Applied Math II Module 1 PDF

Applied Mathematics II Module Introduction: This module consists of four units. The first unit deals with real sequence and infinite series. In this unit we will look briefly at many terms and concepts related to the real sequences and infinite series. The second unit deals with power series, which...

Applied Mathematics II Module Introduction: This module consists of four units. The first unit deals with real sequence and infinite series. In this unit we will look briefly at many terms and concepts related to the real sequences and infinite series. The second unit deals with power series, which is one of the most useful types of infinite series, and their applications. In particular we will also discuss the two special types of power series named Taylor series and Maclaurin series. The third unit discusses on calculus of functions of several variables, specifically focuses on the limit, continuity and partial derivatives of functions of several variables and their applications. The fourth unit deals with multiple integrals particularly, double integrals and triple integrals of functions of two and three variables respectively together with their applications. By doing so students will be able to express terms and concepts related to infinite series, power series, partial derivatives of functions of several variables and multiple integrals. SOME FEATURES OF THE MODULE Visualization: This module makes extensive use of modern computer graphics to clarify concepts and to develop the student’s ability to visualize mathematical objects, particularly those in 3 dimensional space. Quick Check Exercises: Each exercise set begins with approximately five practice exercises that are designed to provide students with an immediate assessment of whether they have mastered key ideas from the section. Applicability of Calculus: One of the good feature, primary goals of this module is to link calculus to the real world and the student’s own experience. This theme is carried through in the examples and exercises. Career Preparation: This module is written at a mathematical level that will prepare students for a wide variety of careers that require a sound mathematics background, including engineering, the various sciences, and business. Historical Notes: Some biographies and historical notes have been included in the module, with the goal of capturing and bringing to life for the student the personalities of history’s greatest mathematicians. Cooperative learning: One of the primary goals of this module is also to promote cooperative learning, so that students share knowledge and skills through a lot of group discussions and group activities given at each of new ideas introduced. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 1  Applied Mathematics II Module Objectives: At the end of this module students will be able to:  Define sequences, types of sequences, infinite series and power series.  Identify basic properties of sequence whether it converges or divergence.  Determine whether or not a given sequence is bounded and monotone.  Identify the relation between sequence and series.  Identify different types of tests for convergence of series and choose appropriate test of convergence  Determine differentiation and integration of a Power Series.  Determine the Taylor’s series representation of a function.  Apply the concept of sequence, real series and Taylor’s formula in solving physical and real life problems.  Determine domain and range of functions of two or three variables  Determine limit and continuity of functions of two or three variables  Determine differentiability of functions of two or three variables  Determine directional derivative of functions of two or three variables  Determine gradient of functions of two or three variables  Determine maximum and minimum (extreme) values of functions of two or three variables on a given region  Apply the concept of differentiability of functions of two or three variables in solving real life problems  Define double and Triple integrals in different coordinates  Determine double and multiple integrals of functions of several variables  Apply multiple integrals in determining volume of a solid region, area of plane region, surface area and so on  Find the mass of a planar lamina using a double integral  Find the center of mass of a planar lamina using double integrals  Find moments of inertia using double integrals Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 2  Applied Mathematics II CHAPTER ONE INFINITE SEQUENCES AND SERIES Unit Introduction In this chapter we will be concerned with infinite sequences and series. This unit is divided into four sections. The first section presents definitions and notations of sequence, convergence and divergence properties of Sequences and the basic properties of sequence, in particular boundedness and monotoness will also be treated in this section. The Second section presents partial sum of a sequence, definition and notation of a series and The third section deals with Different types of tests for convergence, in particular Integral Test, Comparison Test, Root Test, and Ratio Tests. Alternating Series; Absolute and Conditional Convergences will be treated in the fourth section. Unit Objectives: At the end of the unit students will be able to:  Define different types of sequences.  