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4. For what values of a and b is it true that no matter what the initial values x0 , x1 are, the solution of the recurrence relation xn+1 = axn + bxn−1 (n ≥ 1) is guaranteed to be o(1) (n → ∞)? 5. Suppose x0 = 0, x1 = 1, and for all n ≥ 2, it is true that xn+1 ≤ xn + xn−1 . Is it true that ∀n : xn ≤ Fn ? Prove your answer. 34 1. Mathematical Preliminaries 6. Generalize the result of Exercise 5, as follows. Suppose x0 = y0 and x1 = y1 , where yn+1 = ayn + byn−1 (∀n ≥ 1). If furthermore, xn+1 ≤ axn + bxn−1 (∀n ≥ 1), can we conclude that ∀n : xn ≤ yn ?

5 Counting For a given positive integer n, consider the set {1, 2, . . n}. We will denote this set by the symbol [n], and we want to discuss the number of subsets of various kinds that it has. Here is a list of all of the subsets of [2]: ∅, {1}, {2}, {1, 2}. There are 4 of them. We claim that the set [n] has exactly 2n subsets. To see why, notice that we can construct the subsets of [n] in the following way. ’ Each of the n choices that you encountered could have been made in either of 2 ways.

If n is a positive integer. The first definition is nonrecursive, the second is recursive. (1) “n! ” (2) “If n = 1 then n! = 1, else n! ” Let’s concentrate on the second definition. At a glance, it seems illegal, because we’re defining something, and in the definition the same ‘something’ appears. Another glance, however, reveals that the value of n! is defined in terms of the value of the same function at a smaller value of its argument, viz. n − 1. So we’re really only using mathematical induction in order to validate the assertion that a function has indeed been defined for all positive integers n.

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