From: yuen@bhprtc.scpd.oz.au (Daniel Yuen)
Subject: Re: math problem
Organization: BHP Steel, Sheet & Coil, Research & Technology Centre
Date: Mon, 14 Dec 92 06:21:50 GMT

li@sce.carleton.ca (Yao Li) writes:

>                            Math Problem
>                            ------------
>Getting bored? Here is a math problem.
>There is a polynomial:
>       -a*y**n + y**(n-1) + y**(n-2) + ... + y + 1 = 0
>where real number a>0 and the notation "y**n" represents the nth power of y.
>It's conjectured that this polynomial has a unique positive real solution y.
>Can you prove it or find a counter example?
>If succeed, please send me your solution. I will post the results later.
>Happy challenge!
>
>Li Yao
>li@sce.carleton.ca


uphhswh@Msu.oscs.montana.edu writes

>I was a little surprised by the previous several answers given by other
>netters, because it was simply NOT TRUE.  Although I sent my answer to the
>original poster just after he posted, I feel I must post here again.
>
>y**n+ y**(n-1)+...+1 = (1-y**n)/(1-y)
>So the original equation becomes:
>(a+1)y**n = ay**(n+1) +1
>The intersections of LHS curve and RHS curve are solutions.
>at either y=0 or y=infinity, LHS<RHS
>so if there is intersection, there must be EVEN number's intersections.
>so the claim of UNIQUE solution is not true.
>
>W. Huang --- former provincial math contest No.1


It seems interesting that Huang arrives at a conclusion different from the
others albeit taking a seemingly logical approach.  However, we could 
reconcile the difference if we modify Huang's approach as follows :

       -a*y**n + y**(n-1) + y**(n-2) + ... + y + 1 = 0   -------------(1)

can be wriiten as

               (a+1)y**n - ay**(n+1) - 1
       f(y) =  -------------------------  = 0  -----------------------(2)
                      (y - 1)

Note that the denominator [(y-1)] which has been omitted by Huang is
important, as will be seen below.
Note also that

       {lim y->1} f(y) = n-a

and equals 0 only if n=a.  Hence y=1 is a solution of eqn(2), thus also of 
eqn(1), only if n=a.

Now, consider the numerator of eqn (2), take

       g(y) = (a+1)y**n - ay**(n+1) - 1 = 0  -------------------------(3)

which has an even number of solutions (Huang's argument) except under the
limiting case of a=n (see below).  Also, y=1 is always a solution of g(y).
By writing eqn(3) in the form [after Huang]

       (a+1)y**n = ay**(n+1) + 1  ------------------------------------(4)

one could see that the LHS and RHS are monotonically increasing functions
and have matching gradients at two points only, namely, at y=0 and
at y = n(a+1)/(n+1)/a > 0.  Thus, eqn(4) has at least one root and at most
two roots, and one root is always at y=1.

Case (1) : If a!=n, the LHS & RHS of eqn(4) do NOT have matching gradients 
           at y=1.  Hence eqn(4) has two solutions, of which y=1 is not a 
           solution of eqn(1) while the other is.

Case (2) : If a=n, the LHS & RHS of eqn(4) have matching gradients at y=1.
           Hence, the LHS & RHS curves "touch" each other rather than
           "intersect" each other at that point.  Thus eqn(4) has only one
           solution (y=1), which happens to be a solution of eqn(1) as well.

Conclusion :  Ze Yang was right, namely, that eqn(1) has a unique positive
              solution.  (Sorry I did not see the other solutions.)

Declaration :  I did not win any maths contest -- either provincial,
               national or international :-)

Cheers.

Daniel
(yuen@bhprtc.scpd.oz.au)
Newsgroups: soc.culture.china