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I have a polynomial that is the related to a solution of a system of 4 differential equations that represent that density matrix evolution of some quantum system (details not particularly important). More specifically the roots of this polynomial are present as coefficients of the exponentials that arise in the solution. The system is linear and I solve it with Mathematica.

The system (and the polynomial) have three constants that represent the parameters of the quantum system, I'll label them a,b,c. and the coefficients of the polynomial $A,B,C,D$ in the standard convention $$f(x) = Ax^3 + Bx^2 +Cx +D$$

First the constant term.

$$D = F(3,0,0)a^3 + F(2,1,0)a^2 b + F(2,0,1)a^2c + ...+ F(0,0,3)c^3$$

where $F(i,j,k), i+j+k =3$ represents some numerical factor in front of every permutation

For the $x$ term

$$C = G(2,0,0)a^2 + G(1,1,0)ab + G(1,0,1)ac + ...G(0,0,2)c^2$$

Where the similar to the above $G(i,j,k), i+j+k = 2$, with the relation $G(i,j,k) = (k+1)F(i,j,k+1)$

The $x^2$ term follows a similar pattern

$$B = H(1,0,0)a + H(0,1,0)b + H(0,0,1)c$$

where $H(i,j,k) = (k+1)F(i,j,k+2), i+j+k =1$

and the single $x^3$ term has a coefficient of the form $A = 3*F(0,0,3)$

This appears to be some sort of permutation of all three constants, but I can't for the life of me identify any sort of overarching pattern (if there is one). There also doesn't appear to be any rhyme or reason to the pattern of the numerical coefficients themselves, but this is unsurprising.

I admit that this is a Mathematical fishing expedition, if this type of question is inappropriate I understand if it gets closed. My motivation behind asking this is to make this polynomial more tenable, with some 40-odd terms in the polynomial, the exact equation for each root is a quite a beast to behold.

note: In case it wasn't clear - the constants $a,b,c$ appear in each coefficient as the sum of all permutations of their product

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No, it is not clear. Do you want to say that G(1,1,0)=G(1,0,1)? This seems to be the gist of your last sentence, but this does not work well with the relation between F and G. –  Phira Oct 31 '11 at 11:02

1 Answer 1

Consider coefficients $A$, $B$, $C$ as functions of $a,b,c$. Then $$ C(a,b,c)=\frac{\partial}{\partial c}D(a,b,c), $$ $$ B(a,b,c)=\frac{\partial}{\partial c}\frac{D(a,b,c)-D(a,b,0)}c, $$ $$ A(a,b,c)=\frac{C(a,b,c)-C(a,b,0)}{c^2}. $$ It seems that $x$ in the polynomial $f(x)$ takes place of $c$. May be it is possible to obtain something interesting considering $f$ as homogeneous polynomial of $a,b,c,x$.

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