Mathematical Induction


 Norah Fletcher
 9 months ago
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1 Chapter 2 Mathematical Induction 2.1 First Examples Suppose we want to find a simple formula for the sum of the first n odd numbers: (2n 1) = n (2k 1). How might we proceed? The most natural approach is to do the calculation for several small values of n. Doing so, we find n = 1 : n = 2 : n = 3 : n = 4 : 1 (2k 1) = 1 2 (2k 1) = = 4 3 (2k 1) = = 9 4 (2k 1) = = 1. These examples lead us to the conjecture n (2k 1) = n 2. (2.1.1) Is our conjecture correct? It certainly agrees with the results obtained for n = 1, 2, 3, 4 above, but how do we know whether it is true for all n? No amount of computing will answer this question; what we need is a general argument  a proof. 1
2 2 CHAPTER 2. MATHEMATICAL INDUCTION Here is one elementary argument: Let S[n] = n following sum is equal to 2S[n]: (2k 1). Then the (2n 3) + (2n 1). (2n 1) + (2n 3) We have n columns with two numbers each, and each column sums o 2n. Thus 2S[n] = n 2n = 2n 2, and hence S[n] = n 2. This example illustrates a common situation in which we want to establish an infinite number of statements, one for each natural number n. In the above example, a bit of clever grouping established the general result. Not surprisingly, the number of situations in which such an argument is possible is small. The Principle of Mathematical Induction will give us a general approach to obtaining proofs of statements like the above. We will discuss this general principle later in the section, once we have established some of the concepts necessary to formulate and explore a wider range of questions Defining sequences through a formula for the nth term Definition A sequence a of real numbers is a function from the set N 0 of nonnegative integers 0, 1, 2, 3,... to the set R of real numbers. We write either a[n] or a n for the value of the function at n and call this value the nth term of the sequence. We denote the entire sequence by a or {a n }. Note that a[n] and a n are two different notations for the nth term of the sequence a. We will use them interchangeably (though we will be consistent within any example or proposition.) The former notation may help you remember that a is a function, and that each term is really the value of this function at one of the elements of the domain. It is also the notation we ll use when we write Python programs. The latter notation is the traditional notation for terms in a sequence and will be used when we are not developing a program. Remark 1 (zero indexing). Some books define a sequence a to be a function whose domain is the set N of natural numbers 1, 2, 3,.... Perhaps this definition seems more natural, for then the first five terms are a 1, a 2, a 3, a 4 and a 5 rather than a 0, a 1, a 2, a 3 and a 4. Our reason for starting at zero is that we will be programming in Python, and Python (and many other programming languages) use zeroindexed sequences. We want to make the notation in our examples compatible with the syntax we will use in our programs. Were we British, zero indexing might seem more natural; in a multistory building, the ground floor is the zeroth floor and the first floor is the next one up.
3 2.1. FIRST EXAMPLES 3 Example In the previous section, we considered the sequence S, where, for n 1, S[n] is the sum of the n odd numbers 1, 3,..., 2n 1. It is reasonable to define S[0] to be the empty sum, so that S[0] = 0. We showed above that S[n] = n 2 for n 1. This formula is also valid for n = 0, and thus we have a simple formula for the nth term of the sequence S. Sometimes, as above, it is rather easy to guess a formula for the nth term of a sequence by examining the first few terms. What can we try when it is not so obvious? It is sometimes helpful to look at the sequence of ratios or differences of consecutive terms, or the sequence of differences of consecutive terms in the sequences of differences, etc. We illustrate with some examples. Constant differences; arithmetic sequences. Consider the sequence {a n } of all natural numbers for which the remainder when divided by 10 is 3. The first few terms of the sequence are a 0 = 3, a 1 = 13, a 2 = 23, a 3 = 33,.... Consider the sequence {d n } of differences of consecutive terms. Thus d 0 = a 1 a 0 = 13 3 = 10, d 1 = a 2 a 1 = = 10, d 3 = a 4 a 3 = = 10, and in general d n = a n+1 a n = 10. The sequence {a n } is an example of a sequence for which the sequence of differences is the constant sequence whose nth term is d n = 10. One can show (see the next exercise) that the nth term formula for a sequence {a n } is linear in n if and only if the sequence of differences is constant. In this case, a 0 = 3 a 1 = 13 = a 2 = 23 = = = a 3 = 33 = = = a n = 3 + n 10. Exercise Consider the statement: The nth term formula for a sequence {a n } is linear in n if and only if the sequence {d n } of differences is constant. This fact should be familiar to you even if the phrasing in terms of sequences is not. Reconcile the above statement with what you know about linear functions from algebra. Constant ratios; geometric sequences. Consider the sequence whose terms are the successive powers of 2, beginning with 2 0 = 1. The first few terms of