Identify basic properties of sequence.  Determine whether a given sequence converges or not, bounded or not and monotone or not.  Demonstrate how to differentiate increasing and decreasing sequences together with solving exercises.  Apply the concept of sequence in solving real life problems.  Identify the relation between sequence and series.  Define the term series.  Identify the two types of convergence of series.  Choose appropriate test for convergence of infinite series.  Demonstrate the application of different tests together with solving exercises.  Apply the concept of series in solving real life problems Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 3  Applied Mathematics II 1.1 Real Sequences Infinite sequences and series were introduced briefly in A Preview of Calculus in connection with Zeno’s paradoxes and the decimal representation of numbers. Their importance in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. The main objective of this unit is to study about infinite series. To do so it is important to be familiar with the basic concepts of sequences and convergence of sequences primarily. 1.1.1 Notations and Terminology We begin this section with two questions to remind readers their previous study about sequences and motivate (brainstorm) readers for their studies about sequence from the section: 1. Define: i) Arithmetic Sequence ii) Geometric Sequence 2. Give two examples of each sequence. In everyday language, the term “sequence” means a succession of things in a definite order, chronological order, size order, or logical order. In mathematics, the term “sequence” is commonly used to denote a succession of numbers called terms in a definite order: a1 , , , ……,......... The number a1 is called the first term, the number is called the second term is called the third term and in general the nth term is denoted by Definition: A Real sequence is a real valued function whose domain is the set of positive integers greater or equal to a given integer m (usually 0 or 1). Examples: 1. 2,4,6,8,...  1 1 1 1  2. 1, , , , ,...  2 3 4 5  1 2 3 4  3.  , , , ,... 2 3 4 5  Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 4  Applied Mathematics II 1 1 1 1  4.  , , , ,...  2 4 8 16  For example, in the sequence {2,4,6,8,...} of example 1,We have the following: Term 1 2 3 4 … n number Term 2 4 6 8 … 2n Each term is twice the term number; thus the n th term is given by the formula 2. We denote this by writing the sequence as 2,4,6,8,...,2n,.... We call the function f n  2n the general term of the sequence. Similarly sequences of the above types can be defined by th giving a formula for the n -term. Quick check Class Exercises 1.1.1 1: Find the general term of each sequences in example 2-4 above.by relating each term with their respective term numbers.(Group Discussion in a Class) Notations: 1. When the general term of the sequence with elements is known, it is usually denoted by a n nm or  am , am1 , am 2 ,, am n1 , am , am1 , am2 ,, amn1 , an . 2. If m  1, or m  0 the sequence is written as an n1 or an n0.The letter n in  this notation is called the index of the sequence and the element ai is called the i th  term of the sequence. 3. Since sequence an nm is a function, then we may also write f (n)  a n. Graphs of Sequences Since sequences are functions, it makes sense to talk about the graph of a sequence. For  1  example, the graph of the sequence   is the graph of the equation  n  n 1 1 y , n  1,2,3.... n Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 5  Applied Mathematics II Because the right side of this equation is defined only for positive integer values of n, the graph consists of a succession of isolated points (Figure a). This is different from the graph 1 of y  , x  1 which is a continuous curve (Figure b) x Remarks: 1. There are sequences that do not have a simple defining equation. For instance, a) The sequence p n , where p n is the population of the world as of January 1 in the year n. b) Let a n be the digit in the n decimal place of the number e , then a n  is well th defined sequence whose first few terms are 7,1, 8, 2, 8,1, 8, 2, 8, 4, 5, 2. Some sequences also arise from a formula or a set of formulas that specify how to generate each term in a sequence from terms that precede it; such sequences are said to be sequences defined recursively and the formulas are said to be recursion formulas. For instance, a) The Fibonacci Sequence  f n 1 is defined by the recursion formulas:  f1  3, f 2  5 , f n  f n 1  f n  2, n  3. The first few terms of the sequence are: 3, 5 , 8,13, 21, 34, 55, 89,144, A sequence an n  m where  Definition: an  (1) n  bn and either bn  0 or bn  0, n  m is called an oscillating sequence. