solnsA1b

# solnsA1b - C&O 330 - SOLUTIONS #1 PROFESSOR D.M....

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Unformatted text preview: C&O 330 - SOLUTIONS #1 PROFESSOR D.M. JACKSON (1) (20 points) Give an expression for [ x n ] G as an explicit function of n of each of the following two functional equations. (a) (10 points) G = F, where F satisfies the functional equation F = x (1- F ) m , where m is a positive integer. Solution : By Lagrange’s (Implicit Function) Theorem, G = F = X n ≥ 1 x n n λ n- 1 (1- λ )- mn = X n ≥ 1 x n n ( m + 1) n- 2 n- 1 . (b) (10 points) G = e T where T satisfies the functional equation T = xe T . Solution: By Lagrange’s (Implicit Function) Theorem, G = e T = G = [ x n ] e T = 1 + X n ≥ 1 x n n λ n- 1 e λ e nλ = 1 + X n ≥ 1 x n n λ n- 1 e ( n +1) λ = 1 + X n ≥ 1 x n n ! ( n + 1) n- 1 . Comment: Other solutions : i) By Lagrange’s Theorem, T ( x ) = X n ≥ 1 x n n λ n- 1 e nλ = X n ≥ 1 n n- 1 x n n ! = x X n ≥ 1 n n- 1 x n- 1 n ! , so xe T = x X n ≥ 1 n n- 1 x n- 1 n ! . Then, by the Cancellation Law, or by observing that x e T- X n ≥ 1 n n- 1 x n- 1 n ! = 0 1 2 PROFESSOR D.M. JACKSON and that C [[ x ]] has no zero divisors, we conclude that e T- X n ≥ 1 n n- 1 x n- 1 n ! = 0 so e T = X n ≥ 1 n n- 1 x n- 1 n ! = X n ≥ ( n + 1) n- 1 x n n ! . Zero divisors are mentioned in Math 235, in connexion with the ring of square matrices. ii) Alternatively, x- 1 can be used as long as it is clear that the ring being used is the ring of Laurent series over C . This ring was mentioned in the lectures. Half credit should be given for a ‘solution’ in C...
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## This note was uploaded on 11/18/2010 for the course CO 330 taught by Professor R.metzger during the Spring '05 term at Waterloo.

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solnsA1b - C&O 330 - SOLUTIONS #1 PROFESSOR D.M....

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