Computing Limits of Quotients of Multivariate Real Analytic Functions
aa r X i v : . [ c s . S C ] F e b COMPUTING LIMITS OF QUOTIENTS OF MULTIVARIATE REALANALYTIC FUNCTIONS
ADAM STRZEBO ´NSKIA
BSTRACT . We present an algorithm for computing limits of quotients of real analyticfunctions. The algorithm is based on computation of a bound on the Łojasiewicz exponentand requires the denominator to have an isolated zero at the limit point.
Keywords:
Multivariate function limit, symbolic limit computation.1. I
NTRODUCTION
Computation of limits is one of the basic problems of computational calculus. In theunivariate case computing limits of rational functions is easy, and the state of the art limitcomputation algorithms [3, 6] are applicable to large classes of functions. In the multivari-ate case computing limits of real rational functions is a nontrivial problem that has been asubject of recent research [1, 2, 8, 9, 10]. In [7] we compared five methods for computa-tion of limits of real rational functions based on the Cylindrical Algebraic Decomposition(CAD) algorithm. Here we extend the methods to computation of limits of quotients ofmultivariate analytic functions.The limit of a real function may not exist, however the lower limit and the upper limit al-ways exist. A weak version of the limit computation problem consists of deciding whetherthe limit exists and, if it does, finding the value of the limit. A strong version consists offinding the values of the lower limit and the upper limit.Let us state the problems precisely. Denote x = ( x , . . . , x n ) , ¯ R = R ∪ {− ∞ , ∞ } . Let U ⊆ R n be an open set and let A ( U ) denote the set of analytic functions on U . Let g ∈ A ( U ) , h ∈ A ( U ) \ { } , D = { u ∈ U : h ( u ) = } , f : D ∋ u → g ( u ) h ( u ) ∈ R , and let c ∈ U . Problem 1.
Find l ∈ ¯ R such that l = lim u → c f ( u ) or prove that such l does not exist. Problem 2.
Find l , l ∈ ¯ R such that l = lim inf u → c f ( u ) and l = lim sup u → c f ( u ) . Example 3.
Let g = sin ( x + y + z ) and h = − cos x − cos y − cos z . Thenlim ( x , y , z ) → g ( x , y , z ) h ( x , y , z ) = Example 4.
Let g = sin ( xy ) and h = cos x + cos y −
2. Thenlim inf ( x , y ) → g ( x , y ) h ( x , y ) = − ( x , y ) → g ( x , y ) h ( x , y ) = ( x , y ) → g ( x , y ) h ( x , y ) does not exist. F IGURE z = sin ( xy ) cos x + cos y − at 0.Recently two algorithms partially solving Problem 1 for rational functions have beenproposed. The algorithm presented in [9] solves a modified version of the problem, namelyit decides whether the limit exists and is finite. The negative answer includes both thecase when the limit does not exist and the case when the limit exists and is infinite. Thealgorithm uses Wu’s elimination method, rational univariate representations, and requiresadjoining two infinitesimal elements to the field. The algorithm presented in [1] (whichgeneralizes algorithms of [2, 8]) solves Problem 1 under the additional assumption that c is an isolated zero of h . The authors use the theory of Lagrange multipliers to reduce theproblem to computing the limit along a real algebraic set, and solve the reduced problemusing regular chains methods.In [7] we presented five methods based on the CAD algorithm that solve both Problem 1and Problem 2 for arbitrary rational functions. In this note we describe an algorithm that re-duces computation of limits of quotients of multivariate analytic functions to computationof limits of rational functions. The algorithm can be combined with any of the algorithmsdescribed in [7] to solve both Problem 1 and Problem 2 for quotients of analytic functions.2. T HE A LGORITHM
Let f = gh , where g and h are analytic functions in a neighbourhood of c ∈ R n . Withoutloss of generality we will assume that c =
0. If h ( ) = f is continuous at 0, andhence the limit can be obtained by evaluation. Therefore w.l.o.g. we will assume that h ( ) =
0. The only computability assumption we make about functions g and h is that forany d ∈ N we can compute the Taylor polynomials T d g and T d h of total degree d . Thisis a typical case in a computer algebra system, where g and h are given as expressionsobtained by composing analytic functions implemented in the system. Taylor polynomialsof arbitrary degree can be readily computed, but for instance algorithms for testing whetheran expression represents a function that is identically zero may not be available. Date : January 31, 2021.
