Date: 07/23/97 at 12:31:34 From: Wendy Subject: Modern Algebra This problem has brought me to many dead ends - please help! Let n be a positive integer, and define f(n)= 1!+2!+3!+...+n!. Find polynomials P(x) and Q(x) such that f(n+2)=P(n)f(n+1)+Q(n)f(n) for all n > or = 1. A sincere thanks, Wendy
Date: 07/23/97 at 19:36:27 From: Doctor Wilkinson Subject: Re: Modern Algebra This is definitely tricky! Let's give it a try. Substitute the definition of f(n) into the equation you're trying to solve for P(n) and Q(n): 1!+2!+3!+...+n!+(n+1)!+(n+2)! = P(n)(1!+2!+3!+...+n!+(n+1)!) + Q(n)(1!+2!+3!+...+n!) Or (P(n)+Q(n)) (1!+2!+3!+...+n!) + P(n) (n+1)! = 1!+2!+3!+...+n!+(n+1!)+ (n+2)! This suggests that we should have P(n) + Q(n) = 1 and P(n) (n+1)! = (n+1)! + (n+2)! This could be called "let's go for it." We've got to get rid of that big sum! The second equation, which just involves P(n), can be written (n+1)! P(n) = (n+1)! (1 + n + 2) (since (n+2)! = (n+2)(n+1)! and we can factor out the (n+1)! to get P(n) = n + 3 Then you can substitute back in the other equation to get Q(n), after which you'd better check that it all works. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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