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 6  Applied Mathematics II     and  (1)  n Example: Sequences (1) n n 3 are oscillating sequences because of the  3n  4  n 1 fact that the terms of the sequence alternate between positive and negative numbers. Activity 1.1.1 ( Home work) 1.List at least three elements of the sequence given below (Individual Exercises) a) an n1 ,where a  n a ______ , a2  _______, a3  _________ , 1 n 1 n (1) n (n  1) b) a  n n 1 , where an  3n , a1 ______ , a2  _______, a3  _________ c) an n3 , where an  n  3 , a1  ____ , a2  ______, a3  ______ n a  a n  cos a1 _____ , a2  ____, a3  _______ d) n n 0 , where 6 , 2n an  e) a  n n 1 , where (n  1)! , a1  ______ , a2  _____, a3  _____ n an n1 , where an  a1  _____ , a2  ______, a3  ______ f) 2 n 1 , n 1 sn   g) s  n n 1 , where k 1 k , s1  ______ , s2  _____, s3  ______ x n 1 an  h) a  n n 1 , where 2n  1 , a1  ______ , a2  ______, a3  _____ 3. Find the general formula an of the indicated sequence.(Group Discussion) Sequence an a. { 2,9,16,23,30, …} b. {1,8,27,64,125, …} 1 1 1 1 c. {1, , , , , …} 3! 5! 7! 9! 1 1 1 1 1 1 d. 1, 1  , 1   , 1    ,... 3 3 5 3 5 7 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 7  Applied Mathematics II  x3 x5 x7   x, , , , e.  3! 5! 7 !   e e 2 e3 e 4 e5   , , , , ,  f.  2  2 6 8 10 Assessment  Asking an answer for some of the questions.  Check students’ participation in the group activity.  Give feedback to their answers 1.1.2 Convergence and Divergence of Sequence Since sequences are functions, we can inquire about their limits. However, because a sequence a n  is only defined for integer values of , the only limit that makes sense is the limit of as. Definition:(Limits of sequences) 1. A sequence an n m is said to converge to some finite limit L , written as:  lim a n  L ,if and only if n     0, no  N  n  no  an  L  . A sequence that does not converge to some finite limit L is said to diverge. 2. A sequence an n m is said to diverge to  , written as lim an   , if and only if  n   M  0, no  N  n  no  an  M. Similarly , an n m is said to diverge to   if for every negative integer M  3. no  N  n  no  an  M and written as lim a n    n   Examples: 1. Let a n  c , for ,where c is a constant. show that lim a n  c. n   Solution: Given any we need to find N such that an  L   , n  N. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 8  Applied Mathematics II That is we need to find N which satisfies an  L  c  c  0   , n  N Since this is always true we can choose N=1, therefore an  L  0   , n  1. Thus lim a n  c. Limit of constant a n  c sequence is constant. n   1 2. Show that lim 0. n   n Solution: Given any we need to find N such that an  L   , n  N 1 1 That is we need to find N which satisfies a n  L   0    , n  N n n 1 1 We can Choose N  ,so that  . Thus  N 1 1 an  L  0    , n  N n N 1 So by the definition of limit lim 0. n   n 3. Show that lim  n  . n Solution: Here we want to show that for every negative integer no  N  n  no  an   n  M Thus for any number M we can find a number n0  M  12  N such that n  no  an   n   n0   M  1   M  1  M 2 Therefore by the above definition of limit lim  n  . n Theorem 1.1.1: If a sequence an n m converges then its limit is unique.  Proof: Suppose there exists two limits L1 & L2.Therefore by the definition of limit for every positive number there exists N1 , N 2  N such that an  L1   , n  N1 and an  L2   , n  N 2 So if we choose n0  maxN1 , N 2 we have L1  L2  L1  an  an  L2  an  L1  an  L2  2 , n  n0  L1  L2  0  L1  L2 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 9  Applied Mathematics II Remark: 1. Convergence or divergence of a sequence an n m is a property which does not  depend on the initial terms of the sequence rather it is a result of the behavior of the general term eventually i.e. as n   . For instance, see the sequences 1 1 1  100,  200, 400,  800, , , ,  is convergent. 2 3 4 1 1 1 1  1, , , , ,10,  10 , 10,  10,  is divergent. 