Modern Algebra

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/