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# week4 - GENERATING FUNCTIONS Solve an infinite number of...

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July 30, 1996 1 GENERATING FUNCTIONS Solve an infinite number of related problems in one swoop. *Code the problems, manipulate the code, then decode the answer! Really an algebraic concept but can be extended to analytic basis for interesting results. (i) Ordinary Generating Functions {a 0 , a 1 , , a k , } sequence where the kth term is the solution of some problem, for every k. Create the object (“formal power series”) k=0 k k a x where x k is like a place-holder for a k . This looks like an analytic power series but it’s NOT (not yet, anyway). Rules of Operation: just do what comes naturally. ( 29 ± ± a x b x = a b x k k k k k k k ( 29 ( 29 a x b x = c x k k k k k k where c k = j=0 k j k- j a b Examples (I) a k = 1 ,0 k n a k = 0 k > n k=0 k k n n+1 a x = 1 + x + _ + x = 1 - x 1 - x why is last equality true? Because ....... (1 + x + + x n )(1 - x) = 1 + x + + x n = - x - _ - x - x 1 - x n n+1 n+1 (ii) a k = 1 2200 k.

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July 30, 1996 2 k=0 k x = 1 1 - x (1 - x) x x x x - x = 1 k=0 k k 0 k k 0 k+1 k 0 k k 1 k = = NOTE: You don’t need anything about convergence! At the same time, you shouldn’t think of “x” as a variable into which you substitute values (not yet, anyway) but soon it will be OK). (iii) If we have the o.g.f., we can find the sequence: e.g. Suppose the o.g.f. is (1 + x) n then the sequence is found as follows: (1 + x) = n k x n k 0 k Thus, a k = n k (note that a k = 0 for k > n). Exponential Generating Function {a 0 , a 1 , , a k , } k 0 k k a k! x . k k k k k k k k k a k! x b k! x = c k! x , where k j=0 k j k- j c = j k a b’ c’ = a’ b’ k j=0 k j k- j k j=0 k j k c k! = a j! b (k j)! Examples (i) a k = 1, 0 k n; a k = 0 for k > n
July 30, 1996 3 k=0 n k x k! (ii) a k = 1 2200 k k 0 k x x k! = e (iii) a k = n k ( number of k-perms of an n-set) (iv) If we have the e.g.f. sin x then sin x = k 0 k 2 k+1 (-1) (2 k+1)! x so a 2k + 1 = (-1) k k 0 a 2k = 0 k 0 Some Generating Function Manipulations Suppose A(x) = k k k k k a x , B(x) = x - Then A(x) B)(x) = k k k d x where k j=0 k j j=0 k j c = a 1 = a - Similarly, A 2 (x) = k k k d x where k j=0 k j k- j d = a a e.g. a k = n k then 2n n = n k k=0 n 2

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July 30, 1996 4 - Also, 1 x [A(x) - a ] = a x 0 k 0 k+1 k , which is the o.g.f. for the sequence {a 1 ,a 2 , ,a k , } which is the original sequence shifted one place to the left (and the first term dropped off). By contrast, xA(x) = k 0 k k+1 a x which is 0 + k 0 k-1 k a x or the o.g.f. for the sequence {0, a 0 , a 1 , , a k , } (k + 1) th place. d dx a x ka x k 0 k k k 1 k k-1 which is the o.g.f. for {a 1, 2a 2 , 3a 3 , , ka k , All of those ideas carry over to e.g.f. in an analogous way.
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week4 - GENERATING FUNCTIONS Solve an infinite number of...

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