\documentclass{article} \usepackage{axiom,amsthm,amsmath} \newtheorem{ToDo}{ToDo}[section] \begin{document} \title{solve.spad} \author{Martin Rubey} \maketitle \begin{abstract} \end{abstract} \tableofcontents \section{domain SUPEXPR SparseUnivariatePolynomialExpressions} This domain is a hack, in some sense. What I'd really like to do - automatically - is to provide all operations supported by the coefficient domain, as long as the polynomials can be retracted to that domain, i.e., as long as they are just constants. I don't see another way to do this, unfortunately. <>= )abb domain SUPEXPR SparseUnivariatePolynomialExpressions SparseUnivariatePolynomialExpressions(R: Ring): Exports == Implementation where Exports == UnivariatePolynomialCategory R with if R has TranscendentalFunctionCategory then TranscendentalFunctionCategory Implementation == SparseUnivariatePolynomial R add if R has TranscendentalFunctionCategory then exp(p: %): % == ground? p => coerce exp ground p output(hconcat("exp p for p= ", p::OutputForm))$OutputPackage error "SUPTRAFUN: exp only defined for elements of the coefficient ring" sin(p: %): % == ground? p => coerce sin ground p output(hconcat("sin p for p= ", p::OutputForm))$OutputPackage error "SUPTRAFUN: sin only defined for elements of the coefficient ring" asin(p: %): % == ground? p => coerce asin ground p output(hconcat("asin p for p= ", p::OutputForm))$OutputPackage error "SUPTRAFUN: asin only defined for elements of the coefficient ring" cos(p: %): % == ground? p => coerce cos ground p output(hconcat("cos p for p= ", p::OutputForm))$OutputPackage error "SUPTRAFUN: cos only defined for elements of the coefficient ring" acos(p: %): % == ground? p => coerce acos ground p output(hconcat("acos p for p= ", p::OutputForm))$OutputPackage error "SUPTRAFUN: acos only defined for elements of the coefficient ring" @ \section{package UTSSOL TaylorSolve} [[UTSSOL]] is a facility to compute the first few coefficients of a Taylor series given only implicitely by a function [[f]] that vanishes when applied to the Taylor series. It uses the method of undetermined coefficients. \begin{ToDo} Could I either \begin{itemize} \item take a function [[UTSCAT F -> UTSCAT F]] and still be able to compute with undetermined coefficients, or \item take a function [[F -> F]], and do likewise? \end{itemize} Let's see. Try to compute the equation without resorting to power series. I.e., % [[c: SUP SUP F]], and [[f: SUP SUP F -> SUP SUP F]]. Won't this make the computation of coefficients terribly slow? I could also try to replace transcendental kernels with variables\dots Unfortunately, [[SUP F]] does not have [[TRANFUN]] -- well, it can't, of course. However, I'd like to be able to compute % [[sin(1+monomial(1,1)$UFPS SUP EXPR INT)]]. \end{ToDo} <>= )abb package UTSSOL TaylorSolve TaylorSolve(F, UTSF, UTSSUPF): Exports == Implementation where F: Field SUP ==> SparseUnivariatePolynomialExpressions UTSF: UnivariateTaylorSeriesCategory F UTSSUPF: UnivariateTaylorSeriesCategory SUP F NNI ==> NonNegativeInteger Exports == with seriesSolve: (UTSSUPF -> UTSSUPF, List F) -> UTSF Implementation == add <> @ <>= seriesSolve(f, l) == c1 := map(#1::(SUP F), l)$ListFunctions2(F, SUP F)::(Stream SUP F) coeffs: Stream SUP F := concat(c1, generate(monomial(1$F,1$NNI))) -- coeffs: Stream SUP F := concat(c1, monomial(1$F,1$NNI)) @ [[coeffs]] is the stream of the already computed coefficients of the solution, plus one which is so far undetermined. We store in [[st.2]] the complete stream and in [[st.1]] the stream starting with the first coefficient that has possibly not yet been computed. \begin{ToDo} The mathematics is not quite worked out. If [[coeffs]] is initialized as stream with all coefficients set to the \emph{same} transcendental value, and not enough initial values are given, then the missing ones are implicitely assumed to be all identical. It may well happen that a solution is produced, although it is not uniquely determined\dots \end{ToDo} <>= st: List Stream SUP F := [coeffs, coeffs] @ Consider an arbitrary equation $f\big(x, y(x)\big)=0$. When setting $x=0$, we obtain $f\big(0, y(0)\big)=0$. It is not necessarily the case that this determines $y(0)$ uniquely, so we need one initial value that satisfies this equation. \begin{ToDo} [[seriesSolve]] should check that the given initial values satisfy $f\big(0, y(0), y'(0),...