2 3 4 5 Quick check Class Activity 1.1.2 : 1. Use the definition of limit of sequences to show that a) lim  n   n  b)  1  n  n 1 diverges c) lim n 2   n  n 2. Given that lim n  1  1. By using the definition of limit, find the smallest value n  of N for the given value of  in each part. a.   0.25 b.   0.1 c)   0.001 Instructor’s Role  Check and give feedback to their answers The above definitions of limit could not help us to evaluate the limit of a sequence, thus we seek for further properties of convergent sequences to evaluate their limiting value. Theorem 1.1.2 ( Properties of Convergent sequences) Let an n m and bn n m be convergent sequences. Then   an  bn nm lim an  bn   lim an  lim bn a. converges and n n n r. an nm lim r. an   r. lim an b. converges and n n ,where r is a constant. an.bn nm lim an. bn   lim an. lim bn c. converges and n n n  lim a  an  an n   n d.   converges and lim  , provided that lim bn  0 n   n n  m b n  b n lim n  bn Proof: Direct consequence of the above definitions of limits of sequences. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 10  Applied Mathematics II The above theorem which we stated without proof ensure that the algebraic techniques used to find limits of functions can also be applied to find limits of sequences. Theorem 1.1.3: Let an n m be a sequence and let f be a function defined on  [m, ) such that f (n)  an , n  m.Then a) If lim f ( x)  L  R , then an n m converges and lim an  L.  x  n   b) If lim f ( x)   or   , then an n m diverges and  x  lim a n   or   n  Proof: (Reading Assignment) Examples: ln (n  1) 1. Find the limit of the sequence a n n 1 where a n  . n ln( x  1) Solution: Let f ( x)  for x in [1,  ). x ln( x  1)  Since lim f ( x)  lim is   form, then, by using L’Hopitals rule, x   x   x  1 lim ln( x  1) x x  1 1 lim f ( x)  lim   lim  0, x  x  x lim 1 x  x 1 x  ln( n  1) which implies lim 0. n   n 2. Find the limit of the sequence a n n 1 where a n   n. 2n  1 Solution: Dividing numerator and denominator by n and applying the above theorem: n 1 lim 1 1 1 lim  lim  n    n  2 n  1 n  1 1 20 2 2 lim 2  lim n n  n  n 1 Thus the sequence converges. 2 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 11  Applied Mathematics II Theorem 1.1.4: Suppose that lim an  L and that for each n , a n is in the domain of a n   function f. If f is continuous at L then lim f (a n )  f ( L). n   Proof: Exercise. Find the limit of the sequence a n n 1 where  Example 1:    n  a. an  cos   b. an  ln   n  n 1 Solution: a. Since  lim  0, n n   and the cosine function is continuous at 0 ,     lim cos    cos lim   cos0  1. n   n  n n  b. Since n 1 lim 1 1 lim  lim  n   1 n  n  1 n 1 1 1 0 1 lim 1  lim n n  n  n and logarithmic function is continuous at 1.  n   n  lim ln    ln  lim   ln 1  0. n  n  1  n n  1  Quick Check Class activity 1.1.3: Evaluate the limits of the following Sequences ln n  2n2  8 a. a n  c. an  tan n 16n 2 2 1 5n 2  1 b. an  4    d. an  n 4  3n 2 Instructor’s Role:  Check their answers and Give feedback for their answers Theorem 1.1.5: (The Version of Squeezing Theorem for Sequences) Suppose a n nm , bn nm and cn nm are sequences such that an  bn  cn ,  n  m    and, lim an  lim cn  L.Then lim an  lim bn  lim cn  L. n   n   n   n   n   Proof: Exercise Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 12  Applied Mathematics II Example 1: Find the limit of the sequence a n n 1 where  sin n n! a. a n  b. a n  n nn Solutions:  1 sin n 1 a. Since   ,and n n n  1 1 lim    0  lim , n   n  n  n sin n Then, the squeezing theorem implies, lim  0. n   n n! 1 2  3 ...  n  1  2  3   n  1 b. We have 0       ....   (Why?) n n n  n  n ...  n  n  n  n   n  n 1 Thus 0  a n  However, n 1 lim 0  lim  0. n  n   n n! Thus by squeezing theorem lim  0.. n 0 n n Theorem 1.1.6: If lim a n  0 then lim a n  0. n   n   Proof: Depending on the size of a n either a n  a n or an   an. Thus in both cases we have:  an  an  an. However the limit of the two outside terms is 0, hence the limit of a n is 0 by squeezing theorem. Example 1: Show that    1  n 1  a)  1 b)  1 n  converges to 0. n  converges to 0.  n  n 1  2  n 1 1 Solution: a) Since  1 1 1  and n converges to 0 the result follows by the above n n n theorem. 1 b) Since  1 1 1  n and n converges to 0 the result follows by the above theorem. n n 2 2 2 Quick Check Class activity 1.1.4: Evaluate the limits the following Sequences sin 2 n 1 cos n a. a n  b. an  n n Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 13  Applied Mathematics II cos 2n 1 c. an  d. lim n n  n! Instructor’s Role:  Check their answers and Give feedback for their answers Group Activity 1.1.2 1. (Home Takes Group Assignment) a. The current in an electric circuit is measured after each minute and found to be approximated by in 10.(1  e n ).If the limit of this value is the steady state current, what is the steady state current? b. The height of an electronic “bouncing ball” is described by 7n  2 hn  5n  5 What is the limiting value of the height? c. Suppose the number of bacteria in a culture is growing exponentially, with a doubling time of 10 hours. Suppose also that there are 1000 bacteria in the culture. Find a formula for the number, an of bacteria in the culture after n hours.. 