4 4 CHAPTER 2. MATHEMATICAL INDUCTION the sequence are a 0 = 2 0 = 1 a 1 = 2 1 = 2 a 2 = 2 2 = 4 a 3 = 2 3 = 8 a n = 2 n. Observe that this sequence is not arithmetic; the differences of consecutive terms are d 1 = a 1 a 0 = 2 1 = 1 d 2 = a 2 a 1 = 4 2 = 2. Let s look at ratios of consecutive terms: r 1 = a 1 a 0 = 2 1 = 2 r 2 = a 2 a 1 = 4 2 = 2 r 3 = a 3 = 8 a 2 4 = 2 r n = a n = 2n = 2. a n 1 2n 1 Such a sequence, in which the ratio of consecutive terms is constant, is called a geometric sequence Defining sequences recursively For n N 0, the nth triangular number T n is the number associated with a triangular array of dots in which there are n rows, and each row has one more dot than the preceding row. Thus T 0 = 0 (for we have no rows of dots) and T 1, T 2, and T 3 are represented as follows: Thus the first few triangular numbers are T 0 = 0, T 1 = 1, T 2 = 3, T 3 =, T 4 = 10. We can define the sequence for arbitrary n as follows: T 0 = 0, and for n 1, T n = T n 1 + n. In such a recursive definition for a sequence, the nth term is defined to be a function of one or more of the terms that precede it. With a recursive definition, one must also separately specify the first few terms; if the nth term is a function of the preceding k terms, one must specify the first k terms of the sequences..
5 2.1. FIRST EXAMPLES 5 Exercise Write a recursive definition for the arithmetic sequence {3, 13, 23, 33,...} considered above. Exercise Write a recursive definition for the geometric sequence {1, 2, 4, 8,...} considered above. Let us return to our exploration of the triangular numbers. Although we have a recursive definition, we would still like an explicit formula for the nth term; indeed, with a recursive definition, if we want to know, say, T 1000, we must first generate T 1, T 2,..., T 999. One can obtain an nth term formula in a number of ways. One straightforward way involves iterating the recursive formula for T n : T n = T n 1 + n = T n 2 + (n 1) + n = T n 3 + (n 2) + (n 1) + n = (n 2) + (n 1) + n n = k. k=0 We can now obtain a simple formula for T n in the same way we obtained a n(n + 1) formula for the sum of the first n odd numbers. We find T n =. 2 Exercise Verify the last assertion. Exercise Give a picture proof of the formula T n = n(n+1) 2 by considering an n by n + 1 rectangular array of dots. We consider another approach; as we did with the arithmetic sequence in the last subsection, let us look at the sequence {d n } of differences of consecutive terms. d n = T n+1 T n = (T n + n + 1) T n = n + 1. This sequence of differences is not constant. However, we may consider the sequence {d (2) n } of differences for the sequence {d n }. We find d (2) n = d n+1 d n = (n + 2) (n + 1) = 1. In other words, the sequence of second differences is now constant. We saw above that when the sequence of first differences of a is constant, then the nth term formula for a is a linear function of n. Suppose we knew that whenever the second differences are constant, the nth term is a quadratic function of n. We would then have T n = An 2 + Bn + C
6 CHAPTER 2. MATHEMATICAL INDUCTION for constants A, B, and C to be determined. We can solve for these constants. Because we know the first three triangular numbers, 0 = T 0 = C 1 = T 1 = A + B + C 3 = T 2 = 4A + 2B + C. Solving this system gives A = B = 1 2 and C = 0, i.e., T n = 1 2 n n. We obtained the same formula above. It remains to prove that, when the sequence of second differences is constant, the nth term of the sequence is a quadratic function of n. We leave the proof of this claim to the reader. Exercise Show that the sequence {d (2) n } of second differences of {a n } is a nonzero constant sequence if and only if a n = q(n) for some quadratic function q. In other words, you must prove two things: (i) If there are constants A, B, and C with A nonzero such that a n = An 2 + Bn + C for all n 0, then the sequence of second differences is a nonzero constant sequence. (ii) If the sequence of second