ |a8 - L| > ϵ L+ϵ |a9 - L| > ϵ 0.4 L-ϵ |a10 - L| > ϵ |a11 - L| < ϵ |a2 - L| |a12 - L| < ϵ |a3 - L| 0.2 |a13 - L| < ϵ

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