2. Investigate the convergence or divergence of the following sequences by using appropriate method. (Individual Exercises)    2  4n  i. 3   v.  n 6   n  n 1  2  10  n 0   (1) n    1 ii.   vi. an  n 2  n n  n  1  n 1    2n   n4 1  vii.  n  iii.  4   4  7  n 0  n  n  6  n 1    n5    n  viii.  3  iv. tan   n  6  n 0   4n  1  n 1 3. (Assignment). Show that 1 ii. If x 1, then i. If x  0 then lim x 1 n n   lim x n  0 n   Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 14  Applied Mathematics II n n  x  1 iii. lim 1    e x v. lim 1    e n  n n   n x iv. For each x lim 0 n   n! 4. Evaluate the following limits n n10 lim lim n  10 n n  10 n a. b. 5. Consider the sequence: a1  6 , a2  6  6 , a3  6  6  6 ,....Find a recursion formula for a n 1 ? Assessment  Asking an answer for some of the questions.  Check students’ participation in the group activity.  Give feedback to their answers 1.1.3 Bounded and Monotonic Sequences  Bounded Sequences Definition: A sequence an nm is called bounded sequence if there is a positive real  number M such that an  M , for all n  m. Otherwise, it is unbounded.. Examples: a. Consider the sequence an n 1 , where an  sin nx.  Since,  1  sin nx  1, x  R, n  1 then M  1  0  an  M , n  1. b. Since there is no M such that 2 n  M , n  N the sequence is unbounded. But since 0  2 n , n  N the sequence is bounded below by 0 but not bounded above. Remark: Let a n n m be a sequence then,  a) M is called an upper bound if an  M , n (for all n ). b) M is called a lower bound if an  M , n c) A sequence an n m is said to be bounded if it is bounded above and below.  Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 15  Applied Mathematics II Examples:  1 1  1. Since 0  1, for all n 1. The sequence   is bounded both below and n  n  n 1 above. Therefore the sequence is bounded. 2.Consider the sequence 3n  7n  0.Then 0  3n  7, n  N , & 3n  7n 0  increases without bound as increases (not bounded above). Thus the sequence is unbounded. Quick check Exercises 1.1.5: 1. Determine whether or not the following sequences are bounded.    2  (1) n  a. 3   b.    n  n 1  n  1  n 1 Instructor’s Role:  Check their answers and Give feedback for their answers Theorem 1.1.6: Let an n m be a sequence and lim a n  L , where L is a real number.  n  Then an n m is bounded.  Remark: The converse of the above theorem is false. For example, the sequence (1)  n  nm is bounded, since an  1, for all n , but it is divergent.  Monotone Sequences Definition: A sequence an n m is said to be i. Increasing if an  an1 , n  m ii. Decreasing if an  an1 , n  m iii. Strictly increasing if an  an1 , n  m iv. Strictly decreasing if an  an1 , n  m If a sequence an n m is either increasing or decreasing, then it is said to be monotone  sequences and if it is strictly increasing(decreasing) it is said to be strictly monotone sequence. Frequently, one can guess whether a sequence is monotone by writing out some of its initial terms. However, to be certain that the guess is correct, one mustgive a precise mathematical argument. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 16  Applied Mathematics II Testing for Monotonocity: Difference between Ratio between Classification successive terms successive terms a n 1 i) an1  an  0 1 Strictly increasing an a n 1 ii) an1  an  0 1 Strictly decreasing an a n 1 iii) an1  an  0 1 Increasing an a n 1 iv) an1  an  0 1 Decreasing an Examples:1.Identify whether the following sequence increases or decreases.   1   n  a.   b.    n n 1  n  1 n 1   2n  10 n  c.   d.    n!  n 1  n!  n 1 Solutions: a. Since an  0 , we can apply Ratio test. Thus an 1 n  1, n  1 an n 1 That is we have an1  an , for all positive int eger n. Thus the sequence decreases. b. Since an  0 ,using the difference of successive terms we have n 1 n 1 an1  an    2  0, n  1 n  2 n  1 n  3n  1 That is we have an1  an , Thus the sequence increases. c. Since an  0 , we can apply Ratio test. Thus a n 1 2 n 1 n! 2   n  1, n  1 an n  1! 2 n  1 That is we have an1  an , Thus the sequence decreases. d. Since an  0 , we can apply Ratio test. Thus Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 17  Applied Mathematics II a n 1 10 n1 n! 10   n  1, n  9 an n  1! 