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Unformatted text preview: EC3070 FINANCIAL DERIVATIVES THE BINOMIAL THEOREM Pascals Triangle and the Binomial Expansion. To prove the binomial theorem, it is necessary to invoke some fundamental principles of combinato rial calculus. We shall develop the necessary results after derving the general expression for a binomial expansion by a process of induction. Consider the following binomial expansions: ( p + q ) = 1 , ( p + q ) 1 = p + q, ( p + q ) 2 = p 2 + 2 pq + q 2 , ( p + q ) 3 = p 3 + 3 p 2 q + 3 pq 2 + q 3 , ( p + q ) 4 = p 4 + 4 p 3 q + 6 p 2 q 2 + 4 pq 3 + q 4 , ( p + q ) 5 = p 5 + 5 p 4 q + 10 p 3 q 2 + 10 p 2 q 3 + 5 pq 4 + q 5 . The generic expansion is in the form of ( p + q ) n = p n + np n 1 q + n ( n 1) 2 p n 2 q 2 + n ( n 1)( n 2) 3! p n 3 q 3 + + n ( n 1) ( n r + 1) r ! p n r q r + + n ( n 1)( n 2) 3! p 3 q n 3 + n ( n 1) 2 p 2 q n 2 + npq n 1 + p n . In a tidier notation, this becomes ( p + q ) n = n X x =0 n ! ( n x )! x ! p x q n x . We can find the coecient of the binomial expansions of successive degrees by the simple device known as Pascals triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 THE BINOMIAL THEOREM The numbers in each row but the first are obtained by adding two adjacent numbers in the row above. The rule is true even for the units that border the triangle if we suppose that there are some invisible zeros extending indefinitely on either side of each row. Instead of relying merely upon observation to establish the formula for the binomial expansion, we should prefer to derive the formula by algebraic meth ods. Before we do so, we must rearm some notions concerning permutations and combinations that are essential to a proper derivation. Permutations. Let us consider a set of three letters { a, b, c } and let us find the number of ways in which they can be can arranged in a distinct order. We may pick any one of the three to put in the first position. Either of the two remaining letters may be placed in the second position. The third position must be filled by the unused letter. With three ways of filling the first place, two of filling the second and only one way of filling the third, there are altogether 3 2 1 = 6 different arrangements. These arrangements or permutations are abc, cab, bca, cba, bac acb....
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This note was uploaded on 03/02/2012 for the course EC 3070 taught by Professor D.s.g.pollock during the Spring '12 term at Queen Mary, University of London.
 Spring '12
 D.S.G.Pollock

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