home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!cs.utexas.edu!qt.cs.utexas.edu!yale.edu!ira.uka.de!rz.uni-karlsruhe.de!uni-heidelberg!trillian!afm
- From: afm@trillian (Andreas Mueller)
- Subject: Re: math problem
- Message-ID: <1992Dec14.100625.3445@sun0.urz.uni-heidelberg.de>
- Sender: news@sun0.urz.uni-heidelberg.de (NetNews)
- Organization: University of Heidelberg, Germany
- References: <9212111851.AA14283@mariner.sce.carleton.ca>
- Date: Mon, 14 Dec 92 10:06:25 GMT
- Lines: 37
-
- 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?
- Replace y by y^(-1) (this doesn't change positivity and 0
- isn't a solution to the problem anyway). Then the equation
- becomes (after multiplying by y^n)
-
- y^n + y^(n-1) + ... + y = a
-
- The left hand side tends to 0 as y goes to 0, and it's derivative
- is
- n y^(n-1) + (n-1) y^(n-1) + ... + 1 >= 1
-
- is strictly positive for y>0, so the left hand side is strictly
- increasing. So there is a unique positive y which satisfies the
- equation. Then y^(-1) is the solution You are looking for.
-
- Hope this helps
-
- Andreas Mueller
-
- -------------------------------------------------------
- Dr. Andreas Mueller afm@mathi.uni-heidelberg.de
- Mathematisches Institut
- Im Neuenheimer Feld 288
- W - 6900 Heidelberg 1
- -------------------------------------------------------
-