differences is a nonzero constant sequence, then there exists a quadratic function q such that a n = q(n) for all n. 2.2 First Programs So far, our sequences have been simple enough that we could do all our calculations by hand. As our examples become more complicated, calculating by hand may become tedious or impossible. We therefore put our technology to use. Let s look for a simple formula for S[n] := n k 2 = n 2. We would like our sequence to be zeroindexed (see Remark 1 above), and so we let S[0] = 0. This definition is consistent with our definition of S[n] for n > 0 because, for n = 0, the sum is empty. Inspired by the example of the triangular numbers, we will generate the first terms of the sequence S and then look at the sequence d (1) of first differences, the sequence d (2) of second differences, and so on. Our goal is to figure out whether there is an m so that the sequence d (m) of mth differences is constant. Our experience suggests that if such an m exists, then the nth term formula
7 2.2. FIRST PROGRAMS 7 for S will be S[n] = p(n) where p is a polynomial of degree m. We will still have work to do, for we have only proved this statement for m = 1 and m = 2. But in the next section we will see a powerful technique for proving such conjectures. We use this example to develop our first Python program. If you are new to Python, you should first read the appendix. The first thing the program should do is produce a number of terms in the sequence S. Let s generate the first 15 terms, S[0], S[1],..., S[14]. We can use a simple loop. As above, we will name our list S. Because this sequence is defined recursively, we seed the list with the first term, 0. We now go through a loop 14 times (for 1 n < 15), each time replacing our existing list with the list obtained by appending S[n 1] + n 2 to the old. MAX_ELEMENTS = 15 S = [0] for n in range(1,max_elements): S.append(S[n1]+n**2) Next, we must generate terms in the sequence d (1) of first differences. Recall that d (1) [0] = S[1] S[0],..., d (1) [13] = S[14] S[13]. Thus because we only generated 15 terms in S, we may only generate the first 14 terms in d (1). We do not need to seed the list d (1), so we may initially define it to be an empty list. d1 = [] for n in range(1,len(s)): d1.append(s[n]s[n1]) Similar code generates the first 13 second differences and the first 12 third differences. The full program appears below, with the output included in a comment. MAX_ELEMENTS = 15 S = [0] for n in range(1, MAX_ELEMENTS): S.append( S[n1] + n**2 ) d1 = [] for n in range(1, len(s)): d1.append( S[n]  S[n1] ) d2 = [] for n in range(1, len(d1)): d2.append( d1[n]  d1[n1] ) d3 = [] for n in range(1, len(d2)): d3.append( d2[n]  d2[n1] )
8 8 CHAPTER 2. MATHEMATICAL INDUCTION print("sequence: " + str(s)) print("first Differences: " + str(d1)) print("second Differences: " + str(d2)) print("third Differences: " + str(d3)) # example output #Sequence: [0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 50, 50, 819, 1015] #First Differences: [1, 4, 9, 1, 25, 3, 49, 4, 81, 100, 121, 144, 19, 19] #Second Differences: [3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27] #Third Differences: [2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2] The output of the program suggests (but does not prove) that the sequence d (3) is constant. Our experience thus far leads us to conjecture that the nth term of the sequence S is given by S[n] = An 3 + Bn 2 + Cn + D for appropriate constants A, B, C, and D. Exercise As above, obtain a linear system for A, B, C, and D and solve it to obtain S[n] = 1 3 n n2 + 1 n = n(2n2 + 3n + 1) = n(n + 1)(2n + 1). (2.2.1) Exercise Modify the code to generate the sequence T of triangular numbers from the previous subsection, together with its sequences of first, second, and third differences. 2.3 First Proofs; The Principle of Mathematical Induction Our work in the previous section leads to the conjecture: S n := n j 2 = j=1 n(n + 1)(2n + 1) for all n N. This formula agrees with the output of our program. How can we know that the formula is valid for all n? No amount of computing can settle this question; no matter how long we run our program, we will have only checked our result for finitely many n. What we need is a proof. Although many arguments are possible, we give a proof using the principle of mathematical induction. This principle is quite simple. We have an infinite collection of statements P (1), P (2),..., P (n),..., one for each natural number n. Suppose we know (i) P (1) is true, and (ii) whenever P (k) is true P (k + 1) is true. It then follows that P (n) is true for all n.
