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# How to determine the coefficients of a polynomial when they depend on variables?

Suppose that EXP2 is a polynomial expression in the variables X_1,X_2,...,X_n with coefficients that depend on the variables B_1,B_2,...,B_m. Suppose EXP1 is an expression equal to EXP2, but scrambled.

Given EXP1, how can I automatically obtain EXP2?

For example, if I have

a=var('a')

b=var('b')

c=var('c')

d=var('d')

e=var('e')

eqn=a==bd^2+ce^2+e(b+2c*d),

how can I automatically rewrite a as a polynomial in b and c, so that I get an equation that looks like

a==b(d^2+e)+c(e^2+2ed)?

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## Comments

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## 1 Answer

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The best way to work with variables b_i and x_i as you want is to define multivariate polynomial rings Rb and Rx and inject variables. Then any polynomial expression in the b_i's and x_i's will be transformed as a polynomial in the x_i's with coefficients in polynomials in the b_i's.

sage: Rb = PolynomialRing(ZZ,10,name='b')
sage: Rx = PolynomialRing(Rb,10,name='x')

sage: Rb.inject_variables()
Defining b0, b1, b2, b3, b4, b5, b6, b7, b8, b9
sage: Rx.inject_variables()
Defining x0, x1, x2, x3, x4, x5, x6, x7, x8, x9

sage: b1 * (x2 + x3) + b2 * (x3 + x8)
b1*x2 + (b1 + b2)*x3 + b2*x8


For your baby example, you could work in the symbolic ring and just ask coefficients,

sage: var('b c d e')
(b, c, d, e)
sage: a = b*d^2+c*e^2+e*(b+2*c*d)
sage: a
b*d^2 + c*e^2 + (2*c*d + b)*e
sage: a.coefficient(b)
d^2 + e
sage: a.coefficient(c)
2*d*e + e^2


or you could mix the two in the following way.

sage: S.<d,e> = PolynomialRing(QQ)
sage: R.<b,c> = PolynomialRing(S)
sage: R
Multivariate Polynomial Ring in b, c over Multivariate Polynomial Ring in d, e over Rational Field
sage: var('b c d e')
(b, c, d, e)
sage: a = b*d^2+c*e^2+e*(b+2*c*d)
sage: a
b*d^2 + c*e^2 + (2*c*d + b)*e
sage: R(a)
(d^2 + e)*b + (2*d*e + e^2)*c


But as a rule, it is best to avoid the symbolic ring altogether, whenever possible. And the way your main question is formulated, the most sensible approach is the polynomial rings one.

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## Comments

Thank you very much! This answer provides a good solution both to my general question and a simple solution to my specific problem (since in my case I'm really just dealing with a scalar linear function of one set of variables with coefficients depending on another set of variables).

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Asked: 2014-07-14 05:58:46 +0200

Seen: 164 times

Last updated: Jul 14 '14