rational expression simplification with algebraic side

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MITACS Symbolic Analysis Project – Supervisor Peter Borwein Centre for Experimental and Constructive Mathematics Computer Algebra Group Simon Fraser University Ideal Quotients and Simplification The ideal quotient operation is the key to our method. Once again, we use Gr ¨ obner bases for the actual computations. Definition Let I and J be ideals of k [x 1 ,...,x n ]. The ideal quotient I : J is the ideal {f k [x 1 ,...,x n ]: fh I for all h J }. It may not be immediately obvious what this does, so consider this example. Suppose a and b are univariate polynomials over a field and we compute a : b. What do we get ? We must get the ideal of polynomials whose product with b is divisible by a. A minimal element will be lcm(a, b)/b = a/gcd(a,b), which is a numerator for a simplified fraction equivalent to a b . An initial method is the following: Given a/b, construct a, I and b, I in k [x 1 ,...,x n ]. Compute the ideal quotient a, I : b, I using Gr ¨ obner bases, and choose a minimal element c I to be a numerator. Divide bc by a modulo I using Gr ¨ obner bases. This will always succeed. Obtain d with bc ad mod I . Then a/b reduces to c/d modulo I . Reduced Canonical Forms Unfortunately the method above does not always produce a desirable result. Our solution is to construct the module of equivalent fractions, building on the quotient construction above. Given a/b, we generate M = {(c, d): ad - bc 0 mod I } as a module over k [x 1 ,...,x n ]. Using Gr¨ obner bases for modules and a term over position module order, we compute a fraction whose greatest monomial is minimal with respect to a given monomial order. The result is unique if the numerator and denominator of the fraction are reduced modulo I , where I is a prime ideal. Minimizing the Total Degree To address the simplification problem, we must construct fractions with minimal total degree. To wit: If the polynomials and the ideal are homogeneous, then the initial method above will produces a fraction with minimal total degree, provided the chosen numerator has minimal total degree. For the general case we conduct a search. Gr ¨ obner basis methods can not be applied, since they operate on only one term at a time. Future Work It is often possible to recover factorizations of f modulo I from a prime decomposition of f,I . The idea is promising and it is consistent with our understanding, however major technical obstacles remain. We are working to address these issues with the goal of developing a robust algorithm for computing factorizations in k [x 1 ,...,x n ] /I . Introduction Our problem was to simplify rational expressions in the presence of algebraic side relations. This problem can occur quite easily, such as when an expression involves functions. For example: (1 + cos(x)) sin(x) - cos 2 (x)+1 cos 4 (x) - 2 cos 2 (x) + sin(x)+1 -→ sin(x) + cos(x)+1 (1 - cos 2 (x)) sin(x)+1 The difficulty of this problem is threefold: A single expression can have many equivalent representations. Expressions may not factor uniquely, e.g. sin 2 (x) = (1 + cos(x)) (1 - cos(x)) Sometimes we must simplify in a way that does not correspond to cancelling a common divisor. i.e. a b c d but c a and d b. Our Idea Rather than work with the expressions themselves, we had to construct classes of equivalent expressions and compute with them. We first replace any functions by variables. The result is a rational expression a b where a and b are in a polynomial ring k [x 1 ,...,x n ]. The side relations are written in the form f (x 1 ,...,x n )=0. The set of all expressions equivalent to zero is an ideal of k [x 1 ,...,x n ]. Definition Let R be a ring, a set I R is an ideal if i) f + g I for all f,g I ii) fh I for all f I and all h R We let I = f 1 ,...,f s be the ideal generated by our side relations, then we have the following correspondence: Elements of k [x 1 ,...,x n ] /I Ideals of k [x 1 ,...,x n ] f ←→ f,I So that instead of computing with a polynomial f , we will compute with the ideal generated by f together with all of the side relations. Making it Work To compute with ideals of k [x 1 ,...,x n ] we use Gr¨ obner bases. Simply put, a Gr¨ obner basis contains all of the minimal elements of an ideal. We can use them to cancel out the parts of an expression which are in the ideal, and simplify what is left to a canonical form. Perhaps most importantly, they allow us to divide by an ideal’s gener- ators in much the same way as with ordinary univariate polynomials. Example The polynomial y 2 +1 is in the ideal xy +1,x 2 +1. Using Gr¨ obner bases we can confirm this fact, and also obtain y 2 + 1 = (1 - xy )(xy + 1) + y 2 (x 2 + 1) We can apply this technique to perform polynomial division modulo I . In the example above, we could have asked whether xy +1 divides y 2 +1 modulo x 2 +1. We would obtain the answer yes, along with a quotient (1 - xy ). Roman Pearce Michael Monagan Rational Expression Simplification with Algebraic Side Relations

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Page 1: Rational Expression Simplification with Algebraic Side

MITACS Symbolic Analysis Project – Supervisor Peter BorweinCentre for Experimental and Constructive MathematicsComputer Algebra GroupSimon Fraser University

Ideal Quotients and SimplificationThe ideal quotient operation is the key to our method. Once again, weuse Grobner bases for the actual computations.

