From: li@sce.carleton.ca (Yao Li)
Subject: anwser to "math problem" 2
Organization: Indiana University
Date: Tue, 15 Dec 1992 00:12:20 GMT
X-Mailer: ELM [version 2.3 PL11]

I did not check scc closely. The following is a more complete list of people
who shared their wits with us

Victor Adamchik <victor@wri.com>
W. Huang <uphhswh%msu.dnet@terra.oscs.montana.edu>
Shidong Tong <stong@irus.rri.uwo.ca>
Bob Silverman <bs@linus.mitre.org>
ZHIGANG XI <XI@BINAH.CC.BRANDEIS.EDU>
Michael Somos <somos@alpha.ces.cwru.edu>
xyl00@pteng.amdahl.com (Xiaoyang Luo)
cheekai@ncb.gov.sg (Chin Chee-Kai)
<erick@sfu.ca>
Torben Mogensen (torbenm@diku.dk)
Andreas Mueller afm@mathi.uni-heidelberg.de
zg11@ellis.uchicago.edu (zening  ge)
yucao@cco.caltech.edu (Yu Cao)
zzuo@iastate.edu (Zhiqi Zuo)
yangze@acf3.nyu.edu (Ze Yang)
wang@math.psu.edu (Feng Wang)
Duane <B7D@psuvm.psu.edu>
wja@math.psu.edu (Jianan Wu)
ding@eng.auburn.edu (Z. Ding)
z_yan@math.nwu.edu (Z. Yan)
yuen@bhprtc.scpd.oz.au (Daniel Yuen)
echeng@phys.psu.edu (Eddie Cheng)
yding+@pitt.edu (Yijun Ding)

Hopefully I got all.

Let me also include all answers I received through my personal email.
You can also check sci.math

Li Yao
Ottawa




From prince@ruccs.rutgers.edu Fri Dec 11 16:15:15 1992
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From: prince@ruccs.rutgers.edu (Alan Prince)
Message-Id: <9212112117.AA00694@ruccs.rutgers.edu>
To: li@sce.carleton.ca
Subject: Re: math problem
Newsgroups: sci.math
Status: OR

In sci.math you write:


>                            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


Let P(y) be the polynomial in question.
	First, notice that, since there is one sign-change in
the coefficients, P(y) has AT MOST one positive zero, by the 
Harriot-Descartes rule of signs.
	Second, notice that P(0) = 1 and P(y) -> -infty as y ->
+infty, so that P(y) must become negative for some y>0.  Therefore
there is AT LEAST one positive zero. 
	QED.



From victor@wri.com Fri Dec 11 16:45:09 1992
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From: Victor Adamchik <victor@wri.com>
Date: Fri, 11 Dec 92 15:44:20 CST
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Received: by NeXT Mailer (1.63)
To: li@sce.carleton.ca (Yao Li)
Subject: Re: math problem
Status: OR

In article <9212111851.AA14283@mariner.sce.carleton.ca> 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


0 == -a y^n + y^(n-1) + .. + 1  ==

	-a y^n + (y^n-1)/(y-1)  == 


	(y^n - 1 - a y^n y + a y^n) / ( y-1 ) 


=>

0 == a y^n y - (a+1) y^n + 1

=>

0 == y^n ( y - (a+1)/a ) + 1/a

=>

(a+1)/a - y == 1/(a y^n)

Last equation exactly has one solution y = 1.
 

So, your equation has a solution y0,
y0 == 1,     if a == n
y0 > 1,      if a < n     (or arctan(-n/a) < arctan(-1))
0 < y0 < 1,  if a > n     (or arctan(-n/a) > arctan(-1))
--
Victor Adamchik
victor@wri.com


From uphhswh%msu.dnet@terra.oscs.montana.edu Fri Dec 11 17:24:32 1992
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From: uphhswh%msu.dnet@terra.oscs.montana.edu
To: "li@sce.carleton.ca"%TERRA.dnet@terra.oscs.montana.edu
Subject: math problem
Status: OR

hi, I like math. 
y**n-1 +...+1 = (1-y**n)/(1-y)
so the equation becomes:
a*y**(n+1) +1 =(a+1)*y**n
the intersection of LHS curve and RHS curve is solution.
at y=0 and y=infinity, LHS>RHS
so if there is intersection (y>0), there must be even number's intersections.
so the claim of unique solution is not true.


you reminds me the days when I was so fascinated by math, (which resulted in
no.1 in provincial math contest)
have a nice day.
W. Huang

From stong@irus.rri.uwo.ca Fri Dec 11 16:03:37 1992
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Message-Id: <9212112103.AA06536@sun.irus.rri.uwo.ca>
Date: Fri, 11 Dec 92 16:03:50 EST
From: "Shidong Tong" <stong@irus.rri.uwo.ca>
To: li@sce.carleton.ca
Subject: math problem
Status: OR

Hi, Li Yao,

here is a simple counter example:

	f(y) = -y**6 + y**5 + y**4 + y**3 + y**2 + y + 1

you will find f(y) = 0 in the region [-0.85, -0.84] and [1.98, 1.99].