OMPUTING LIMITS OF QUOTIENTS OF MULTIVARIATE REAL ANALYTIC FUNCTIONS 3
Remark . This specification is not sufficient to always solve Problem 1 or 2. For instance,let ϕ k = x + y k , let ϕ ∞ = x , and suppose that g and h are expressions that as functions (butnot as expressions) satisfy equalities g = ϕ k and h = ϕ m for some (unknown) k , m ∈ N ∪{ ∞ } . If k = m then lim ( x , y ) → g ( x , y ) h ( x , y ) = g and h to explicit polynomials. Computationof the Taylor polynomials T d g and T d h for any finite d does not allow to decide whether k and m are finite, but greater than d /
2, or are infinite. Hence Problem 1 or 2 cannot alwaysbe solved. Similarly, we cannot always decide whether the zero of h at ( , ) is isolated.The algorithm we present here solves Problem 1 and 2 if h has an isolated zero at 0.Otherwise the algorithm does not terminate. Considering Remark 5, this is the best we canhope for with the given input specification.Let U ⊆ R n be an open neighbourhood of 0, and let k u k denote the Euclidean norm of u ∈ R n . The algorithm is based on the following theorem, which is a special case of theŁojasiewicz inequality [5]. Theorem 6.
Let h ∈ A ( U ) and suppose that { u : k u k < ρ ∧ h ( u ) = } = { } for some ρ > . Then there exist positive constants r, c, and α such that if k u k < r then | h ( u ) | ≥ c k u k α . Algorithm 7. (MLIM)Input: g ∈ A ( U ) , h ∈ A ( U ) \ { } , such that h ( ) = .Output: lim inf u → g ( u ) h ( u ) . (1) Set d = . (2) Compute lim u → u d + ··· + u dn T d − h ( u ) . (3) If the limit does not exist or is not zero, set d = d + and go to step ( ) . (4) Return lim inf u → T d − g ( u ) T d − h ( u ) . Theorem 8.
If h has an isolated zero at then Algorithm 7 terminates and returns lim inf u → g ( u ) h ( u ) Otherwise the algorithm does not terminate.Proof.
Suppose that h has an isolated zero at 0. By Theorem 6 there exist positive constants r , c , and α such that if k u k < r then | h ( u ) | ≥ c k u k α . To prove that Algorithm 7 terminatesit suffices to show that if 2 d > α thenlim u → u d + · · · + u dn T d − h ( u ) = R d − h ( u ) = h ( u ) − T d − h ( u ) . We have u d + · · · + u dn | T d − h ( u ) | ≤ k u k d | h ( u ) − R d − h ( u ) | ≤ | | h ( u ) |k u k d − | R d − h ( u ) |k u k d | Since R d − h ( u ) is an analytic function whose Taylor series does not contain terms ofdegree lower than 2 d , | R d − h ( u ) |k u k d is bounded in a neighbourhood of 0. Moreover, if k u k < r , | h ( u ) |k u k d ≥ c k u k α k u k d = c k u k α − d −→ u → ∞ ADAM STRZEBO ´NSKI hence lim u → | | h ( u ) |k u k d − | R d − h ( u ) |k u k d | = u → u d + · · · + u dn T d − h ( u ) = u → g ( u ) h ( u ) note that g ( u ) h ( u ) = T d − g ( u ) + R d − g ( u ) T d − h ( u ) + R d − h ( u ) = T d − g ( u ) T d − h ( u ) + R d − g ( u ) T d − h ( u ) + R d − h ( u ) T d − h ( u ) We have R d − g ( u ) T d − h ( u ) = R d − g ( u ) k u k d k u k d T d − h ( u ) Since R d − g ( u ) is an analytic function whose Taylor series does not contain terms ofdegree lower than 2 d , | R d − g ( u ) |k u k d is bounded in a neighbourhood of 0. Moreover, k u k d | T d − h ( u ) | ≤ n d u d + · · · + u dn | T d − h ( u ) | −→ u → u → R d − g ( u ) T d − h ( u ) = u → R d − h ( u ) T d − h ( u ) = u → g ( u ) h ( u ) = lim inf u → T d − g ( u ) T d − h ( u ) Suppose now that the zero of h at 0 is not isolated. Let d ≥ T d − h at 0 is not isolated, then u d + · · · + u dn T d − h ( u ) attains arbitrarily large values in any neighbourhood of 0, and hencelim u → u d + · · · + u dn T d − h ( u ) does not exist or is not zero.If the zero of T d − h at 0 is isolated, then { u : k u k ≤ ρ ∧ T d − h ( u ) = } = { } forsome ρ >
