chap5_6 - 5. Review of Useful Series and Sums The preceding...

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5. Review of Useful Series and Sums The preceding chapters have introduced ways to calculate the probabilities of random events, based on various assumptions. You may have noticed that many of the problems you’ve encountered are actually similar, despite the contexts being different. Our approach in the next few chapters will be to classify some of these common types of problems and develop general methods for handling them. Rather than working each problem out as if we’d never seen one like it before, our emphasis will now shift to checking whether the problem is one of these general types. 5.1 Series and Sums Before starting on the probability models of the next chapter, it’s worth reviewing some useful results for series, and for summing certain series algebraically. We’ll be making use of them in the next few chapters, and many, such as the geometric series, have already been used. 1. Geometric Series: a + ar + ar 2 + ··· + ar n 1 = a (1 r n ) 1 r = a ( r n 1) r 1 for r 6 =1 If | r | < 1 ,then a + ar + ar 2 + = a 1 r 2. Binomial Theorem: There are various forms of this theorem. We’ll use the form (1 + a ) n = n X x =0 μ n x a x . In a more general version, when n is not a positive integer, (1 + a ) n = X x =0 μ n x a x if | a | < 1 . (Wehavealreadydefined ¡ n x ¢ when n is not a positive integer.) While the binomial theorem may not look like a summation result, we’ll use it to evaluate series of the form P x =0 ¡ n x ¢ a x . 51
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52 3. Multinomial Theorem: A generalization of the binomial theorem is ( a 1 + a 2 + ··· + a k ) n = X n ! x 1 ! x 2 ! x k ! a x 1 1 a x 2 2 a x k k . with the summation over all x 1 ,x 2 , k with P x i = n . The case k =2 gives the binomial theorem in the form ( a 1 + a 2 ) n = n X x 1 =0 μ n x 1 a x 1 1 a n x 1 2 . 4. Hypergeometric Identity: X x =0 μ a x ¶μ b n x = μ a + b n . There will not be an infinite number of terms if a and b are positive integers since the terms become 0 eventually. For example μ 4 5 = 4 5! (5) = (4)(3)(2)(1)(0) 5! =0 Sketch of Proof: (1 + y ) a + b =(1+ y ) a (1 + y ) b . Expand each term by the binomial theorem and equate the coefficients of y n on each side. 5. Exponential series: e x = x 0 0! + x 1 1! + x 2 2! + x 3 3! + = X n =0 x n n ! (Recall, also that e x =l im n →∞ ³ 1+ x n ´ n ) 6. Special series involving integers: 1+2+3+ + n = n ( n +1) 2 1 2 +2 2 +3 2 + + n 2 = n ( n +1)(2 n 6 1 3 3 3 + + n 3 = n ( n 2 ¸ 2
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53 Example: Find X x =0 x ( x 1) μ a x ¶μ b n x Solution: For x =0 or 1 the term becomes 0, so we can start summing at x =2 .Fo r x 2 ,we can expand x ! as x ( x 1)( x 2)! X x =0 x ( x 1) μ a x ¶μ b n x = X x =2 x ( x 1) a ! x ( x 1)( x 2)!( a x )! μ b n x . Cancel the x ( x 1) terms and try to re-group the factorial terms as “something choose something”. a ! ( x 2)!( a x )! = a ( a 1)( a 2)!
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This note was uploaded on 12/01/2010 for the course MATH Stat 230 taught by Professor A during the Spring '07 term at Waterloo.

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chap5_6 - 5. Review of Useful Series and Sums The preceding...

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