{"paper_meta":{"paper_id":"arxiv:0710.4508","title":"0710.4508","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0710.4508v2 [cs.CC] 19 Mar 2008\nA Numerical Algorithm for Zero Counting.\nI: Complexity and Accuracy\nFelipe Cucker ∗\nDept. of Mathematics\nCity University of Hong Kong\nHONG KONG\ne-mail: macucker@cityu.edu.hk\nTeresa Krick †\nDepartamento de Matem ́atica\nUniv. de Buenos Aires & CONICET\nARGENTINA\ne-mail: krick@dm.uba.ar\nGregorio Malajovich‡\nDepto. de Matem ́atica Aplicada\nUniv. Federal do Rio de Janeiro\nBRASIL\ne-mail: gregorio@ufrj.br\nMario Wschebor\nCentro de Matem ́atica\nUniversidad de la Rep ́ublica\nURUGUAY\ne-mail: wschebor@cmat.edu.uy\nAbstract. We describe an algorithm to count the number of distinct real zeros of\na polynomial (square) system f. The algorithm performs O(log(nDκ(f))) iterations\n(grid refinements) where n is the number of polynomials (as well as the dimension of\nthe ambient space), D is a bound on the polynomials’ degree, and κ(f) is a condition\nnumber for the system. Each iteration uses an exponential number of operations. The\nalgorithm uses finite-precision arithmetic and a major feature in our results is a bound\nfor the precision required to ensure the returned output is correct which is polynomial\nin n and D and logarithmic in κ(f). The algorithm parallelizes well in the sense that\neach iteration can be computed in parallel time polynomial in n, log D and log(κ(f)).\n1\nIntroduction\nIn recent years considerable attention was put on the complexity of counting problems over\nthe reals. The counting complexity class #PR was introduced [20] and completeness results\nfor #PR were established [3] for natural geometric problems notably, for the computation\nof the Euler characteristic of semialgebraic sets. As one could expect, the “basic” #PR-\ncomplete problem consists of counting the real zeros of a system of polynomial equations.\nAlgorithms for counting real zeros have existed since long. One such algorithm follows\nfrom the work of Tarski [25] on quantifier elimination for the theory of the reals. Its complex-\nity is hyperexponential. Algorithms with improved complexity (doubly exponential) were\ndevised in the 70s by Collins [5] and W ̈utrich [27]. A breakthrough was reached a decade\n∗Partially supported by City University SRG grant 7002106.\n†Partially supported by grants UBACyT X112/06-09, CONICET PIP 2461/00 and ANPCyT 33671/05.\n‡Partially supported by CNPq grants 304504/2004-1, 472486/2004-7, 470031/2007-7, 303565/2007-1, and\nby FAPERJ grant E26/170.734/2004.\n1\n\nlater with the introduction of the critical points method by Grigoriev and Vorobjov [13, 12]\nwhich uses exponential time. Algorithms counting connected components (and hence, in\nthe zero-dimensional case, solutions) based on this method can be found in [14, 16], and in\nthe straight-line program model of computation in [1]. These algorithms parallelise well in\nthe sense that one can devise versions of them working in parallel polynomial time when\nan exponential number of processors is available. The #PR-completeness of the problem\nstrongly indicates that this is the best we can hope for.\nAll the algorithms mentioned above are “symbolic algorithms.” They have been devised\nupon the premise that no perturbation or round-offerror is present. Were this not the\ncase, it is not difficult to see that errors would accumulate quite badly. Roughly speaking,\nthese algorithms construct some object of exponential size on which some basic computation\n(e.g., linear algebra) is eventually performed. A question is posed, can one devise “numerical\nalgorithms” (maybe iterative, which need not terminate for ill-posed inputs) with a better\nbehavior viz the accumulation of round-offerrors? For the problem of deciding the existence\nof (or computing) a zero of a polynomial system such algorithms were given in [8, 6, 18].\nThe goal of this article is to describe and analyze a numerical algorithm for zero counting.\nWe will do so by developping appropriate versions of the tools used in [8, 6].\nLet d1, . . . , dn ∈N and d = (d1, . . . , dn). We will denote by Hd the space of polynomial\nsystems f = (f1, . . . , fn) with fi ∈R[X0, . . . , Xn] homogeneous of degree di.\nZero rays of polynomial systems f ∈Hd are associated to pairs of zeros (−ζ, ζ) of the\nrestriction f|Sn of f to the n-dimensional unit sphere Sn ⊂Rn+1. Thus, it will be convenient\nto consider a system f ∈Hd as a (central symmetric, analytic) mapping of Sn into Rn. If\nwe denote by Z(f) = {ζ ∈Sn : f(ζ) = 0} the zero-set of f in Sn then the number #R(f) of\nzero rays of the system f is half the cardinality of Z(f).