Definition Let I andJ be ideals ofk[x1, . . . , xn]. The ideal quotientI : J is the ideal{f ∈ k[x1, . . . , xn] : f h ∈ I for all h ∈ J}.

It may not be immediately obvious what this does, so consider thisexample. Supposea andb are univariate polynomials over a field andwe compute〈a〉 : 〈b〉. What do we get ?

We must get the ideal of polynomials whose product withb is divisibleby a. A minimal element will belcm(a, b)/b = a/gcd(a,b), which is anumerator for a simplified fraction equivalent toa

b.

An initial method is the following:

•Givena/b, construct〈a, I〉 and〈b, I〉 in k[x1, . . . , xn].

•Compute the ideal quotient〈a, I〉 : 〈b, I〉 using Grobner bases,and choose a minimal elementc 6∈ I to be a numerator.

•Divide b c by a moduloI using Grobner bases. This willalways succeed. Obtaind with b c ≡ a d mod I.

•Thena/b reduces toc/d moduloI. �

Reduced Canonical FormsUnfortunately the method above does not always produce a desirableresult. Our solution is to construct themoduleof equivalent fractions,building on the quotient construction above. Givena/b, we generate

M = {(c, d) : ad− bc ≡ 0 mod I}

as a module overk[x1, . . . , xn]. Using Grobner bases for modules and aterm over positionmodule order, we compute a fraction whose greatestmonomial is minimal with respect to a given monomial order. Theresult is unique if the numerator and denominator of the fraction arereduced moduloI, whereI is a prime ideal.

Minimizing the Total DegreeTo address the simplification problem, we must construct fractions withminimal total degree. To wit:

• If the polynomials and the ideal arehomogeneous, then the initialmethod above will produces a fraction with minimal total degree,provided the chosen numerator has minimal total degree.

•For the general case we conduct a search. Grobner basis methods cannot be applied, since they operate on only one term at a time.

Future Work• It is often possible to recover factorizations off modulo I from a

prime decomposition of√〈f, I〉.

•The idea is promising and it is consistent with our understanding,however major technical obstacles remain.

•We are working to address these issues with the goal of developing arobust algorithm for computing factorizations ink[x1, . . . , xn] / I.

IntroductionOur problem was to simplify rational expressions in the presence ofalgebraic side relations. This problem can occur quite easily, such aswhen an expression involves functions. For example:

(1 + cos(x)) sin(x)− cos2(x) + 1

cos4(x)− 2 cos2(x) + sin(x) + 1−→ sin(x) + cos(x) + 1

(1− cos2(x)) sin(x) + 1

The difficulty of this problem is threefold:

•A single expression can have many equivalent representations.

•Expressions may not factor uniquely, e.g.

sin2(x) = (1 + cos(x)) (1− cos(x))

•Sometimes we must simplify in a way that does not correspond tocancelling a common divisor. i.e.a

b →cd but c - a andd - b.

Our IdeaRather than work with the expressions themselves, we had to constructclassesof equivalent expressions and compute with them.

•We first replace any functions by variables. The result is a rationalexpressionab wherea andb are in a polynomial ringk[x1, . . . , xn].

•The side relations are written in the formf (x1, . . . , xn) = 0.

•The set of all expressions equivalent to zero is anidealof k[x1, . . . , xn].

Definition Let R be a ring, a setI ⊆ R is an ideal if

i) f + g ∈ I for all f, g ∈ I

ii) f h ∈ I for all f ∈ I and allh ∈ R

We let I = 〈f1, . . . , fs〉 be the ideal generated by our side relations,then we have the following correspondence:

Elements ofk[x1, . . . , xn] / I Ideals ofk[x1, . . . , xn]

f ←→ 〈f, I〉

So that instead of computing with a polynomialf , we will computewith the ideal generated byf together with all of the side relations.

Making it Work•To compute with ideals ofk[x1, . . . , xn] we useGrobner bases.

•Simply put, a Grobner basis contains all of the minimal elements ofan ideal. We can use them to cancel out the parts of an expressionwhich are in the ideal, and simplify what is left to a canonical form.

•Perhaps most importantly, they allow us to divide by an ideal’s gener-ators in much the same way as with ordinary univariate polynomials.

Example The polynomialy2+1 is in the ideal〈xy+1, x2+1〉. UsingGrobner bases we can confirm this fact, and also obtain

y2 + 1 = (1− xy) (xy + 1) + y2 (x2 + 1)

We can apply this technique to perform polynomial division moduloI.In the example above, we could have asked whetherxy + 1 dividesy2 + 1 modulo〈x2 + 1〉. We would obtain the answeryes, along with aquotient(1− xy).

Roman Pearce Michael Monagan

Rational Expression Simplification withAlgebraic Side Relations