The Binomial Theorem and Consequences

Transcription

1 The Binomial Theorem and Consequences Juris Steprāns York University November 17, 2011

2 Fermat s Theorem Pierre de Fermat claimed the following theorem in 1640, but the first published proof (by Leonhard Euler) appeared only about a century later: If p is prime and a is any integer then p (a p a). So, for example, 31 (7 31 7) and 19 (3 19 3) even though checking this hand would be difficult and tedious. In spite of the long time it took for a proof to appear, there is an elementary argument establishing this equality. However, we need some preliminary concepts before we can get to the proof. We will first establish the binomial theorem and this needs the notion of a binomial coefficient.

3 Factorial notation Recall that the symbol k! represents the product k. The number k! represents the number of ways of rearranging k ordered items for example 52! represents the number of ways a deck of 52 cards can be rearranged. To see this suppose we have k items and they are listed as a 1, a 2, a 3,... a k. The number of choices for a 1 is k. Given a 1, the number of choices for a 2 is ust k 1. Given a 1, a 2, the number of choices for a 3 is ust k 2. Given a 1, a 2, a 3, the number of choices for a 4 is ust k 3. Continuing like this we see that in the end that there is ust one choice for a k.

4 So the total number of rearrangements of the k element set is: k (k 1) (k 2) How many ways are there of listing the letter A, B, C and D?

5 So the total number of rearrangements of the k element set is: k (k 1) (k 2) How many ways are there of listing the letter A, B, C and D? 4! = 24.

6 Binomial coefficients The binomial coefficients are defined by n n! = m (n m)!m! but what are they good for? n The first thing to note is that the number represents the m number of ways of choosing m elements from an n element set.

7 To see this we argue as for the factorial. Suppose we have n items and m of them are listed as a 1, a 2, a 3,... a m. The number of choices for a 1 is n. Given a 1, the number of choices for a 2 is ust n 1. Given a 1, a 2, the number of choices for a 3 is ust n 2. Given a 1, a 2, a 3, the number of choices for a 4 is ust n 3. Continuing like this we see that in the end that there are n (m 1) choices left for a m. In other words, the number of ways of listing m items a 1, a 2, a 3,... a m is n (n 1) (n 2) (n 3)... (n (m 2)) (n (m 1)) and this is equal to...

8 n (n 1) (n 2) (n 3) (n m) (n (m 1)) (n (m 2)) = n! (n m)! However, we are interested only in the number of m element subsets not lists. The number above counts each enumeration of a set, not ust the set. hence we should divide this number by the number of all enumerations of an m element set. But we know this is m! and so the number of ways of choosing m elements from an n element set is n! m!(n m)! = n m

9 Theorem (Binomial Theorem) For any positive integer n (a + b) n = =0 n a n b or, equivalently, (a + b) n is equal to a n + na n 1 n b + a n 2 b 2 n a n b nab n 1 + b n 2

10 Examples (a + b) 2 = a 2 + 2ab + b 2 (x + y) 5 = x 5 + 5x 4 y + 10x 3 y x 2 y 3 + 5xy 4 + y 5 (x + 1) 5 = x 5 + 5x x x 2 + 5x + 1 (x 1) 5 = x 5 5x x 3 10x 2 + 5x 1 (1 1) 5 = = 0 (x 1) 6 = x 5 6x x 4 20x x 2 6x + 1 (1 1) 6 = = 0

11 The Pascal Triangle gives the binomial coefficients: Each entry of the triangle is the sum of the two adacent entries of the previous row. In other words, n + 1 n n = + 1

12 Another view of Pascal s Triangle is useful: and points to the identity we will need.

13 There are various ways of proving this. One is to use the fact that n + 1 is the number of -element subsets of an {1, 2, 3,..., n, n + 1}. Each such set is either: n a -element of {1, 2, 3,..., n} and there are such sets or {n + 1} Z where Z is a ( 1-element ) subset of n {1, 2, 3,..., n} and there are such sets. 1 Hence n + 1 n = + 1 n.

14 Another proof is algebraic: n n n! + = 1 ( 1)!(n ( 1))! + n!!(n )! = n! ( 1)!(n + 1)! + n! n! + n!(n + 1) =!(n )!!(n + 1)! n!(n + 1) (n + 1)! n + 1 = =!(n + 1 )!!(n + 1 )! =

15 To prove the binomial theorem mathematical induction will be used. Let B be the set of all n such that the binomial theorem fails; in other words, B is the set of all n such that (a + b) n =0 n a n b We know that 1 and 2 do not belong to B. If B is empty we are done so let k be the least element of B and note that k > 2 so that it is possible to write k = n + 1 for some positive integer n. Then (a + b) k = (a + b) n+1 = (a + b) n (a + b) and the minimality of k ensure that n / B. Hence (a + b) n = =0 n a n b

16 But then (a + b) n (a + b) = n a n b (a + b) = =0 =0 n a n +1 b + n a n b +1 = =0 =1 =0 n n+1 a n ( 1) b + n a n ( 1) b = 1 n a n+1 b ( ) n n + a n ( 1) b n + a n n b n+1 = 1 n =1

17 a n+1 + =1 ( ) n n + a n ( 1) b + b n+1 1 Now recall from the Pascal Triangle calculation that n n n = 1 and hence (a + b) k is equal to (a + b) n (a + b) = a n+1 + =1 n + 1 a n ( 1) b + b n+1 = =1 k 1 a k k + a k b + b k = k =0 k a k b