\big) = 0$. \end{ToDo} Now consider the derivatives of $f$, where we write $y$ instead of $y(x)$ for better readability: \begin{equation*} \frac{d}{dx}f(x, y)=f_1(x, y) + f_2(x, y)y^\prime \end{equation*} and \begin{align*} \frac{d^2}{dx^2}f(x, y)&=f_{1,1}(x, y)\\ &+f_{1,2}(x, y)y^\prime\\ &+f_{2,1}(x, y)y^\prime\\ &+f_{2,2}(x, y)(y^\prime)^2\\ &+f_2(x, y)y^{\prime\prime}. \end{align*} In general, $\frac{d^2}{dx^2}f(x, y)$ depends only linearly on $y^{\prime\prime}$. \begin{ToDo} This points to another possibility: Since we know that we only need to solve linear equations, we could compute two values and then use interpolation. This might be a bit slower, but more importantly: can we still check that we have enough initial values? Furthermore, we then really need that $f$ is analytic, i.e., operators are not necessarily allowed anymore. However, it seems that composition is allowed. \end{ToDo} <>= next: () -> F := nr := st.1 res: F if ground?(coeff: SUP F := nr.1)$(SUP F) @ %$ If the next element was already calculated, we can simply return it: <>= then res := ground coeff st.1 := rest nr else @ Otherwise, we have to find the first non-satisfied relation and solve it. It should be linear, or a single non-constant monomial. That is, the solution should be unique. <>= ns := st.2 eqs: Stream SUP F := coefficients f series ns while zero? first eqs repeat eqs := rest eqs eq: SUP F := first eqs if degree eq > 1 then if monomial? eq then res := 0 else output(hconcat("The equation is: ", eq::OutputForm)) $OutputPackage error "seriesSolve: equation for coefficient not linear" else res := (-coefficient(eq, 0$NNI)$(SUP F) /coefficient(eq, 1$NNI)$(SUP F)) nr.1 := res::SUP F -- concat!(st.2, monomial(1$F,1$NNI)) st.1 := rest nr res series generate next @ %$ \section{package EXPRSOL ExpressionSolve} \begin{ToDo} I'd really like to be able to specify a function that works for all domains in a category. For example, [[x +-> y(x)^2 + sin x + x]] should \lq work\rq\ for [[EXPR INT]] as well as for [[UTS INT]], both being domains having [[TranscendentalFunctionCategory]]. \end{ToDo} <>= )abb package EXPRSOL ExpressionSolve ExpressionSolve(R, F, UTSF, UTSSUPF): Exports == Implementation where R: Join(OrderedSet, IntegralDomain, ConvertibleTo InputForm) F: FunctionSpace R UTSF: UnivariateTaylorSeriesCategory F SUP ==> SparseUnivariatePolynomialExpressions UTSSUPF: UnivariateTaylorSeriesCategory SUP F OP ==> BasicOperator SY ==> Symbol NNI ==> NonNegativeInteger MKF ==> MakeBinaryCompiledFunction(F, UTSSUPF, UTSSUPF, UTSSUPF) Exports == with seriesSolve: (F, OP, SY, List F) -> UTSF Implementation == add <> @ The general method is to transform the given expression into a form which can then be compiled. There is currently no other way in Axiom to transform an expression into a function. We need to replace the differentiation operator by the corresponding function in the power series category, and make composition explicit. Furthermore, we need to replace the variable by the corresponding variable in the power series. It turns out that the compiler doesn't find the right definition of [[monomial(1,1)]]. Thus we introduce it as a second argument. In fact, maybe that's even cleaner. Also, we need to tell the compiler that kernels that are independent of the main variable should be coerced to elements of the coefficient ring, since it will complain otherwise. <>= opelt := operator("elt"::Symbol)$OP opdiff := operator("D"::Symbol)$OP opcoerce := operator("coerce"::Symbol)$OP replaceDiffs: (F, OP, Symbol) -> F replaceDiffs (expr, op, sy) == lk := kernels expr for k in lk repeat if freeOf?(coerce k, sy) then expr := subst(expr, [k], [opcoerce [coerce k]]) if is?(k, op) then arg := first argument k if arg = sy::F then expr := subst(expr, [k], [(name op)::F]) else expr := subst(expr, [k], [opelt [(name op)::F, replaceDiffs(arg, op, sy)]]) -- => "iterate" if is?(k, %diff) then args := argument k differentiand := replaceDiffs(subst(args.1, args.2 = args.3), op, sy) expr := subst(expr, [k], [opdiff differentiand]) -- => "iterate" expr seriesSolve(expr, op, sy, l) == ex := replaceDiffs(expr, op, sy) f := compiledFunction(ex, name op, sy)$MKF seriesSolve(f(#1, monomial(1,1)$UTSSUPF), l)$TaylorSolve(F, UTSF, UTSSUPF) @ %$ \section{Bugs} <>= seriesSolve(sin f x / cos x, f, x, [1])$EXPRSOL(INT, EXPR INT, UFPS EXPR INT, UFPS SUPEXPR EXPR INT) @ returns \begin{verbatim} (((0 . 1) 0 . 1) NonNullStream # . UNPRINTABLE) \end{verbatim} but <>= U ==> UFPS SUPEXPR EXPR INT seriesSolve(s +-> sin s *((cos monomial(1,1)$U)**-1)$U, f, x, [0])$EXPRSOL(INT, EXPR INT, UFPS EXPR INT, UFPS SUPEXPR EXPR INT) @ works. This is probably due to missing [[/]] in [[UFPS]]. \section{License} <>= -- Copyright (C) 2006 Martin Rubey -- -- This program is free software; you can redistribute it and/or -- modify it under the terms of the GNU General Public License as -- published by the Free Software Foundation; either version 2 of -- the License, or (at your option) any later version. -- -- This program is distributed in the hope that it will be -- useful, but WITHOUT ANY WARRANTY; without even the implied -- warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR -- PURPOSE. See the GNU General Public License for more details. @ <<*>>= <> <> <> <> @ \end{document}