\nIn this paper we describe a finite-precision algorithm computing #R(f), given f ∈Hd.\nTo analyze its complexity and accuracy, besides the number n of polynomials, we will rely on\ntwo more additional parameters. One is D = maxi≤n di. The other is a condition measure\nκ(f) for the system f. We will describe this measure in detail in Section 2 below. We\nwill also let S = max Si where Si is the number of non-zero coefficients of fi. Note that\nS is bounded by a simple expression in terms of n and D, namely, S =\n n+D\nD\n \n. Yet, we\nwill express dependancy on S since this may be relevant for the case of sparse systems of\npolynomials. Our main result is the following.\nTheorem 1.1. There exists an iterative algorithm which, with input f ∈Hd,\n(1) Returns #R(f).\n(2) Performs O(log(nDκ(f))) iterations and has a total cost (number of arithmetic opera-\ntions) of\nO\n \nlog(nDκ(f))(n + 1)2\n 2(n + 1)D2κ(f)2\nα∗\n 2n!\n,\nwhere α∗≈0.0384629388 . . . is a universal constant.\n(3) Can be well-parallelized in the sense that it admits a parallel version running in time\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2))\nwith a number of processors exponential in this quantity.\n2\n\n(4) Can be implemented with finite precision (both versions, sequential and parallel). The\nrunning time remains the same (with α∗replaced by α• ≈0.028268 · · ·) and the re-\nturned value is #R(f) as long as the machine precision (i.e., the round-offunit) u\nsatisfies\nu ≤\n1\nO\n D2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n .\n(5) It can be modified to return, in addition and for each real zero ζ ∈Sn of f, an approx-\nimate zero x of f in the sense that Newton’s iteration, starting at x, converges to ζ\nquadratically fast.\nRemark 1.2. (i)\nA system f for which arbitrarily small perturbations may change\nthe value #R(f) is considered ill-posed in our context since for arbitrarily small machine\nprecisions finite precision algorithms may return an incorrect value.\nConsequently, the\ncondition number κ(f) is infinite in these cases (and only then). This happens when f has\nmultiple real zeros and, in particular, when f has infinitely many real zeros. In these cases\nthe algorithm of Theorem 1.1 may not halt.\n(ii)\nNumerical algorithms compute functions φ on real data. Error analysis for algorithms\ncomputing (vectors of) real numbers —i.e. for which the image of φ has non-empty interior—\nare usually expressed in terms of bounds for the relative error of the computed quantities.\nThat is, for data d, bounds in\n∥φ(d) −fl(φ(d)∥\n∥φ(d)∥\nwhere fl(φ(d)) is the vector actually computed with finite precision. This relative error\nvaries continuously with d and depends on the condition of d and on the precision u. Such a\nform of analysis, however, becomes meaningless when computing quantities taking a finite\nnumber of values. Indeed, if Ra denotes the set of input data d for which φ(d) = a the\nfollowing happens. When d is in the interior of Ra we have that the relative error above\nis 0 for sufficiently small u. In contrast, when d is on the boundary of Ra, that error may\nremain constant for all u > 0. Because of this, error analysis for this kind of discrete-valued\nproblems has a different form, as in Theorem 1.1. One bounds how small u needs to be to\nguarantee a correct answer. Such a bound, needless to say, also depends on the condition\nof the data d. Examples of this type of analysis can be found in [4, 6, 7, 8]. In each of these\nreferences a condition number for the problem at hand occurs in the error analysis. We note\nthat the one in [6] is essentially our κ(f).\nThe rest of the paper is organized as follows. In Section 2 we describe the basic objects we\nwill deal with as well as fixing the notation. In Sections 3 and 4 we prove the two technical\nresults our algorithm relies on. In Section5 we describe the algorithm under the assumption\nof infinite precision and we prove parts (1), (2), and (3) of Theorem 1.1. The geometric\nideas making the algorithm work are best seen in this context. Section 6 then describes the\nnecessary modifications to make the algorithm work as well under finite precision. These\nmodifications are simple and can be summarized by saying that we relax a bit the inequalities\ntested in the algorithms to make room for the finite-precision errors to fit in.