-- 
Shidong Tong
London, Ontario



Try this program, have fun!

------------------------ CUT HERE ----------------------------
/* scc.c
 * math problem
 * 1992-12-11	Shidong Tong
 */

#include <stdio.h>
#include <math.h>
main() {
	double	a, b, c;
	double	y, y2, y3, y4, y5, y6;
	double	f;

	puts( "--------------------------------------------" );
	puts( "f = -y**6 + y**5 + y**4 + y**3 +y**2 + y + 1" );
	puts( "--------------------------------------------" );
	puts( "   y      f(y)" );

	a = -0.9;
	b = -0.8;
	c = 0.01;
	for ( y=a; y<b; y+=c ) {
		y2 = y * y;
		y3 = y * y2;
		y4 = y * y3;
		y5 = y * y4;
		y6 = y * y5;
		f = - y6 + y5 + y4 + y3 + y2 + y + 1;
		printf( "%7.2f %7.3f\n", y, f );
	}

	puts( "" );
	a = 1.95;
	b = 2.05;
	c = 0.01;
	for ( y=a; y<b; y+=c ) {
		y2 = y * y;
		y3 = y * y2;
		y4 = y * y3;
		y5 = y * y4;
		y6 = y * y5;
		f = - y6 + y5 + y4 + y3 + y2 + y + 1;
		printf( "%7.2f %7.3f\n", y, f );
	}
}

From bs@linus.mitre.org Fri Dec 11 15:00:54 1992
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From: bs@linus.mitre.org
Message-Id: <9212111956.AA07985@gauss.mitre.org>
To: li%sce.carleton.CA@NNSC.NSF.NET
Subject: Re: math problem
Newsgroups: sci.math
Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
Cc: 
Status: OR


Your problem is trivial. There is indeed a unique positive solution.
The proof dates back to Descartes.

There is only one change of sign in the coefficients. By Descartes'
rule of signs the number of positive roots must be 1 or 0.


-- 
Bob Silverman
These are my opinions and not MITRE's.
Mitre Corporation, Bedford, MA 01730
"You can lead a horse's ass to knowledge, but you can't make him think"

From XI@BINAH.CC.BRANDEIS.EDU Thu Dec 10 23:16:57 1992
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Date: Thu, 10 Dec 1992 23:18 EDT
From: ZHIGANG XI <XI@BINAH.CC.BRANDEIS.EDU>
Subject: Re: math problem
To: li@sce.carleton.ca
Message-Id: <01GS62GBIWW0920FAE@BINAH.CC.BRANDEIS.EDU>
X-Vms-To: IN%"li@sce.carleton.ca"
Status: OR

In article <9212110333.AA13492@mariner.sce.carleton.ca>, 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

let x = 1/y, you get another face of your equation:
    a = x + x**2 + x**3 + ... + x**n
the function f(x) = x+ x**2 + x**3 + ... has value from 0 to positive
infinity, I guess your conjecture is right.


From somos@alpha.ces.cwru.edu Sat Dec 12 09:11:46 1992
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From: Michael Somos <somos@alpha.ces.cwru.edu>
Message-Id: <9212121412.AA05744@lychee.CES.CWRU.Edu>
To: li@sce.carleton.ca
Subject: Re: math problem
Status: OR

Yao Li, you posted a challenge problem to sci.math recently.  I think
that if x is positive, then y=x**(n+1) is convex, and hence can only
intersect a line at most twice.  y=x=1 is one point on the curve and
hence any line thru that point will have at most one other intersection.
A bit of thought shows that the line does have another intersection,
hence there exists a unique root.  Shalom, Michael Somos 
<somos@alpha.ces.cwru.edu>

From @amdahl.uts.amdahl.com,@juts.ccc.amdahl.com:xyl00@pteng.amdahl.com Sat Dec 12 12:56:01 1992
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From: xyl00@pteng.amdahl.com (Xiaoyang Luo)
Message-Id: <9212121754.AA09122@pteng.amdahl.com>
To: li@sce.carleton.ca
Subject: Math solution ...
Status: OR



     O.K. Here is the proof that the conjecture is correct.

