notes-110228

# notes-110228 - MATH 311-02 Notes Introduction to Higher...

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MATH 311-02 Notes Introduction to Higher Math Even when we don’t have an explicit formula for a recurrence, we can often work out valuable information about it using inductive methods: Proposition 1. Let b n be given by a recurrence relation: b 1 = 1 , b 2 = 2 , and b n = nb n - 1 + b n - 2 for n > 2 . For positive integers n , 2 | b n if and only if 2 | n . Proof. Let P ( n ) be the assertion that 2 | b n if and only if 2 | n ; alternatively, it could be phrased as the assertion that b n and n have the same parity. The base cases P (1) and P (2) are easily veriﬁed: P (1) is odd as is 1, and P (2) is even as is 2. Now let us ﬁx a k 2, and inductively assume that P (1), P (2), . . . , P ( k ) are all true. Speciﬁcally, we will use the facts that P ( k ) and P ( k - 1) are true below. Now we have two cases: either 2 | ( k +1), in which case our goal is to show that 2 | b k +1 , or 2 - ( k + 1), in which case our goal is to show that 2 - b k +1 . We address these casewise: Case I: 2 | k + 1 (a.k.a. k + 1 is even). Since k + 1 is even, k is odd and k - 1 is even, and since we presumed P ( k ) and P ( k - 1) to be true, it thus respectively follows that b k is odd and b k - 1 is even. Then, we note that b k +1 = ( k + 1) b k + b k - 1 ; this is a sum of a product of an even and odd number with an even number; the result of such a computation is even, so 2

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notes-110228 - MATH 311-02 Notes Introduction to Higher...

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