0. Since R d − h ( u ) is an analytic function whose Taylor series does not containterms of degree lower than 2 d , there exists M > | R d − h ( u ) |k u k d ≤ M for all k u k ≤ ρ .Let Z = { u : k u k ≤ ρ ∧ h ( u ) = } . For u ∈ Z \ { } we have T d − h ( u ) = − R d − h ( u ) and hence u d + · · · + u dn | T d − h ( u ) | ≥ n − d k u k d | R d − h ( u ) | ≥ n − d M − > OMPUTING LIMITS OF QUOTIENTS OF MULTIVARIATE REAL ANALYTIC FUNCTIONS 5
Since 0 is a limit point of Z \ { } , lim u → u d + · · · + u dn T d − h ( u ) does not exist or is not zero. Therefore Algorithm 7 does not terminate. (cid:3) Remark . Algorithm 7 can be adapted to compute the upper limit or the limit of f ( u ) byreturning the upper limit or the limit of T d − g ( u ) T d − h ( u ) in step ( ) . Example 10.
Algorithm 7 may not stop for the first d such that T d − h has an isolatedzero at 0, even if T d − h = h . Let h ( x , y ) = ( x n ) + ( x − y n ) (cf. [4], Example 1). Then T d − h = h for d ≥ n +
1. However, since h ( t n , t ) = t n lim u → x d + y d T d − h ( x , y ) will not be zero for any d ≤ n . Example 11.
The second part of the proof does need two cases, that is the zero of T d − h at 0 may be isolated even if the zero of h at 0 is not isolated. Let h ( x , y ) = y + ( y − x ) − x − y . Then T h = y + ( y − x ) has an isolated zero at 0. However, the zero of h at 0is not isolated. In a neighbourhood of zero there are two analytic solutions of h ( x , y ) = y ( x ) = x − x + x − x + x + · · · y ( x ) = x + x − x − x − x + · · · Remark . In some cases it is possible to detect that the zero of h at 0 is not isolated andterminate the algorithm. For instance if h m is the lowest degree nonzero form of the Taylorseries of h and h m ( a ) > h m ( b ) < a , b ∈ R n , then h ( ta ) > h ( tb ) < t ∈ ( , ε ) with some ε >
0, and hence the zero of h at 0 is not isolated.The number of limit computations in step ( ) can be reduced by using fast negativecriteria to decide that the zero of T d − h ( u ) at 0 is not isolated.3. E XAMPLE
Let us compute the lower limit and the upper limit of g ( x , y , z ) h ( x , y , z ) at 0, where g = exp ( sin ( x + y + z )) − h = q cos ( x ) − sin ( y ) − z − d = ( ) we have T d − h = − x − y . Since T d − h ( , , z ) =
0, the limitlim ( x , y , z ) → x + y + z T d − h does not exist, and so in step ( ) we set d = ( ) . Now T d − h = − x − x y − x − y − y − z ( x , y , z ) → x + y + z T d − h = ADAM STRZEBO ´NSKI hence we move on to step ( ) . We have T d − g = x + x + y and the returned value islim inf ( x , y , z ) → g ( x , y , z ) h ( x , y , z ) = lim inf ( x , y , z ) → T d − g ( x , y , z ) T d − h ( x , y , z ) = − ( ) lim sup ( x , y , z ) → g ( x , y , z ) h ( x , y , z ) = lim sup ( x , y , z ) → T d − g ( x , y , z ) T d − h ( x , y , z ) = Mathematica . In thisexample methods based on topological properties performed better. Algorithm 15 (TLIM)took 77 seconds when using Algorithm 14 (ZCQ2) and 323 seconds when using Algorithm13 (ZCQ1). Methods based on optimization were not able to complete the computation in12 hours. The most time-consuming part of the computation is the limit in step ( ) with d =
3. R
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OLFRAM R ESEARCH I NC ., 100 T RADE C ENTRE D RIVE , C
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