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

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

     if a = n;

	it is obvious y = 1 is the unique solution. otherwise each item
	will have same sign.

     if a < n; 

	the solution will lie in [ (n/a)**(1/n), n/a ] 

      Look back at the equation, we have,

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

=>       (y**n - 1)/(y-1) = a *y**n 

=>       y-1 = (y**n - 1) / (a*y**n)   ( because y != 0 and y != 1 )

=>       y + 1/(a*y**n) - 1 + 1/a = 0   (2)
    
    that is the solutions for (1) and (2) are equivalent within 

	[(n/a)**(1/n), n/a] 

         let f(y) =  y + 1/(a*y**n) - 1 + 1/a 


	 f'(y) = 1-n/(a*y**(n+1),  f'(y) > 0, within [(n/a)**(1/n), n/a] 

	 so f(y) is monotonous.


    similarly, we could say same thing when a > n, except the range is
        [(n/a), (n/a)**(1/n)] 

     
    BTW, my major is computer science. so some terms might be wrong.
	 

       xyl00  at  pteng.amdahl.com



From cheekai@ncb.gov.sg Sun Dec 13 03:16:13 1992
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From: cheekai@ncb.gov.sg (Chin Chee-Kai)
Message-Id: <9212130815.AA00411@ncb.gov.sg>
To: li@sce.carleton.ca
Subject: Re:  math problem
Status: OR

>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?

Yes, the conjecture is true.
My younger brother gave a proof as follows:
-----------------------------------------------------------------------
Call the conjecture P.
Rearraging, one gets:
	y^{n-1} + ... + y + 1 = a*y^n		---(1)

=>	(y^n - 1)/(y-1) = a*y^n
=>	y^n - 1 = a*y^(n+1) - a*y^n
Let	f(y) = a*y^(n+1) - (a+1)*y^n + 1
So	f'(y) = a(n+1)*y^n - (a+1)n*y^(n-1)

Observe that f(0)=1, & f(1)=0.  Also, f'(0)=0, & f'(1)=a-n
Case 1:	a-n = 0
	=> y=1 is a multiple-root of f(y) and also satisfies eqn (1)
	Thus, P is true in this case.
Case 2: a-n < 0
	y=1 does not satisfy eqn (1) (although it is a root of f(y)=0)
	But f(1+) < 0 (where 1+ means the positive neighborhood of 1)
	because f'(1) < 0 and f'(y) is continuous.
	Since f(y) --> +infinity as y --> +infinity, this means that
	f(y) = 0 for some 1 < y < +infinity.
	=> P is true in this case.
Case 3: a-n > 0
	y=1 does not satisfy eqn (1) (although it is a root of f(y)=0)
	Since f'(1) > 0, and f'(y) is continuous, so f'(1-) > 0
	=> f(1-) < 0.  But f(0) > 0 (= 1).  Hence there exists a
	0 < y < 1 such that f(y) = 0.
	=> P is also true.
--------------------------------------------------------------------------

Chin Chee-Kai

From erick@sfu.ca Sun Dec 13 19:59:18 1992
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From: erick@sfu.ca
Message-Id: <9212140059.AA07148@fraser.sfu.ca>
To: li@sce.carleton.ca
Subject: Re: math problem
Newsgroups: sci.math
Organization: Simon Fraser University, Burnaby, B.C., Canada
Cc: 
Status: OR

In article <9212111851.AA14283@mariner.sce.carleton.ca> you write:
>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.

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

So, (-a*y**n)(y-1) + y**n = 1
    (-a*y**(n+1)) + (1+a)y**n = 1

Call the LHS f(y).  Consider the derivative of f(y):
    -a(n+1)y**n + n(1+a)y**(n-1) = y**(n-1) ( n(1+a) - a(n+1)y )

Therefore on the positive real axis, f(y) is increasing for y <= (an+n)/(an+a),
and decreasing for y >= (an+n)/(an+a).  Also, f(0)=0 and f(y) -> -oo as y -> oo.
Since f(1) = 1, either f'(1) = 0 and f(y) <> 1 for y <> 1, or there is exactly
one other y with f(y)=1.

But if f'(1) = 0, then (an+n)/(an+a) = 1, so a = n.  But then y=1 is a solution
(and the unique solution, because f attains a maximum at y=1) of the original
polynomial.

Thus the conjecture is true.  QED

-- 
 -- Erick, the only non-trivial solution in the integers to c = a^b = b^a :-)


From torbenm@thor.diku.dk Mon Dec 14 07:13:54 1992
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From: Torben AEgidius Mogensen <torbenm@diku.dk>
Message-Id: <9212141214.AA05601@thor.diku.dk>
To: li@sce.carleton.ca
Subject: Re: math problem
Newsgroups: sci.math
Status: OR

In sci.math you write:

>                            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?

First part: there is a positive real solution.

Note that the value for y=0 is 1. Note that y**n grows faster than the
remainder of the polynomium. Thus the polynomium will eventually
become negative. Thus there is a y>0 where it is 0.

Second part: there is only one solution.

The polynomium will grow until a*y**n grows faster than y**(n-1) +
y**(n-2) + ... + y + 1. After this it falls monotonically, as -a*y**n
increases dominance. Since the polynomium starts at 1 in y=0, it will
remain positive until it starts to decrease. After this point there
can only be one intersection with the y-axis. Note that there may be
negative roots.

	Torben Mogensen (torbenm@diku.dk)

Newsgroups: soc.culture.china