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Unformatted text preview: MATH 118, LECTURE 29: BINOMIAL EXPANSION 1 BINOMIAL EXPANSION We know by binomial expansion that for nonnegative integers m = 0 , 1 , 2 , 3 ,... we have ( a + b ) m = m summationdisplay n =0 parenleftbigg m n parenrightbigg a n b m n where we have the m choose n operator given by parenleftbigg m n parenrightbigg = m ! ( m n )! n ! . This corresponds to the number of distinct subsets of n elements from an overall set of m elements, disregarding the order with which we choose the subset. So for instance, if we have a set of 5 items and are asked how many distinct ways we could select 2 elements we would find that we have parenleftbigg 5 2 parenrightbigg = 5! 3! 2! = 5 4 3 2 (3 2) 2 = 10 possibilities. This corresponds to the 10 element set { (1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (2 , 3) , (2 , 4) , (2 , 5) , (3 , 4) , (3 , 5) , (4 , 5) } . Of particular interest for us is the fact that (1 + x ) m = m summationdisplay n =0 parenleftbigg m n parenrightbigg x n = 1 + m summationdisplay n =1 m ( m 1) ( m n + 1) n ! x n . This looks like a Taylor series! There is, however, a subtle difference in that the series is finite. We can exactly determine the expansion with a finite number of monomial terms x n . It is, in that sense, a trivial Taylor series expansion. Also, since the equation is exact, the series converges for < x < ....
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 Spring '09
 Soares

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