\n2\nPreliminaries\nDenote by Hd the subspace of R[X0, . . . , Xn] of homogeneous polynomials of degree d. Then,\nHd = Hd1 × · · · × Hdn.\n3\n\nIf g ∈Hd we write\ng(X) =\nX\nJ\ngJXJ\nwhere J = (J0, . . . , Jn) is assumed to range over all multi-indices such that |J| = Pn\nk=0 Jk =\nd, XJ = XJ0\n0 XJ1\n1 . . . XJn\nn\nand gJ ∈R. Multinomial coefficients are defined by:\n d\nJ\n \n=\nd!\nJ0!J1! · · · Jn!.\nThe space Hd is endowed with the inner product\n⟨g, h⟩=\nX\n|J|=d\ngJhJ\n d\nJ\n \nwhich gives rise to the norm ∥g∥=\np\n⟨g, g⟩. These norms, for d1, . . . , dn, induce a norm in\nHd by taking for f = (f1, . . . , fn) ∈Hd:\n∥f∥= ∥(f1, . . . , fn)∥= max\n1≤i≤n ∥fi∥.\nLet O(n + 1) be the orthogonal group. The inner product above is known to be O(n + 1)-\ninvariant: for all Q ∈O(n + 1) and all g, h ∈Hd,\n⟨g ◦Q, h ◦Q⟩= ⟨g, h⟩.\n(This is a direct consequence of [26, III-7] or [2, Theorem 1 p. 218], by considering O(n+ 1)\nas subgroup of U(n + 1)).\nThe associated norm ∥f∥on Hd is therefore also O(n + 1)-\ninvariant. We will use this norm on Hd all along this paper. For x = (x1, . . . , xn) ∈Rn we\nrecall that ∥x∥2 = (x2\n1 + · · · + x2\nn)1/2 and ∥x∥∞= max{|x1|, . . . , |xn|}. We will often denote\n∥x∥2 simply by ∥x∥.\nFor f ∈Hd and x ∈Sn define\nμnorm(f, x) = ∥f∥√n\n\nDf(x)−1\n|TxSn\n \n \n√d1\n√d2\n...\n√dn\n \n \n\n(1)\nwhere Df(x)|TxSn is the restriction to the tangent space of x at Sn of the derivative of f at\nx and the norm is the spectral norm, i.e. the operator norm with respect to ∥∥2. We now\ndefine the condition number κ(f) of f ∈Hd:\nκ(f) = max\nx∈Sn min\n \nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\n \n.\nRemark 2.1. The quantity κ(f) is closely related to other condition numbers for similar\nproblems.\nA version of the quantity μnorm(f, ζ) was introduced in [21, 22, 23] (see also [2, Chap-\nter 12]) for a complex polynomial system f and a zero ζ of f in the complex unit sphere\nSn\nC ⊂Cn+1. The normalized condition number of such a system f was then defined to be\nμnorm(f) :=\nmax\nζ∈Sn\nC |f(ζ)=0 μnorm(f, ζ).\n(2)\n4\n\nActually, the version of μnorm(f, ζ) introduced in [21, 22, 23] differs from (1) in the fact\nthat ∥f∥is defined as (P ∥fi∥2)1/2 (and there is no √n factor). It is bounded above by the\nexpression in (1).\nOver the reals, the right-hand side in (2) may not be well-defined since the zero set\nof f may be empty.\nIn [8] real systems were considered (as in the present paper) and\nan algorithm deciding feasibility of f (i.e., whether f has a real zero) was proposed. Its\ncomplexity was analyzed in terms of a condition number which, using our notation and\nmodulo minor details, is defined as follows\n \n \n \n \n \nmin\nζ∈Sn|f(ζ)=0 μnorm(f, ζ)\nif f is feasible\nmax\nζ∈Sn\n∥f∥\n∥f(ζ)∥∞\nif f is infeasible.\nNote the use of min (instead of max) in the first line above. This is due to the fact that the\ntime needed for the algorithm in [8] to detect the existence of a zero depends on the best\nconditioned zero of f. The existence of other, poorly conditioned (or even singular), zeros\nof f is irrelevant.\nShortly after, the algorithm in [8] was extended to an algorithm which would, in addition\nand if f is feasible, return a zero of f [6]. The complexity of this extension was studied\nin terms of a condition number (denoted ̺(f) in [6]) which, essentially, coincides with our\nκ(f).\nProposition 2.2. For all f ∈Hd, κ(f) ≥1.\nProof.\nLet x ∈Sn. Because of orthogonal invariance, we may assume without loss of\ngenerality that x = e0 := (1, 0, . . . , 0).\nIt is then immediate that ∥f(x)∥∞≤∥f∥. This shows that the second expression in the\ndefinition of κ is at least 1.\nFor the first expression, i.e., μnorm(f, x), define g = (g1, . . . , gn) ∈Hd by gi(X) = fi(X)−\nfi(e0)Xdi\n0 . Then g(e0) = 0 and [2, Corollary 3 p. 234], μnorm(g, e0) ≥1 (this is shown for\nthe version of μnorm with the 2-norm for ∥f∥, which is bounded above by the expression (1)).\nSince Df(e0) = Dg(e0) and ∥g∥≤∥f∥, we can conclude μnorm(f, e0) ≥μnorm(g, e0) ≥1.\n3\nThe exclusion Lemma\nIn this article, d( , ) denotes the Riemannian (angular) distance in Sn (which satisfies\n0 ≤d(x, y) ≤π, ∀x, y ∈Sn) and for x ∈Sn, r > 0, we set B(x, r) := {y ∈Sn : d(y, x) < r}\nand B(x, r) := {y ∈Sn : d(y, x) ≤r}.\nThe following result can be used to support an exclusion test.\nLemma 3.1. Let f ∈Hd and let x, y ∈Sn such that d(x, y) ≤\n√\n2. Then,\n∥f(x) −f(y)∥∞≤∥f∥\n√\nD d(x, y)\nIn particular, if f(x) ̸= 0, there is no zero of f in B(x, min{∥f(x)∥∞/(∥f∥\n√\nD),\n√\n2}).