Fx j n f n n x n 1 x j n f n 2 n 2 x n 2 1 x j n f n

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F(x) = j n F n n! x n = 1 + x + j n $ 0 F n % 2 (n % 2)! x n % 2 = 1 + x + j n $ 0 F n % 1 % F n (n % 2)! x n % 2 ² F N (x) = 1 + j n $ 0 F n % 1 % F n (n % 1)! x n % 1 = 1 % j n $ 0 F n % 1 (n % 1)! x n % 1 % j n $ 0 F n (n % 1)! x n % 1 = 1 + (F(x) - 1) + j n $ 0 F n (n % 1)! x n % 1 ² F O (x) = F N (x) + j n $ 0 F n n! x n = F N (x) + F(x) This can be solved using ODE approach for FPS (same as if F(x) a real (or complex) function). F(x) = c 1 e R + x + c 2 e R - x where 1= F 0 = F(0) = c 1 + c 2 1 = F 1 = F N (0) = c 1 R + + c 2 R -
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S. Tanny MAT 344 Spring 1999 86 Solve for c 1 , c 2 : c 1 = R + c 2 = - R - 1 5 1 5 ² F(x) = ( - ) 1 5 R % e R % x R & e R & x ² = (R + n + 1 - R - n + 1 ) x n j n $ 0 F n n! x n 1 5 j n $ 0 1 n! DERANGEMENT RECURSION D n / number of perms in S n with no fixed points. D n + 1 = n (D n + D n + 1 ) D 1 = 0, D 2 = 1 8 LINEAR BUT NOT CONSTANT COEFF. Define D 0 = 1 to make recursion hold for n = 1. Let D(x) = j n $ 0 D n n! x n = 1 + j n $ 2 D n n! x n = 1 + j n $ 0 D n % 2 (n % 2)! x n % 2 ² D N (x) = j n $ 0 D n % 2 (n % 1)! x n % 1 = j n $ 0 D n % 1 % D n n! x n % 1 = j n $ 0 D n % 1 n! x n % 1 % xD(x)
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S. Tanny MAT 344 Spring 1999 87 Doesn’t look too promising? What now? = x j n $ 0 D n % 1 n! x n % xD(x) D N (x) = x (D(x) - D 0 ) N + xD(x) = xD N (x) + xD(x) ² D N (x) (1 - x) = xD(x) ² D N (x) D(x) x 1 & x 1 1 & x & 1 ² R n D(x) = R n(1 - x) - x + c D(x) = 1 1 & x e & x e c But D(0) = D 0 = 1 Y e c = 1 ( Y c = 0.) ² D(x) = e & x 1 & x If you don’t know ODE, there is another way to get this result: (See Roberts, pp. 224 ff) D(x) = 1 + j n $ 0 D n % 2 (n % 2)! x n % 2 D n + 2 = (n + 1) (D n + 1 + D n ) 8 If only this were (n + 2) instead!? We can show (see Roberts, p.225) D n + 2 = (n + 2) D n + 1 + (-1) n + 2
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S. Tanny MAT 344 Spring 1999 88 ² D(x) = 1 + + j n $ 0 D n % 1 (n % 1)! x n % 2 j n $ 0 ( & 1) n % 2 x n % 2 (n % 2)! = 1 + x [D(x) - 1] + (e -x - 1 + x) ² D(x) = 1 + xD(x) - x + e -x - 1 + x. = x D(x) + e -x ² D(x) = e -x /(1 - x) = 1 & x 1! % x 2 2! & ³ 1 % x % x 2 % ³ j n $ 0 D k k! x k j 4 k 0 1 & 1 1! % 1 2! ³ % ( & 1) k 1 k! x k ² D k = k! 1 & 1 1! % 1 2! ³ % ( & 1) k 1 k! Counting Bracketings in Products In how many different ways can the “product” x 1 ,x 2 ³ x n be parenthesized?
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  • Fall '06
  • miller
  • Negative and non-negative numbers, Pallavolo Modena, Sisley Volley Treviso, Fabrique Nationale de Herstal, S. Tanny

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