10 n  1 That is we have an1  an , for all positive int eger n  9. Thus the sequence decreases after the first nine terms, but notice that the first nine terms show that the sequence is increasing. We call such sequences Eventually decreasing. Another third technique for testing monotonocity is using the derivative of the function obtained by replacing n by x in the general term of the sequence. Derivative of Conclusion for the sequence f for x  1 with an  f n f ' x   0 Strictly increasing f ' x   0 Strictly decreasing f ' x   0 Increasing f ' x   0 Decreasing n Example 1: Show that a n  is decreasing sequence. n 1 2 Solution: Consider the function f x   x x 1 2 x 2  1  2x 2 1 x2 f ' x     0, x  1 x 2 1 2 x 2 1 2 Thus f is decreasing on 1,   and so f n  f n  1 Therefore a n n 1 is decreasing.  Theorem 1.1.7:a. Every bounded and increasing sequence converges. (to the least upper bound of its range) b. Every bounded and decreasing sequence converges. (to the greatest lower bound of its range) Proof: Exercise. Examples:1. Show that the sequence a n nm converges, where  2n 1 1 1 1 a. an  b. an 1       n! 1! 2! 3! n! a n 1 2 Solution: a. i) Since  1 an n 1 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 18  Applied Mathematics II we have an1  an , for all positive int eger n.Thus the sequence is decreasing. ii) Again since 2n  2, for all n 1 , n! the sequence is bounded. Therefore, the above theorem implies the sequence converges b. i) Since an1  an  0 , the sequence is increasing. ii) But since there is no M ,  an  M the sequence is unbounded. Therefore, the above theorem implies the sequence diverges. Group Activity 1.1.3: (They will discuss some of the questions in their respective groups and present the result for the whole group) 1. Determine whether or not the indicated sequences are bounded, monotonic or strictly monotonic. (Group Discussion)    (1) n   2n  a.   f.  n   n  n 1  4  10,000  n 5 b. (0.09)  n  n 1 g.  (n  1) 2   2     n  n 1  n  (1) n  c.   n   n 1 h. (1) n n   n 0   n  1      d.   i. sin    n  n 1   n  1  n 0 e.  n  1 2  n 0 n n.  n    e  n 1 2. State whether or not the sequence converges, if it converges, find its limit.    (1) n   n2  c.   f.  4   n  n 1  7n  12 n  0     n   1  n  d. tan  g. e    4n  3   n  0   n 1    2n    1  h. e. ln   h.  4     5n  1  n 1 n  n 1  Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 19  Applied Mathematics II   n    1 j.  4 5 n i. n    n 0 3 State whether the following sequence converges or not, if it does find the limit.    2  n   1  n     1     n   n 1  n   n 1 a. e.    log 10 n  n x    f.  e dx  b.  n  n 1 0 n  0  n 1 2   g. n 2 sin(n )n0  n  c.   n 1  5 n 1     2 n 1   n dx  h. 4  n 1 d.  2  n1  x n  0   x   5n 1     n   n 1 i. 4 a) For convergent sequences, if lim a n  L then what is lim a n 1 ? n   n   b ) Assuming the sequence defined recursively by a1  6 , a2  6  6 , a3  6  6  6 ,... converges find its limit. Assessment  Asking an answer for some of the questions.  Check students participation in the group activity.  Give feedback to their answers Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 20  Applied Mathematics II 1.2 INFINITE SERIES In this section we will be concerned with infinite series, which are sums that involve infinitely many terms. Since it is impossible to add up infinitely many numbers directly, one goal will be to define exactly what we mean by the sum of an infinite series and identify the basic properties about convergence and divergence of a series. However, unlike finite sums, it turns out that not all infinite series actually have a sum, so we will need to develop tools for determining which infinite series have sums and which do not. To do so it is important be familiar with the basic concepts of partial sums of infinite series and convergence and divergence properties of a series. 1.2.1 SUMS OF INFINITE SERIES The most familiar examples of such sums occur in the decimal representations of real numbers. For example, when we write in the decimal form 0.3333..., we mean.0.333…..=0.3 + 0.03 + 0.003 + 0.0003 + which suggests that the decimal representation of can be viewed as a sum of infinitely many terms.  Definition: A sum a n 0 n  a0  a1  a3 ... of infinitely many terms of a sequence is called an infinite series. For instance,  1 1 1 1 1 a.  n     , n 1 3 3 9 27 81  b.  (1) n 0 n 1  (1) 1 (1)   are examples of infinite series. " " NB: The symbol  is called sigma notation. Sums of infinitely many terms of a sequence are defined and computed by indirect limiting process as follows. For a sequence ak k m ,  n a k m k = am  am 1  am  2      an Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 21  Applied Mathematics II is the sum of the first (n  m)  1 terms of the sequence. In particular, for m  0 , n a k 0 k = a0  a1  a2      an , which is called the n th partial sum of the sequence, and is usually denoted by sn. Thus 0 s0  a0   a k , k 0 1 s1  a0  a1   a k , k 0 2 s 2  a0  a1  a 2   a k , k 0 3 s3  a0  a1  a 2  a3   a k , k 0  n s n  a0  a1  a 2   a n   a k , k 0  For instance, 3 a. s3   (3k  1) 1  4  7  10 K 0 5 b. s5   2 k 1  2  2 2  2 3  2 4  2 5 K 0 5 (1) k  1 1 1 c.     