\n5\n\nProof.\nAn immediate consequence of the definition of the O(n + 1)-invariant inner\nproduct is that Hd endowed with this inner product is a reproducing kernel Hilbert space [9,\nProp. 2.21]. This implies that, for all g ∈Hd and x ∈Rn+1,\ng(x) = ⟨g(X), (xT X)deg g⟩.\n(3)\nBecause of orthogonal invariance, we can assume that x = e0 and y = e0 cos θ + e1 sin θ,\nwhere θ = d(x, y). Equation (3) implies that\nfi(x) −fi(y)\n=\n⟨fi(X), (xT X)di⟩−⟨fi(X), (yT X)di⟩= ⟨fi(X), (xT X)di −(yT X)di⟩\n=\n⟨fi(X), Xdi\n0 −(X0 cos θ + X1 sin θ)di⟩.\nHence, Cauchy-Schwarz-Bunyakowsky implies:\n|fi(x) −fi(y)| ≤∥fi∥∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥.\nSince\nXdi\n0 −(X0 cos θ + X1 sin θ)di = Xdi\n0 (1 −(cos θ)di) +\ndi\nX\nk=1\n di\nk\n \n(cos θ)di−k(sin θ)kXdi−k\n0\nXk\n1 ,\nwe have:\n∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥2\n=\n(1 −(cos θ)di)2 +\ndi\nX\nk=1\n di\nk\n \n(cos θ)2(di−k)(sin θ)2k\n=\n(1 −(cos θ)di)2 + 1 −(cos θ)2di\n=\n2(1 −(cos θ)di)\n≤\n2(1 −(1 −θ2\n2 )di)\n(4)\n≤\n2(1 −(1 −di\nθ2\n2 ))\n(5)\n≤\ndiθ2,\nwhere the inequality in line (4) is obtained from Taylor expanding cos θ around 0, and the\ninequality in line (5) is due to the fact that (1 −a)d ≥1 −da for a ≤1.\nWe conclude that\n|fi(x) −fi(y)| ≤∥fi∥θ\np\ndi\nand hence\n∥f(x) −f(y)∥∞≤∥f∥θ\nq\nmax\ni\ndi.\nFor the second assertion, we have\n∥f(y)∥∞\n≥\n∥f(x)∥∞−∥f(x) −f(y)∥∞\n≥\n∥f(x)∥∞−∥f∥\n√\nD d(x, y)\nsince d(x, y) ≤\n√\n2\n>\n∥f(x)∥∞−∥f∥\n√\nD ∥f(x)∥∞/(∥f∥\n√\nD)\n=\n0.\n6\n\n4\nThe proximity Theorem\n4.1\nNewton and Smale\nNewton iteration on the sphere Sn is defined by\nNf :\nSn\n→\nSn\nx\n7→\nNf(x) = expx\n \n−Df(x)−1\n|TxSnf(x)\n \nwhere expx is the exponential map at x,\nexpx h = cos(∥h∥)x + sin(∥h∥)\n∥h∥\nh.\nFurthermore, the standard invariants of α-theory, introduced by Smale in [24], can be\ndefined as:\nβ(f, x)\n=\n\nDf(x)−1\n|TxSnf(x)\n\n ,\nγ(f, x)\n=\nsup\nk≥2\n\nDf(x)−1\n|TxSnDkf(x)|(TxSn)k\nk!\n\n1/(k−1)\n,\nα(f, x)\n=\nβ(f, x)γ(f, x).\nRemark 4.1.\n(i) It is easy to see that β(f, x) = d(x, Nf(x)).\n(ii) We will not use Newton’s method in our algorithm. We are instead interested in its\nalpha theory which guarantees existence of zeros near points x with α(f, x) small\nenough.\n(iii) The Newton iteration presented above is not the iteration known as ‘projective New-\nton’. There is an alpha theory for that method, available in [19].\nHere we use slight modifications of the quantities α, β and γ, more adapted to our\npurposes. We set\nβ(f, x)\n:=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\nγ(f, x)\n:=\nD3/2\n2\nμnorm(f, x)\nα(f, x)\n:=\nβ(f, x)γ(f, x).\nThe definition of γ is motivated by the estimate of γ [2, Theorem 2 p. 267].\nγ(f, x) ≤γ(f, x).\nwhich yields the lower bound\nκ(f) ≥\nmax\nζ|f(ξ)=0 2D−3/2γ(f, ζ).\n(6)\n7\n\nWe also observe that γ(f, x) ≥D3/2\n2\nsince μnorm(f, x) ≥1 and that β(f, x) ≤β(f, x)\nsince\nβ(f, x) =\n\nDf(x)−1\n|TxSnf(x)\n\n ≤√n∥f(x)∥∞\n\nDf(x)−1\n|TxSn\n\n ≤μnorm(f, x)∥f(x)∥∞\n∥f∥\n= β(f, x).\nTherefore α(f, x) ≤α(f, x).\n4.2\nProximity and unicity from data at a point\nDefinition 4.2. We say that x ∈Sn is an approximate zero for f if and only if the Newton\nsequence {xk}k∈N, where x0 := x and xk+1 := Nf(xk), is defined for all k and moreover\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nd(x0, x1).\nThe limit point ζ = limk→∞xk is a fixed point for Newton iteration and a zero of f. It is\ncalled the associated zero to x.\nIn what follows we denote σ := P\nk≥0 2−2k+1 = 1.632843018 . . . and we set\nBf(x) := {y ∈Sn | d(x, y) ≤σβ(f, x)}.\nThe main technical tool in our algorithm is provided by the following result.\nTheorem 4.3. There exists an universal constant α∗:= 0.0384629388 . . . such that for all\nx ∈Sn, if α(f, x) < α∗, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z in Bf(x) the Newton sequence starting at z converges\nto ζ.\n4.3\nBackground material\nTheorem 4.3 is a consequence of the following two results, which are restatements of results\nproved in [10]. While [10] deals with Newton iteration on arbitrary complete real analytic\nRiemannian manifolds, here we reword them in terms of Newton iteration on the unit\nsphere Sn (Example 1 in [10]). The γ-Theorem for mappings [10, Theorem 1.3] becomes\nthe following.\nTheorem 4.4. Let f : Sn →Rn be analytic. Suppose that f(ζ) = 0 and Df(ζ) is an\nisomorphism. Let\nR(f, ζ) := min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nIf d(x, ζ) ≤R(f, ζ), then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and\nd(xk, ζ) ≤\n 1\n2\n 2k−1 d(x, ζ). In particular, {xk} converges to ζ.