Re: A Math Challenge (kind of) Message #5 Posted by C.Ret on 16 Aug 2011, 7:39 p.m., in response to message #4 by Namir
Namir,
I have to repeat the question of [u][b]Marcus[/u][/b] which is of primary importance. If the three roots you give are the only one of the polymer, there is no solution to your problem.
If 1, 1/8 and 1 are the only and simple roots of the polymer P(x), then P(x) is exactly of degree 3.
We have to found the four coefficients a, b, c, and d of P(x):
P(x) = a.x^{3} + b.x^{2} + c.x + d.
P’(x) = 3a.x^{2} + 2b.x + c
P"(x) = 6a.x + 2b
For the three roots x_{1}= 1, x_{2} = 1/8 and x_{3} = 1;
P( 1 ) = 0 a + b – c + d == 0 (eq.1)
P( 1/8 ) = 0 a/512 + b/64 + c/8 + d == 0 (eq.2)
P( +1 ) = 0 a + b + c + d == 0 (eq.3)
From (eq.1) and (eq.3) we have 2b + 2d == 0, thus d = b.
As explain by [u][b]Marcus[/u][/b] at the extremum (minimum and maximum), the first derivate P’(x) is zero, thus :
P’( 0 ) = 0 c == 0 (eq.4)
P’( 1/2 ) = 0 3a/4 + b + c == 0 (eq.5)
From (eq.4) and (eq.5) b = 3a/4
The point 0 is a maximum if P"(0) negative.
P"( 0 ) < 0 b < 0 (eq.6)
The point ½ is a minimum if P"(1/2) is positive.
P"( 1/2 ) > 0 3a + 2b > 0 (eq.6)
3a  3a/2 > 0
3a/2 > 0
At this point , we have to found a set of coefficients :
a > 0, b= 3a/4, c=0 and d=b=3a/4.
From (eq.2) :
a/512  3a/128  3a/4 == 0
a == 0
Thus, no such a polymer exist ! It may only be a solution if the polymer is of a greater degree.
OK, I found no evident solution, but my approch clearly show that an appropriate algorithm may be lead by linear algebra, system of equations and inequations.
Edited: 16 Aug 2011, 7:59 p.m.