9 2.3. FIRST PROOFS; THE PRINCIPLE OF MATHEMATICAL INDUCTION9 In our example, the statement P (n) is S n, the sum of the first n perfect squares, equals n(n+1)(2n+1). Consider the base case, i.e., the statement when n = 1. Clearly S 1 = 1 2 = 1. Because 1(1 + 1)(2(1) + 1) = =, it is indeed the case that S 1 = 1(1+1)(2(1)+1). Thus P (1) is true. For the induction step, we suppose that P (k) is true for some k 1 and we consider P (k + 1). In other words, we consider k+1 k S k+1 = j 2 = j 2 + (k + 1) 2. j=1 By the induction hypothesis, P (k) is true, and so the sum of the first k squares can be replaced by k(k+1)(2k+1). We find S k+1 = = = = = j=1 k(k + 1)(2k + 1) + (k + 1) 2 (k + 1) [k(2k + 1) + (k + 1)] (k + 1) [ 2k 2 + 7k + ] (k + 1) [(k + 2)(2k + 3)] (k + 1)[(k + 1) + 1][2(k + 1) + 1]. Thus P (k+1) is true if P (k) is true. By the principle of mathematical induction, P (n) is true for all n. You may have noticed that we did not prove that the principle of mathematical induction is indeed a valid proof technique. The full justification requires us to give a more precise definition of the natural numbers than we did in the first chapter. We defer this discussion to Chapter??. We note, however, that if one accepts the principle of mathematical induction, one can use it to prove a useful generalization: Proposition Consider a collection of statements P (a), P (a + 1), P (a + 2),... for some integer a. Suppose (i) P (a) is true and (ii) whenever P (k) is true for some integer k, P (k + 1) is true. Then P (n) is true for all n a. We leave the proof to the reader. Exercise Prove Proposition The point of the proposition is that we may use the technique of induction even when the base case is some integer other than 1. In the above example of
10 10 CHAPTER 2. MATHEMATICAL INDUCTION summing the squares, our proof was a bit more verbose than we will tend to be in the future. For example, we will seldom say explicitly what the statement P (n) is because it is usually obvious. We illustrate by proving another proposition by induction. Note also that our base case will be n = 0 and not n = 1. Proposition (Bernoulli s inequality) Let x 1. Then for every integer n 0, (1 + x) n 1 + nx. (2.3.1) Proof. The proof is by induction on n with base case n = 0. When n = 0, (1 + x) 0 = 1 = 1 + 0x and the result holds. Now suppose the result holds for some k 0 and consider k + 1. Then (1 + x) k+1 = (1 + x)(1 + x) k. By the induction hypothesis, (1+x) k 1+kx. Also, because x 1, 1+x 0. Thus (1 + x) k+1 = (1 + x)(1 + x) k (1 + x)(1 + kx) = 1 + kx + x + kx (k + 1)x + 0, and the result indeed holds for k+1. By the principle of mathematical induction, it holds for all integers n 0. Exercise Induction can be used to prove some divisibility results. Consider, for example, a n = 11 n for n 1. By writing down the prime factorizations of the first few, make a conjecture about a factor they all have in common. Prove your claim. 2.4 Strong Induction Some proofs require the strong principle of mathematical induction: Suppose we have a collection of statements P (n), one for each n N. Suppose (i) P (1) is true and (ii) for k 1, if P (i) is true for all 1 i k, then P (k + 1) is true. Then P (n) is true for all n. Although it seems as if we are assuming more at the induction step, the strong principle of mathematical induction and the usual principle of mathematical induction are, in fact, equivalent. We illustrate with an example why we need this formulation as well. Example Suppose we have 3cent and 5cent stamps. What postage can be made?
