Consider an infinite row of dominoes, labeled by 1, 2, 3,, where each domino is standing up. What should one do to knock over all dominoes?

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1 1 Section 4.1 Mathematical Induction Consider an infinite row of dominoes, labeled by 1,, 3,, where each domino is standing up. What should one do to knock over all dominoes? Principle of Mathematical Induction To prove that P (n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: Basis Step: We verify that P (1) is true. Inductive Step: We show that the conditional statement P (k) P (k + 1) is true for all positive integers k. Remark 1. In general showing P (1) is true is easy.. To complete Inductive Step, we assume that P (k) is true and with this assumption we prove that P (k + 1) is true. 3. If we want to prove that a certain property is true for n 3, for example, then the Basis Step should be replaced by: We verify that P (3) is true. Similarly the Inductive Step should be replaced by: We show that the conditional statement P (k) P (k + 1) is true for all integers k 3. Example 1 Use mathematical induction to prove that n = n(n+1) for all n Z +. Proof Let P (n) denote n = n(n+1). We want to show that P (n) is true for all positive integers n. We will do so by completing the Basis Step and the Inductive Step: Basis Step: Is P (1) true? Yes, because P (1) means 1 = 1, which is definitely true. Inductive Step: Suppose P (k) is true. To complete this step, we need to show that P (k + 1) is true. Since we are assuming that P (k) is true, it means that k = Using this assumption, we want to show that k(k + 1). ( ) k + (k + 1) = (k + 1)(k + ). ( ) Indeed, k + (k + 1) = k(k+1) + (k + 1) (since we assumed that ( ) is true), and this in turn equals (k + 1) ( k + 1) = (k + 1) ( ) k+ = (k+1)(k+), proving ( ). This means that P (k + 1) is true.

2 Example Use mathematical induction to show that (n 1) = n for all positive integers n. Example 3 Use mathematical induction to show that n nonnegative integers n. = n+1 1 for all Example 4 Use mathematical induction to prove the inequality n < n for all positive integers n.

3 3 Example 5 Use mathematical induction to prove the inequality n < n! for every integer n with n 4. Example 6 Use mathematical induction to prove that n 3 n is divisible by 3 whenever n is a positive integer. Example 7 Use mathematical induction to prove that the derivative of f(x) = x n equals nx n 1 whenever n is a positive integer. You can freely use the fact that (fg) = f g + fg. Example 8 Let n be a positive integer. Show that every n n checkerboard with one square removed can be tiled using L-triominoes.

4 4 Section 4. Strong Mathematical Induction and Well-Ordering In this section, we will study a modified version of mathematical induction called Strong Mathematical Induction. When applying mathematical induction, to show that P (n) is true for all positive integers n, we need to show two steps: Basis Step: we verify that P (1) is true. Inductive Step: we show that if we assume that P (k) is true, then P (k + 1) is also true for all positive integers k. In the Strong Mathematical Induction, to show that P (n) is true for all positive integers n, we complete the following two steps: Basis Step: we verify that P (1) is true. Strong Inductive Step: we show that if we assume that P (1), P (), P (3),, P (k) are all true, then P (k + 1) is also true for all positive integers k. Remark 1. Why does the strong mathematical induction work? In other words, why completing the basis step and the strong inductive step guarantees that P (n) is true for all positive integers n? Obviously, P (1) is true. Since P (1) is true, by the strong inductive step, P () is also true. Now we know that P (1) and P () are true, so again by the strong inductive step, P (3) is also true. Continuing in this fashion, it follows that P (n) is true for all positive integers n.. When applying mathematical induction, to show that P (k + 1) is true we assume only P (k) is true. In strong mathematical induction, we assume that P (1), P (), P (3),, P (k) are all true (more assumptions!), so showing P (k + 1) is true becomes easier. Example 1 Show that if n is an integer greater than 1, then n can be written as the product of one or more primes.

5 5 Example Show that every polygon with n vertices, where n 3, can be triangulated into n triangles. Example 3 Consider a game in which two players take turns removing any positive number of matches they want from one of two piles of matches. The player who removes the last match wins the game. Show that if the two piles contain the same number of matches initially, the second player can always guarantee a win.

6 6 Section 4.3 Recursive Definitions and Structural Induction Consider a sequence {a n } whose initial terms are 1,, 6, 4, 10, 70,. In this case, it is easy to see that a n = n!. However, there is another way to describe {a n } which goes as follows. First of all, a 1 = 1. Note that a = 1 = a 1. Similarly, a 3 = 3 = 3a, a 4 = 4a 3, and so on. In general, a n = na n 1. Conclusion: { a 1 = 1, a n = na n 1 for n describes exactly the same sequence. The definition used in ( ) is called an inductive definition (or a recursive definition). Example 1 Consider an inductively defined sequence { a 1 = 1, Find a, a 3, and a 4. What is a n in general? a n = a n for n ( ) Example Consider an inductively defined sequence { b 1 =, Find b, b 3, and b 4. What is b n in general? b n = 3b n 1 for n Example 3 For a sequence {a n }, consider a new sequence given by { S 1 = a 1, What is the relation between {a n } and {S n }? S n = S n 1 + a n for n

7 7 Example 4 Consider a sequence defined by a 1 = 1, a n+1 = a n a n + 1 for n 1 Find a formula for a n by evaluating a, a 3, a 4 and so on. Prove that your guess is correct. Example 5 Consider a sequence defined by { b 1 = 1, b n+1 = b n + 1 for n 1 Find a formula for b n by evaluating b, b 3, b 4 and so on. Prove that your guess is correct.