K 3 k! 3! 4! 5! 5 1 1 1 1 d. r K 3 k    where r is a constant, are sequences of partial sums. r3 r4 r5 In the sequence of partial sums, if , includes more and more terms of the series and we can conclude that:  lim S n   a n n  n 0 Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series. For determining which infinite series have sums Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 22  Applied Mathematics II and which do not, it is important be familiar with the basic concepts of convergence of infinite series. 1.2.2 Convergence and Divergence of Infinite Series  Definition: An infinite series  an , with the sequence of partial n 1 sum sn n 1 , is said to be convergent if lim s.n exists.  n   Otherwise the series diverges. Remark: If the sequence of partial sums s n n 1 converges to L , then the series  n  lim s.n  lim n   n    ak   ak  L. k 0 k 1 The number L is called the sum of the series. Example: 1. Show that the series  1 a.  k (k  1) k 1 ,Known as Telescoping series converges and find its sum.  1 b.  k , Known as Harmonic series, diverges. k 1    1 k 2. Determine whether the series converges or diverges (exercise!!!) k 1 Solutions: 1. We know first write in closed form that means we need an expression for in which the number of terms in its expression do not vary. 1 1 1 a. Since   ,by partial fractions k (k  1) K k  1 we can see that: 1 1 1 1 sn     1 2 2  3 (n  1)n n(n  1) 1 1   1 1   1 1 1 1                1 2   2 3   n 1 n   n n  1 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 23  Applied Mathematics II 1 1 1 1 1 1 1 1          1 2 2 3 n 1 n n n 1 1 1 . n 1 Now,  1  lim s.n  lim 1    1. n  n   n  1 This means that the series converges to 1 and  1  k (k  1) 1. k 1 b. s1  1 1 s2  1  2 1 1 1 s 22  s 4  1    2 3 4 1 1 1 1  1    1  2.  2  4 4 2 1 2 1 1 1 1 1 1 1 s 23  s8  1        2 3 4 5 6 7 8 1 1 1 1 1 1 1 1  1        1  3.   2  4 4  8  8 88  2 1 1 2 2  1 s 2n  1  n.   2   1  1  lim s 2n  lim 1  n.    1    lim n   , that is, the series is not n  n    2   2  n  bounded above. Thus the series diverges. Quick check Class Exercises 1.2.1: 1. Determine whether the series converges and if so find its sum.   1 1 a)  k 3 ( k  1)(k  2) b) k k 3 2 k Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 24  Applied Mathematics II Instructor’s Role:  Check their answers and Give feedback for their answers One important example of an infinite series is the geometric series which is useful for expressing repeating decimals as fractions.  Definition: - A series of the form c r nm n , where r and c are constants and c  0 , is called a geometric series. Theorem 1.2. 1: Let r be a real number and c  0. Then the geometric series c r m   if r  1  c r n  1  r nm diverges if r  1  Proof: To be discussed in the class Note that the number r in the above theorem is called the ratio of the geometric series. Example: 1. Determine the convergence or divergence of the following series.  n 4 a.    n2  7    230.7 n b. n 2 Solutions: 1. 4 a. Taking c  1, r  and m  2 , we have 7 2 4  n   4  7   16     n2  7  4 21. 1 7 b. Taking c  23, r  0.7 and m  2 , we have  23. 0.7  2  230.7  =  7.889. n n 2 1  0.7 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 25  Applied Mathematics II Geometric series allows us to express any repeating decimal as an infinite series and hence as a rational number. Examples: a) =3 3 1  3 n 1   3      10 1 n 1  10  1 3 1 10 b) 0.45454545…. =0.45+0.0045+0.000045+….  1   45n   1   100  45   45    n 1  100  1 1 99 100 Quick check Class Exercises 1.2.2: 1.Find the rational number represented by the following repeating decimals a) 0.99999....... c) 0.44444...... b) 5.373737...... d) 0.451141414...... 2. Suppose that a ball dropped from a height h hits the floor and rebounds to a height proportional to h , that is, to the height  h (assume   1 ). It then falls from the height  h , hits the floor, and rebounds to the height  ( ( h))   2 h , and so on. Find the total distance traveled by the ball.(Exercise) Teachers’ role:  Observe while they work and answer for the raised questions.  Check and give feedback for their answers   Theorem 1.2.2: If the series a n 1 n and b n 1 n converge, then  i.  (a n 1 n  bn ) converges and     (an  bn )  n 1  an  n 1 b n 1 n  ii. For a constant  ,  a n 1 n converges and   . an  .  an n 1 n 1 Proof: Exercise. Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 26  Applied Mathematics II   5 7  Example 1: - Show that the series   3 n   converges, and find its sum. n(n  1)  n 1  Solutions: Since 1  5   n 1   5      , 5 3 5  n 1 3 n n 1  3  1 2 1 3 and  7  n(n 1)  7 , n 1 then   5 7   5  7 5 9   3 n   n(n 1)  n 1 3n   n(n 1)  7 n 1  n 1 2 2 2. Find the sum of the following series  3k  4 k  2 k 3 a)  b)  k 0 5k k 0 3 k Solution: 3k  4 k  3   4  k k a. Since,     5k 5  5 3k  4 k    k k  3 4  k 0 5 k     k 0  5      k 0  5  1 1   4 3 1 1 5 5 5 15  5 2 2 2 k 3   2   k 8 b. We have  k   2 3     24 k 0 3 k 0  3  1 2 3 Remark: (Change of Base):    For a series  a n , let bn  am  n , n  0 ,  an   bn nm nm n0    1 1 1 Example: - For the series  ,    (n  5)! n  5 n !, n  5 n! n0 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 27  Applied Mathematics II  Theorem 1.2.3: Let m be a positive integer. The series a n0 n converges if and only if the   series  an converges. Moreover, if nm a n0 n  L , then  a nm n  L  (a0  a1  a 2   a m1 ); Or  If a nm n  M , then  a n0 n  a0  a1  a 2   a m1  M Remark: i. Notice that the convergence or divergence of an infinite series is not affected by where you start the summation. ii. From the above theorem ; if the series is convergent, then the sum does depend on where you begin the summation.  n 3 Example: Observe that     4 , n0 4  but  n 3 27     n3 4  16.  Theorem 1.2.4: If a n 1 n converges, then lim a n  0. n  Proof: By using the sequence a n n0 , sn  a1  a2    an1  an and n sn1  a1  a2  an1.  Since the series a n 1 n converges and  lim sn  lim sn1   an n n n 1 Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 28  Applied Mathematics II Therefore an  s n  s n1 lim an  lim s n  s n1  n n lim an  lim sn  lim s n1 n n n lim an  0 n  Remark: 1. The contra positive of the above theorem is important, that is, if lim a n  0 , then n   a n 1 n diverges (sometimes called divergence test). For instance, a. Since n lim  1  0, n 1 n   n the series  diverges. n 1 n  1 b. Since n  1 lim 1    e  0 , n   n  n  1  1   diverges. n 1  n the series 2. The converse of the above theorem is false, that is, “If lim a n  0 , then the series n   a n 1 n converges” is false. 1 For instance, lim  0 , but the series n  n  1 n n 1 a divergent harmonic series. Quick check Class Exercises 1.2.3: Test for divergence of the following series.   1 1 a)  (1  n ) n 1 b)  n sin n n 1 Instructor’s Role  Observe while they work  Check and give feedback for their answers Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 29  Applied Mathematics II Group Activity 1.2.1. (Group Discussion and assignment) 1. Find the sum of the following series, if it converges.   25 6  3 9 27 81 d.    100 n   100 n  a. 1     n 0 2 4 8 16  (1) n  3 e.  n b.  k  0 10 k n0 5   n  1  1 2 n f.  ln  c.  n0 3 n n 1 n  2. Express the following decimals as an infinite series and find its sum if it converges. a. 0.5555555 d. 0.112112112 b. 0.898989 e. 0.314231423142 c. 12.273273273 f. 0.62454545 3. Show that:  1 a.  (1) k 0 k xk  1 x , x  1.  1 b.  (1) k 0 k x 2k  1 x2 , x  1. 4. Find a series expansion for the given expression. x x a. for x  1. c. for x  1. 1 x2 1 x x b. for x  1. 1 x 5. Let d k 1 be a sequence of real numbers that converges to 0. Show that   (d k 1 k  d k 1 )  d1  6. Prove that the series  (a k 1  a k ) converges if and only if the sequence ak 1  k 1 converges. Assessment  Asking an answer for some of the questions  Give feedback to answers  Check students participation in the group activity Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 30  Applied Mathematics II 1.3 Tests for Convergence of Non-Negative Term Series Unlike finite sums, it turns out that not all infinite series actually have a sum as seen in the previous section, so we will need to develop tools for determining which infinite series have sums and which do not. So in this section we will define Non-negative term series and discuss some techniques (tests) for determining their convergence and divergence.  Definition: A series a n m n is said to be a non-negative terms series if and only if an  0, n  m.  Remark: For a positive term series  a n , it holds that n m sm  sm 1  sm  2 ...  s j ... That is, the sequence of partial sum s j m is an increasing.  Theorem 1.3.1 : A series with non-negative terms converges if and only if its sequence of partial sums is bounded. Proof: Exercise I. The Integral Test Theorem 1.3.2 : (The Integral Test) If f is continuous, decreasing and positive on m ,   , then   the series  k m f (k ) converges iff  f ( x)dx converges, where m f (k )  ak. Proof: Reading Assignment Kassahun Nigatu (MSc) and Yitagesu Daba (MSc) 31  Applied Mathematics II Examples:  ln k 1. Show that  k 1 k diverges. Solution: 1  ln x Let f x    f ' x    0, x  1. So f x  is positive, decreasing on 1,   ln x x x2 and since      t dx  lim ln x  1  lim ln t   ln 1 ln x  f ( x)dx  lim  2 t 2 2 t  x t  t  1 1  lim ln t    2 t    ln k Therefore by Integral Test  k 1 k diverges.  1 2. Show that The Harmonic Series,  k , diverges. k 1 Solution:. Clearly f is continuous, decreasing and positive on 1,   , and since 1 Let f ( x)  x  t  lim ln t  ln 1 1  f ( x)dx   x dx  t lim lim ln x x 1 t  