11 2.5. PROBLEMS 11 Clearly we can not make postage of 1, 2, 4, or 7 cents. It appears that we can make all the others. Indeed, postage of 3 and 5 cents is trivial, and = = = = = , and so on. It thus appears that any postage of 8 cents or more can be made. We prove this conjecture using the strong induction principle. We have just shown explicitly how to make postage of 8, 9, 10, or 11 cents. Suppose now that k 11 and that, for all 8 i k, we can make postage of i cents using 3 and 5cent stamps. Consider postage of k + 1 cents. Because k 11, (k +1) 3 = k 2 9. By the strong induction hypothesis, we can make postage of k 2 cents, and by adding an additional 3cent stamp, we can make postage of k + 1 cents. Thus by the strong principle of mathematical induction, we can make any postage of 8 cents or more with 3 and 5cent stamps. Exercise Determine all postage that can be made using 4 and 7cent stamps. 2.5 Problems Whenever possible, we pose problems in which you need to explore, conjecture, and prove. You may, of course, explore however you like, using trial and error, doing concrete examples via paperandpencil calculations, or using your developing programming skills to write a short program to do many examples quickly for you. There is no right way to explore. Needless to say, most of the problems posed ask you to discover a known result. Thus were you to open other books or go online, you could find formulas and proofs. The point, of course, is to learn how to discover results for yourself. A final note: Just because the title of the chapter is Mathematical Induction does not mean you must or should use induction to do all of these problems. For many of them, induction will work. But for many, other proofs exist as well, and you should look for them. Each new proof gives new insight. 1. Find and prove a formula for the sum of the first n even natural numbers. 2. Find and prove a formula for the sum of the first n perfect cubes, i.e., for n k Find and prove a formula for the sum of the first n powers of 2, beginning n 1 with 2 0. That is, find and prove a formula for 2 k. See if you can give more than one proof. k=0
12 12 CHAPTER 2. MATHEMATICAL INDUCTION n 1 4. Find and prove a formula for 3 k. k=0 n 1 5. A finite geometric series is a series of the form ar k for nonzero a and r. Find and prove a formula for this sum.. Consider the sequence {a n } defined recursively by a 0 = 1 and a n = a n for n 1. Find an explicit formula for a n and an explicit formula for n 1 a k. k=0 7. Recall that a sequence {c n } is called arithmetic if it is given recursively by c 0 = c and c n = c n 1 + d for n 1 and for fixed constants c and d. n 1 Find a formula for c n and for c k. k=0 8. A sequence {a n } is bounded above if there exists an M such that a n M for all n. Let a 0 = 1 and, for n > 0, let a n = 2a n (a) Generate enough terms of the sequence to make a conjecture about whether or not this sequence is bounded above. Prove your conjecture. (b) A sequence {a n } is nondecreasing if, for all n, a n a n+1. We may similarly define non increasing sequences. A sequence is said to be monotone if it is either nondecreasing or nonincreasing. Determine whether the sequence in this problem is monotone. 9. Explore the way in which the boundedness and monotonicity of the sequence in the previous problem depend on the value a 0. Give proofs of all your claims. 10. Our experience with polynomials and exponential functions suggests to us that each of the following inequalities should be true, at least for n sufficiently large. Find all n for which each inequality holds. (a) 2 n 2n + 1. (b) 2 n n 2. (c) n! 2 n A game of Poison begins with a row of 10 pennies and two players. On each player s turn, he or she may remove one penny or two pennies. The last penny is poison, and thus the object is to force one s opponent to take the last penny. Would you prefer to be the first player or the second player? Generalize to an npenny game, with proof, of course. k=0
13 2.5. PROBLEMS Suppose we have postage stamps in denominations of a cents and b cents and that a and b have no common factors. What is the largest postage that can not be made? 13. The purpose of this problem is to prove a number of useful inequalities. (a) Suppose x and y are real numbers. Show that 2xy x 2 + y 2. (b) Suppose a and b are nonnegative real numbers. Show that ab a + b 2. (This inequality is often called the ArithmeticGeometric Mean (AGM) inequality.) (c) Let n be a natural number, and suppose a k > 0 for 1 k 2 n. Prove (a 1 a 2 a 2 n) 1 2 n a 1 + a a 2 n 2 n. Programming Project. In this chapter, we introduced arithmetic sequences (where there is a common difference d between consecutive terms) and geometric sequences (where there is a common ratio r between consecutive terms). We also saw examples of sequences for which we had to look at second and third differences in order to guess a formula for the nth term. Write a program that takes a list consisting of the first six terms of a sequence and finds the first few terms in the sequence of differences or ratios. Then use it to find nth term formulas for sequences agreeing with the first six terms of each of the following: { a = 3, 3 2, 1 2, 1 8, 1 } 40, 1 240,... { 7 b = 3, 3, 3, 7 }, 1, 1,.... 3
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