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Symbolic Integration Lecture 14 the Problem of Integration in Finite Terms

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Symbolic Integration Lecture 14 the Problem of Integration in Finite Terms The problem of integration in finite terms 1. The Freshman Calculus Problem 2. The Artificial Intelligence Problem Symbolic Integration 3. The Algebraic (Rational Function) Problem 4. The Decision Problem (Liouville-Risch...) 5. The Computer Algebra System problem. Lecture 14 6. Extensions to definite integration, numerical integration, approximation by Taylor series, Laurent series, asymptotic series. Richard Fateman CS 282 Lecture 14 1 Richard Fateman CS 282 Lecture 14 2 The Freshman Calculus Problem. Find the Freshman Calc Æ AI: James Slagle’s flaws. approach 1. Humans are intelligent • AND-OR trees for evaluation of difficulty of 2. MIT freshman are especially intelligent humans. a path (do we abandon this route or plow on?) 3. Freshman study integral calculus (this is in 1960’s before Advanced Placement). 4. Therefore solving freshman calculus problems or or requires intelligence. 5. If a computer can solve freshman calculus problems, and then it is intelligent (An example of Artificial Intelligence). 6. To solve these problems, imitate freshman students. 7. Future work: do the rest of math, and the rest of • Game playing: could be any complex task. AI. Richard Fateman CS 282 Lecture 14 3 Richard Fateman CS 282 Lecture 14 4 Slagle’s program SAINT Symbolic Automatic The Algebraic (Rational Function) Problem INTegrator • ELINST: elementary instance expression • Many attempts to do this right, e.g. Theses of pattern matching (in assembler). R.Tobey, E. Horowitz, and with extensions, M. • Simplification program (in assembler). Rothstein, B. Trager. • Used lisp prefix expressions (* x (sin x)) • Main idea is to take a rational expression • First major lisp program (in Lisp 1.5). q(x)/r(x), express it as a polynomial + “proper” • Took about a minute per problem (about the fraction P+Q/R with deg(Q)<= deg(R). same time as a student!) running uncompiled on Integrate P, a polynomial, trivially. an IBM 7090 (32k word x 36 bits/word). • Could not do rational function integration (out Expand Q/R in partial fractions. of space.) Richard Fateman CS 282 Lecture 14 5 Richard Fateman CS 282 Lecture 14 6 1 Partial fractions “trivialized” an example of the general cubic... • From Q/R Find all the complex roots of R, a0, ...,an. Express Q/R as ∑ ci/(x-ai) which integrates to ∑ ci log(x-ai). Can we always find {ai}? Complex roots can be found numerically. Can we find ci? It turns out that we can expand the summation and solve everywhere... with a big problem.. Richard Fateman CS 282 Lecture 14 7 Richard Fateman CS 282 Lecture 14 8 The answer is unmanageable even if we can What’s the problem? compute it. 3 2 • For the cubic, each c looks something like this • If the denominator R= s3*x +s2*x +s1*x+s0, i then the each of the roots ai look like this: • and so it is ugly. Often enough these expressions can be simplified, and so the search has been to find a minimal algebraic extension in which the ci can be expressed. We DON’T need to see the ai, but can use facts like a1a2a3 is constant term of polynomial R; we have neat forms for all symmetric functions of roots. Richard Fateman CS 282 Lecture 14 9 Richard Fateman CS 282 Lecture 14 10 This trick works for numeric denominators For a general quartic, it gets worse. too • And for order 5 or more there is in general no formula “in radicals” for the roots. That is, we can’t compute the ai in general for symbolic polynomials • So we can find approximate numerical roots in C if the problem has a denominator solely in Q[x], all bets are off “symbolically”. • The solution is “RootSum(f(x),poly,x)” expressions where we have ∑ f(ai) where {ai} are the roots of the polynomial poly. Richard Fateman CS 282 Lecture 14 11 Richard Fateman CS 282 Lecture 14 12 2 Rothstein (etal), find the ci easier Squarefree isn’t a problem • Let B(x) be square free and of degree >= A(x), • Let the problem be ∫A(x)/B(x)n with deg(A)<=deg(B). • Since B itself is square free, its GCD with B’=dB/dx is •then ∫ (A/B)dx = ci log(vi) 1. Thus we can compute c,e such that c*B+e*B’=1. •where ci, are the distinct roots of the • Now multiply through by A to get cAB+eB’A=A, and polynomial R(c)=Resultant(x,A(x)-cB’(x),B(x)) substitute in the original ∫A(x)/B(x)n to get n and v =gcd(A(x)-c B’(x),B(x)) for i = 1,...n. ∫(cAB+eB’A)/B(x) . Now dividing through we get i i ∫ cA/Bn-1 + ∫ eA* (B’/Bn) where the second term can be integrated by parts to lower the power of Bn in the • We need Square-free computation and denominator. n Resultant wrt x (reminder: ∫ u dv = uv-∫ v du. Let u =eA, dv=B’/B , v=1/((1-n)Bn-1), du=(eA)’, so ∫ v du comes out as <something>/Bn-1+ ∫ <something>/Bn-1 . Iterate until Bn = B1 .) Richard Fateman CS 282 Lecture 14 13 Richard Fateman CS 282 Lecture 14 14 Resultant isn’t a problem The decision problem • Well, actually it is something we’d prefer not • Liouville [1809-1882] During the period (1833-1841) presented a theory of integration; proved elliptic to do since it takes a while, but we have integrals cannot have elementary expressions. algorithms for it. • Various other writers advanced the subject in late 1800’s • J. Ritt (1948) Integration in Finite Terms Columbia • Conclusion: we can do rational function Univ. Press integration pretty well. But it took us into the • M. Rosenlicht (AMM. 1972) Integration in Finite mid-1980s to do it right. Terms • R. Risch (1968) [unreadable] • B. Trager, J. Davenport, M.Bronstein [implementation] Richard Fateman CS 282 Lecture 14 15 Richard Fateman CS 282 Lecture 14 16 Risch’s result What’s a Differential Field? Theorem: Let K be a differential field and f be from K . Then an elementary extension of the field K which has the • Equip a field with an additional operator D same field of constants as K and contains an element g which satisfies identities parallel to the rules such that g’ = f exists if and only if there exist constants for differentiation of functions. These c1..cn from K and functions u0,..,un from K such that structures obviously include various fields of f= u0’+ ∑i=1,...,nci(ui’/ui) functions (e.g. the field of rational functions or in one variable over the real field). g=∫ fdx = u0+ ∑ i=1..,n ci log(ui) • The goal is to provide a setting to answer concretely such questions as, "Does this note: this allows additional logs in the answer, but that’s all. The structure of the integral is specified. function have an elementary antiderivative?” Richard Fateman CS 282 Lecture 14 17 Richard Fateman CS 282 Lecture 14 18 3 Formally, start with a field F and a Also derivation map • Derivation is a map of F into itself •If a, b are in a differential field F with a a # a’ such that being nonzero, we call a an exponential of b or (a+b)’ = a’ + b’ b a logarithm of a if b’ = a’/a. (ab)’=a’b+ab’ CONSEQUENTLY • For algebraic extensions of F, there is a theorem that says that F is of characteristic 2 ≠ (a/b)’= (a’b-ab’)/b if a,b, ∈ F and b 0 zero and K is an algebraic extension field of F (an)’ = nan-1a’ for all integers n then the derivation on F can be extended 1’ =0; the constants F are a subfield of F uniquely to K. Richard Fateman CS 282 Lecture 14 19 Richard Fateman CS 282 Lecture 14 20 We must also deal with algebraic pieces examples • sometimes we prefer arctan (if a<0 below), sometimes logs. • Algebraic extensions z where p(z)=0 in F • Monomial extensions exp() and log(). Independent extensions are required e.g. log(x) and log(x2) don’t cut it since (log(x2))- • Risch converts ∫ xnsin x to exponentials (2log(x)) is a constant (perhaps 0) • and then can’t do it.. Richard Fateman CS 282 Lecture 14 21 Richard Fateman CS 282 Lecture 14 22 Extensions of the Risch algorithm The Risch Algorithm: two steps back. • Allow certain functions beyond log, exp such • There is a fallacy in claiming the Risch "algorithm" is an algorithm at all: it depends, for solution of that the form of the integral still holds. For subproblems, on heuristics to tell if certain π x 2 example, erf(x) = (2/ )∫0 exp(-t )dt has the expressions are equivalent to zero. appropriate structure. • We have gone over Richardson’s arguments to show that the zero-equivalence problem over a class of expressions much smaller than that of interest for integration is recursively unsolvable. • But that is not the problem with most implementations of the Risch algorithm. They mostly have not been programmed completely, because, even assuming you can solve the zero-equivalence problem, the procedure is hard to program. Richard Fateman CS 282 Lecture 14 23 Richard Fateman CS 282 Lecture 14 24 4 The Risch Algorithm: answers are not The Risch Algorithm: answering wrong ncessarily what you expect problem • It is not nearly as useful as you might think, because • Also, most people are interested in definite, it returns algebraic antiderivatives whose validity not indefinite integrals, at least once they've may be on a set of measure zero.
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