\n8\n\nNow let α0 := 0.130716944 . . . denote the smallest positive root of the polynomial ψ(u)2−\n2u, and\ns0 :=\n1\nσ + (1−σα0)2\nψ(σα0)\n \n1 +\nσ\n1−σα0\n = 0.103621842 . . .\nWe state the α-Theorem for mappings [10, Theorem 1.4] for the sphere Sn.\nTheorem 4.5. Let f : Sn →Rn be analytic. Let x ∈Sn be such that β(f, x) ≤s0π and\nα(f, x) ≤α0. Then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and converges\nto a zero ζ of f. Moreover,\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nβ(f, x)\nand\nd(xk, ζ) ≤σβ(f, x).\nFinally we introduce ψ(u) := 1−4u+2u2, which is positive and decreasing for 0 < u < 1−\n√\n2\n2 ,\nand state [10, Lemma 4.3]:\nLemma 4.6. Let x, y ∈Sn with d(x, y) < π. Suppose that Df(x) is nonsingular and\nν := d(x, y)γ(f, x) < 1 −\n√\n2\n2 .\nThen\nγ(f, y) ≤\nγ(f, x)\n(1 −ν)ψ(ν).\n4.4\nProof of Theorem 4.3\nSet ν∗:= 0.0628039411 . . . for the only real root of the polynomial\nΨ(u) := (3 −\n√\n7)(1 −u)ψ(u) −4u,\nand α∗:= ν∗\nσ = 0.0384629388 . . .. Note that α∗≤min{α0, s0π}.\nSince γ(f, x) ≥D3/2\n2\n, the hypothesis of Theorem 4.5 hold from α(f, x) ≤α(f, x) < α∗≤\nα0 and β(f, x) ≤β(f, x) ≤2α(f,x)\nD3/2\n<\n2α∗\nD3/2 < s0π.\nUsing Remark 4.1(i) it follows that x is an approximate zero of f, and that the associated\nzero ζ satisfies:\nd(x, ζ) ≤σβ(f, x) ≤σβ(f, x).\nThis already proves Parts (i) and (ii) of Theorem 4.3.\nWe show (iii). Since d(x, ζ) ≤σβ(f, x) < σs0π < π,\nν = d(x, ζ)γ(f, x) ≤σβ(f, x)γ(f, x) ≤σα(f, x) ≤σα∗= ν∗< 1 −\n√\n2\n2 ,\nand we can apply Lemma 4.6. Therefore\n4σβ(f, x)γ(f, ζ) ≤4σβ(f, x)γ(f, x)\n1\n(1 −ν)ψ(ν) ≤4ν∗\n1\n(1 −ν∗)ψ(ν∗) = 3 −\n√\n7,\n9\n\nbecause (1−u)ψ(u) decreases for 0 < u < 1−\n√\n2\n2 , and ν∗is a zero of (3−\n√\n7)(1−u)ψ(u)−4u.\nThis shows, since 2σβ(f, x) ≤π, that\n2σβ(f, x) ≤R(f, ζ) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nWe conclude applying Theorem 4.4 to z ∈Bf(x), since\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤2σβ(f, x) ≤R(f, ζ).\nIt follows that the Newton sequence {zk}k∈N starting at z converges to ζ.\nRemark 4.7. The hypothesis on the radius of injectivity in [10] was recently found to be\nredundant.\n5\nInfinite precision\n5.1\nGrids and Graphs\nOur algorithm works on a grid on Sn. We easily construct one by projecting onto Sn a\ngrid on the cube Cn = {y | ∥y∥∞= 1}. We make use of the (easy to compute) bijections\nφ : Cn →Sn and φ−1 : Sn →Cn given by φ(y) =\ny\n∥y∥and φ−1(x) =\nx\n∥x∥∞.\nGiven η := 2−k for some k ≥1, we consider the uniform grid Uη of mesh η on Cn. This is\nthe set of points in Cn whose coordinates are of the form i2−k for i ∈{−2k, −2k+1, . . ., 2k},\nwith at least one coordinate equal to 1 or −1. We denote by Gη its image by φ in Sn. Note\nthat, for y1, y2 ∈Cn,\nd(φ(y1), φ(y2)) ≤π\n2 ∥y1 −y2∥2 ≤π\n2\n√\nn + 1 ∥y1 −y2∥∞.\n(7)\nGiven η as above we associate to it a graph Gη as follows. We set A(f) := {x ∈Sn |\nα(f, x) < α∗}. The vertices of the graph are the points in Gη ∩A(f). Two vertices x, y ∈Gη\nare joined by an edge if and only if Bf(x) ∩Bf(y) ̸= ∅.\nNote that as a simple consequence of Theorem 4.3 we obtain the following lemma.\nLemma 5.1.\n(i) For each x ∈A(f) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each point\nz in Bf(x), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈A(f). Then ζx = ζy ⇐⇒Bf(x) ∩Bf(y) ̸= ∅.\nWe define Z(Gη) := S\nx∈Gη Bf(x) ⊂Sn where x ∈Gη has to be understood as x running\nover all the vertices of Gη. Similarly, for a connected component U of Gη, we define\nZ(U) :=\n[\nx∈U\nBf(x).\nLemma 5.2.\n(i) For each component U of Gη, there is a unique zero ζU ∈Z(f) such that ζU ∈Z(U).\nMoreover, ζU ∈∩x∈UBf(x).\n10\n\n(ii) If U and V are different components of Gη, then ζU ̸= ζV .\nProof.\n(i) Let x ∈U. Since x ∈A(f), by Lemma 5.1 (i) there exists a zero ζx of f\nin Bf(x) ⊆Z(U). This shows the existence. For the second assertion and the uniqueness,\nassume that there exist ζ and ξ zeros of f in Z(U). Let x, y ∈U be such that ζ ∈Bf(x),\nand ξ ∈Bf(y). Since U is connected, there exist x0 = x, x1, . . . , xk−1, xk := y in A(f) such\nthat (xi, xi+1) is an edge of Gη for i = 0, . . . , k −1, that is, Bf(xi)∩Bf(xi+1) ̸= ∅. If ζi and\nζi+1 are the associated zeros of xi and xi+1 in Z(f) respectively, then by Lemma 5.1(ii) we\nhave ζi = ζi+1, and thus ζ = ξ ∈Bf(x) ∩Bf(y).\n(ii) Assume ζU = ζV ∈Bf(x) ∩Bf(y) ⊂Z(U) ∩Z(V ), then x and y are joined by an edge\nand belong to the same connected component.