8 8 Example 6 Let f be a differentiable function defined on R. Pick a real number a, and consider a sequence defined by { x1 = a, x n+1 = x n f(xn) f (x n) for n 1 What can you say about lim n x n, provided the sequence is well-defined and the limit exists? Example 7 Consider the following sequence of integers given by the recursive definition Write the first few terms if 1. a 1 = 4. a 1 = 5 3. a 1 = 3 4. a 1 = 7 What do you notice? a n+1 = { an, if a n is even, 3a n + 1, if a n is odd.

9 9 Section 7. Solving Linear Recurrence Relations As seen in the previous section, finding the formula of a given inductively defined sequence is in general very difficult and the techniques used to find the formula strongly depend on the nature of given sequence. However, a certain class of recurrence relations allows a systematic way to solve the sequence. Definition 1. A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c a n + + c k a n k, where c 1, c,, c k are real numbers, and c k 0.. A linear nonhomogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n 1 + c a n + + c k a n k + F (n), where c 1, c,, c k are real numbers with c k 0 and F (n) is a function not identically zero depending only on n. Example 1 1. a n = a n 1 1 a n is a linear homogeneous recurrence relation of degree.. a n = a n 1 + 3a n a n 3 n is a linear nonhomogeneous recurrence relation of degree a n = a n 1 + a n is not a linear homogeneous recurrence relation. 4. a n = a n 1 + na n does not have constant coefficients. Remark Recurrence relation itself does not determine a sequence. For example, in a sequence defined by a recurrence relation a n = a n 1 + 3, we need to specify a 1 to get a, a 3, a 4, and so on. In general, a recurrence relation of degree k requires a 1, a,, a k to determine the sequence completely. Let s begin with the easiest case. Example A linear homogeneous recurrence relation of degree 1 defines is of the form a n = ca n 1, that is, it represents a geometric progression with common ratio c. Definition Consider a n = c 1 a n 1 +c a n, a linear homogeneous recurrence relation of degree. The quadratic equation x c 1 x c = 0 is called the characteristic equation of the recurrence relation. The solutions of this equations are called the characteristic roots of the recurrence relation. Theorem 1 Suppose {a n } satisfies the recurrence relation a n = c 1 a n 1 + c a n. If r 1, r are distinct characteristic roots of the recurrence relation, then a n must be of the form αr n 1 + βr n for some constants α, β.

10 10 Example 3 Find a n when a 1 = 1, a = 5, and a k = 5a k 1 6a k for k 3. Example 4 Find a n when a 0 = 1, a 1 = 4, and a k = 3a k 1 a k for k. Note that the sequence begins with a 0. Example 5 Find f n when f 1 = 1, f = 1, and f k = f k 1 + f k for k 3.

11 11 What if we have a double root? Theorem Suppose {a n } satisfies the recurrence relation a n = c 1 a n 1 + c a n. If r 0 is the only root of the characteristic equation of the recurrence relation, then a n must be of the form αr n 0 + βnr n 0 for some constants α, β. Example 6 Find a n when a 1 = 0, a = 9, and a k = 6a k 1 9a k for k 3. We can generalize the result above. We begin with the generalization of Theorem 1. Theorem 3 Suppose {a n } satisfies the recurrence relation a n = c 1 a n 1 + c a n + + c k a n k. If the equation x k c 1 x k 1 c x k c k = 0 has k distinct roots r 1, r,, r k, then a n must be of the form α 1 r n 1 + α r n + + α k r n k for some constants α 1, α,, α k. Example 7 Find a n when a 0 =, a 1 = 5, a = 15, and a k = 6a k 1 11a k + 6a k 3 for k 3. Note that the sequence begins with a 0.

12 1 We can also generalize Theorem. Theorem 4 Suppose {a n } satisfies the recurrence relation a n = c 1 a n 1 + c a n + + c k a n k. If the equation x k c 1 x k 1 c x k c k = 0 has t distinct roots r 1, r,, r t with multiplicities m 1, m,, m t, respectively, so that m i 1 for i = 1,,, t and m 1 + m + + m t = k, then a n must be of the form (α 1,0 r n 1 + α 1,1 nr n 1 + α 1, n r n α 1,m1 1n m 1 1 r n 1 ) + (α,0 r n + α,1 nr n + α, n r n + + α,m 1n m 1 r n ) + where α i,j are constants. + (α t,0 r n t + α t,1 nr n t + α t, n r n t + + α t,mt 1n mt 1 r n t ), Example 8 Suppose that the roots of the characteristic equation of a linear homogeneous recurrence relation are,, 5, 5, 5, 6, 6, 7. What is the form of the general solution?.