\n5.2\nThe infinite precision algorithm\nCount Roots 1(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet U1, . . . , Ur be the connected components of Gη\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈Ui and all xj ∈Uj, d(xi, xj) > πη√n + 1\nand\n(ii) for all x ∈Gη \\ A(f), ∥f(x)∥∞> π\n2 η\np\n(n + 1)D∥f∥\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\n5.3\nProof of Theorem 1.1(1–3)\nProof of Part (1)\nThis proof requires some arguments of convexity.\nWe can natu-\nrally define spherical convex hulls for sets of points in Hn, an open half-sphere in Sn. If\nx1, . . . , xq ∈Hn we define\nSCH(x1, . . . , xq) := Cone(x1, . . . , xq) ∩Sn\nwhere Cone(x1, . . . , xq) is the smallest convex cone with vertex at the origin and containing\nthe points x1, . . . , xq. Alternatively, we have,\nSCH(x1, . . . , xq) =\n λ1x1 + · · · + λqxq\n∥λ1x1 + · · · + λqxq∥| λ1, . . . , λq ≥0,\nX\nλi = 1\n \n.\nWe will use the following fact.\nLemma 5.3. Let x1, ..., xq ∈Hn ⊂Rn+1. If Tq\ni=1 B(xi, ri) ̸= ∅, then SCH(x1, . . . , xq) ⊂\nSq\ni=1 B(xi, ri).\nProof.\nLet x ∈SCH(x1, . . . , xq) and y ∈Tq\ni=1 B(xi, ri). We will prove that x ∈B(xi, ri)\nfor some i.\nIf x = y, this is obvious.\nIf x ̸= y, let H be the half-space\nH :=\n \nz ∈Rn+1 : ⟨z, y −x⟩< 0\n \n.\n11\n\nSince ∥x∥= ∥y∥= 1, we have ⟨x + y, y −x⟩= 0, and we note that in this case, x + y\ndetermines the mid-line between x and y. Moreover, since x ̸= y, we have x ∈H since\n⟨x, y −x⟩= ⟨x, y⟩−∥x∥2 < ∥x∥∥y∥−∥x∥2 = 0. Therefore the half-space H is the set of\npoints z in Rn+1 such that the Euclidean distance ∥z −x∥< ∥z −y∥.\nOn the other hand, H must contain at least one point of the set {x1, ..., xq} since if\nthis were not the case, the convex set Cone(CH(x1, . . . , xq)) would be contained in {z :\n⟨z, y −x⟩≥0}, contradicting x ∈SCH(x1, . . . , xq). Let, therefore, xi ∈H. It follows that\n∥x −xi∥< ∥y −xi∥\nwhich implies\nd(x, xi) < d(y, xi) ≤ri.\nWe can now proceed. Assume the algorithm halts, we want to show that if r equals\nthe number of connected components of Gη, then #R(f) = #Z(f)/2 = r/2. We already\nknow by Lemma 5.2 that each connected component U of Gη determines uniquely a zero\nζU ∈Z(f). Thus it is enough to prove that Z(f) ⊂Z(Gη).\nAssume that there is a zero ζ of f in Sn such that ζ is not in Z(Gη). Let B∞(φ−1(ζ), η) :=\n{y ∈Uη | ∥y −φ−1(ζ)∥∞≤η} = {y1, . . . , yq}, the set of all neighbors of φ−1(ζ) in Uη, and\nlet xi = φ(yi), i = 1, . . . , q. Clearly, φ−1(ζ) is in the cone spanned by {y1, . . . , yq} and hence\nζ ∈SCH(x1, . . . , xq).\nWe claim that there exists j ≤q such that xj ̸∈A(f). Indeed, assume this is not the\ncase. We consider two cases.\n(a)\nAll the xi belong to the same connected component U of Gη. By Lemma 5.2 there\nexists a unique zero ζU ∈Sn of f in Z(U) and ζU ∈∩iBf(xi). We may apply Lemma 5.3\nto deduce that\nSCH(x1, . . . , xq) ⊆\n[\nBf(xi).\nIt follows that, for some i ∈{1, . . . , q}, ζ ∈Bf(xi) ⊆Z(U), contradicting that ζ ̸∈Z(Gη).\n(b)\nThere exist l̸= s and 1 ≤i < j ≤r such that xl∈Ui and xs ∈Uj. Since condition\n(i) in the algorithm is satisfied, d(xl, xs) > πη√n + 1. But, by (7),\nd(xl, xs) ≤π\n2\n√\nn + 1∥yl−ys∥∞≤π\n2\n√\nn + 1\n ∥yl−φ−1(ζ)∥∞+ ∥φ−1(ζ) −ys∥∞\n \n≤πη\n√\nn + 1,\na contradiction.\nWe have thus proved the claim. Let then 1 ≤j ≤q be such that xj ̸∈A(f). Since\ncondition (ii) in the algorithm is satisfied ∥f(xj)∥∞> π\n2 η\np\n(n + 1)D∥f∥. It follows from\nthe inequality d(xj, ζ) ≤π\n2\n√n + 1η and Lemma 3.1 that ∥f(ζ)∥∞> 0, a contradiction.\nProof of Part (2)\nWe need a few lemmas.\nLemma 5.4. If ζ1 ̸= ζ2 ∈Z(f) then\nd(ζ1, ζ2) ≥2(3 −\n√\n7)D−3/2\nκ(f)\n.\n12\n\nProof.\nFor i = 1, 2, using (6) and Proposition 2.2,\nR(f, ζi) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζi)\n)\n≥min\n(\nπ, (3 −\n√\n7)D−3/2\nκ(f)\n)\n= (3 −\n√\n7)D−3/2\nκ(f)\n.\nNow suppose that d(ζ1, ζ2) < R(f, ζ1) + R(f, ζ2) and choose x ∈Sn such that d(x, ζ1) <\nR(f, ζ1) and d(x, ζ2) < R(f, ζ2). Then Theorem 4.4 implies that ζ1 = ζ2, a contradiction.\nLemma 5.5. Let x1, x2 ∈Gη with associated zeros ζ1 ̸= ζ2. If η ≤\n2(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen\nd(x1, x2) > πη√n + 1.\nProof.\nAssume d(x1, x2) ≤πη√n + 1.\nSince x2 ̸∈Bf(x1), d(x1, x2) > σβ(f, x1).\nConsequently,\nd(x1, ζ1) ≤σβ(f, x1) < d(x1, x2) ≤πη\n√\nn + 1\nand, similarly, d(x2, ζ2) < πη√n + 1. But then,\nd(ζ1, ζ2) ≤d(ζ1, x1) + d(x1, x2) + d(x2, ζ2) < 3πη\n√\nn + 1 ≤2(3 −\n√\n7)D−3/2\nκ(f)\ncontradicting Lemma 5.4.\nLemma 5.6. Let x ∈Sn such that x ̸∈A(f).\nIf η ≤\nα∗\n(n+1)D2κ(f)2 then ∥f(x)∥∞>\nπ\n2 η\np\n(n + 1)D∥f∥.\nProof.\nSince x ̸∈A(f) we have α(f, x) ≥α∗. We divide the proof in two cases.\nCase I. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n=\n∥f∥\n∥f(x)∥∞\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗∥f(x)∥2\n∞\n(n + 1)D2∥f∥2\nwhich implies, since η ≤1\n2 <\n4D\nπ2α∗,\n∥f(x)∥∞≥\n√η√n + 1D∥f∥\n√α∗\n> π\n2 η\np\n(n + 1)D∥f∥.\nCase II. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗\n(n + 1)D2μnorm(f, x)2\nwhich implies α∗≥η(n + 1)D2μnorm(f, x)2. Also,\nα∗≤α(f, x) = 1\n2β(f, x)μnorm(f, x)D3/2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞.\n13\n\nPutting both inequalities together we obtain\nη(n + 1)D2μnorm(f, x)2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞\nor yet,\n∥f(x)∥∞≥2η(n + 1)D1/2∥f∥> π\n2 η\np\n(n + 1)D∥f∥.\nWe can now conclude the proof of Part (2).\nAssume η ≤\nα∗\n(n+1)D2κ(f)2 .\nThen the\nhypotheses of Lemmas 5.5 and 5.6 hold. The first of these lemmas ensures that condition (i)\nin the algorithm is satisfied. The second, that condition (ii) is so. Therefore, the algorithm\nhalts as soon as\nα∗\n2(n+1)D2κ(f)2 < η ≤\nα∗\n(n+1)D2κ(f)2 . This gives a bound of O(ln(nDκ(f)))\nfor the number of iterations. Since the number of grid points considered at this iteration\n(η =\nα∗\n(n+1)D2κ(f)2 ) is at most 2(n + 1)\n \n2(n+1)D2κ(f)2\nα∗\n n\n, the bound for the total complexity\nfollows.\nProof of Parts (3) and (5)\nWe have already seen that the number of iterations is\nbounded by O(ln(nDκ(f))). At each of these iterations, we need to perform a number of\ncomputations on the (at most) 2(n+1)\n \n2(n+1)D2κ(f)2\nα∗\n n\ngrid points to decide whether they\nare in A(f). These can be done independently. Then, we need to compute the number of\nconnected components of Gη. This can be done (see, e.g., [15]) in parallel time O(ln(|Vη|))2\nwhere |Vη| denotes the number of vertices of Gη and therefore, in parallel time at most\nO(n2(ln(nDκ(f))2 + ln(α∗)2)). Since this is the dominant step in the computation at a\ngiven iteration, it follows that the total parallel time consumed by the algorithm is at most\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2)). This shows part (3). For part (5), just note\nthat, for i = 1, . . . , r, any vertex xi of Ui is an approximate zero of the only zero of f in\nZ(Ui).\n6\nFinite Precision\n6.1\nMaking room to allow errors\nOur finite precision algorithm will be a variation of Algorithm Count Roots 1. But since\nfinite precision computations will be affected by errors, we need to make room in the infinite\nprecision algorithm to allow them.\nFor this aim, we state the corresponding version of\nTheorem 4.3.\nTheorem 6.1. There exist a universal constant α• = 0.028268 · · · such that, for all x ∈Sn,\nif α(f, x) < α•, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z s.t. d(x, z) ≤2σβ(f, x) the Newton sequence starting at\nz converges to ζ.\n14\n\nProof.\nParts (i) and (ii) follow from Theorem 4.3 and the fact that α• < α∗. Part (iii)\nis proved by taking ν• = 0.046158 · · · to be the only real root of the polynomial Ψ(u) :=\n(3 −\n√\n7)(1 −u)ψ(u) −6u, and α• = ν•\nσ = 0.028268. Then, one proves as in Theorem 4.3\nthat 3σβ(f, x) ≤R(f, ζ) from which it follows that, for all z s.t. d(x, z) ≤2σβ(f, x),\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤3σβ(f, x) ≤R(f, ζ)\nand hence, that the Newton sequence {zk}k∈N starting at z converges to ζ.\nThe proofs of Lemmas 5.5 and 5.6 yield, mutatis mutandis, the following results.\nLemma 6.2. Let x1, x2 ∈Gη with associated zeros ξ1 and ξ2, ξ1 ̸= ξ2. If η ≤(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen d(x1, x2) > 2πη√n + 1.\nLemma 6.3. Let x ∈Sn such that α(f, x) > α•\n3 . If η ≤\nα•\n4D2(n+1)κ(f)2 then ∥f(x)∥∞>\nπη\np\n(n + 1)D∥f∥.\n6.2\nBasic facts\nWe recall the basics of a floating-point arithmetic which idealizes the usual IEEE standard\narithmetic. This system is defined by a set F ⊂R containing 0 (the floating-point numbers),\na transformation r : R →F (the rounding map), and a constant u ∈R (the round-offunit)\nsatisfying 0 < u < 1. The properties we require for such a system are the following:\n(i) For any x ∈F, r(x) = x. In particular, r(0) = 0.\n(ii) For any x ∈R, r(x) = x(1 + δ) with |δ| ≤u.\nWe also define on F arithmetic operations following the classical scheme\nxe◦y = r(x ◦y)\nfor any x, y ∈F and ◦∈{+, −, ×, /}, so that\ne◦: F × F →F.\nThe following is an immediate consequence of property (ii) above.\nProposition 6.4. For any x, y ∈F we have\nxe◦y = (x ◦y)(1 + δ),\n|δ| ≤u.\nWhen combining many operations in floating-point arithmetic, quantities such as\nQn\ni=1(1 + δi)ρi naturally appear. Our round-offanalysis uses the notations and ideas in\nChapter 3 of [17], from where we quote the following results:\nProposition 6.5. If |δi| ≤u, ρi ∈{−1, 1}, and nu < 1, then\nn\nY\ni=1\n(1 + δi)ρi = 1 + θn,\nwhere\n|θn| ≤γn =\nnu\n1 −nu.\n15\n\nProposition 6.6. For any positive integer k such that ku < 1, let θk, θj be any quantities\nsatisfying\n|θk| ≤γk =\nku\n1 −ku\n|θj| ≤γj =\nju\n1 −ju.\nThe following relations hold.\n1. (1 + θk)(1 + θj) = 1 + θk+j for some |θk+j| ≤γk+j.\n2.\n1 + θk\n1 + θj\n=\n 1 + θk+j\nif j ≤k,\n1 + θk+2j\nif j > k.\nfor some |θk+j| ≤γk+j or some |θk+2j| ≤γk+2j.\n3. If ku, ju ≤1/2, then γkγj ≤γmin{k,j}.\n4. iγk ≤γik.\n5. γk + u ≤γk+1.\n6. γk + γj + γkγj ≤γk+j.\nFrom now on, whenever we write an expression containing θk we mean that the same\nexpression is true for some θk, with |θk| ≤γk.\nWhen computing an arithmetic expression q with a round-offalgorithm, errors will\naccumulate and we will obtain another quantity which we will denote by fl(q). We write\nError(q) = |q −fl(q)|.\nAn example of round-offanalysis which will be useful in what follows is given in the next\nproposition, the proof of which can be found in Section 3.1 of [17].\nProposition 6.7. There is a round-offalgorithm which, with input x, y ∈Rn, computes\nthe dot product of x and y. The computed value fl(⟨x, y⟩) satisfies\nfl(⟨x, y⟩) = ⟨x, y⟩+ θ⌈log2 n⌉+1⟨|x|, |y|⟩,\nwhere |x| = (|x1|, . . . , |xn|).\nIn particular, if x = y, the algorithm computes fl(∥x∥2)\nsatisfying\nfl(∥x∥2) = ∥x∥2(1 + θ⌈log2 n⌉+1).\nWe will also have to deal with square roots and arccosinus. The following result will\nhelp us to do so.\nLemma 6.8. (i) Let θ ∈R such that |θ| ≤1/2. Then,\n√\n1 −θ = 1 −θ′ with |θ′| ≤|θ|.\n(ii) Let 0 < a ≤1 and ε ∈R such that 0 < a + ε < 1. Then, arccos(a + ε) = arccos(a) +\nυ\n1\n√\n1−(a+ε)2 with |υ| ≤|ε|.\nProof.\nAssume θ > 0 (if θ < 0 it is done similarly). By the intermediate value theorem\nwe have that 1 −\n√\n1 −θ = θ(√ξ)′ with ξ ∈(1 −θ, 1). But\n(\np\nξ)′ =\n1\n2√ξ ≤\n1\n√\n2,\n16\n\nthe last since ξ ≥1/2. This proves (i).\nPart (ii) is shown similarly. Again, assume for simplicity that ε > 0. Then, for some\nξ ∈(a, a + ε),\narccos(a + ε) −arccos(a) = ε arccos′(ξ) = ε\n1\np\n1 −ξ2 =\nυ\np\n1 −(a + ε)2 .\nWe assume that, besides the four basic arithmetic operations, we are allowed to compute\nsquare roots and arccosinus with finite precision. That is, if op denotes any of these two\noperators, we compute f\nop such that\nf\nop(x) = op(x)(1 + δ),\n|δ| ≤u.\nFrom Lemma 6.8(i) it follows that, for all a > 0,\n^\np\na(1 + θk) = √a(1 + θk+1).\nRemark 6.9. Our choice of the precision u in Theorem 1.1(4) guarantees that ku < 1/2\nholds whenever we encounter θk in what follows, and consequently, |θk| ≤γk ≤2ku. This\nimplies that in all what follows we have γg = O(ug) for all the expressions g we will\nencounter.\nAccording to the previous remark we will introduce a further notation that will consid-\nerably simplify our exposition. For all expression g, we will write\n[[g]] := O(ug).\nThis notation will avoid we burden ourselves with the consideration of multiplicative con-\nstants.\n6.3\nThe finite precision algorithm\nOur finite precision algorithm is a variation of Algorithm Count Roots 1 in Section 5.3.\nGiven x ∈Sn we define below fl(A′(f)) and fl(B\n′\nf(x)), which are convenient floating\nversions of the sets A′(f) =\n \nx ∈Sn | α(f, x) < 1\n2α•\n \nand B\n′\nf(y) = {z ∈Sn | d(x, y) ≤\n3\n2σβ(f, x)} respectively.\nGiven f ∈Hd and x ∈Sn, we let M ∈Rn×n be a matrix representing\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)|TxSn.\nand we set σmin(M) = ∥M −1∥−1. Therefore\nμnorm(f, x)\n=\n∥f∥√n ∥M −1∥= ∥f∥√n σmin(M)−1,\nβ(f, x)\n=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\n= √n σmin(M)−1∥f(x)∥∞,\nα(f, x)\n=\nβ(f, x)μnorm(f, x)D3/2\n2\n= ∥f∥n σmin(M)−2∥f(x)∥∞\nD3/2\n2\n.\n17\n\nThis implies that\ny ∈B\n′\nf(x)\n⇐⇒\nd(x, y) ≤3\n2σβ(f, x)\n⇐⇒\nσmin(M)d(x, y) ≤3\n2σ√n∥f(x)∥∞,\nx ∈A′(f)\n⇐⇒\nα(f, x) < α•\n2\n⇐⇒\n∥f∥n∥f(x)∥∞D3/2 < α•σmin(M)2.\nThese statements are equivalent under infinite precision, but the expressions at the right-\nhand side are more convenient to handle when working with finite precision. This motivates\nour definitions of\nfl(B\n′\nf(x))\n:=\n \ny ∈Sn | fl(σmin(M)d(x, y)) ≤fl(3\n2σ√n∥f(x)∥∞)\n \nfl(A′(f))\n:=\nn\nx ∈Sn | fl(∥f∥n ∥f(x)∥∞D3/2) < fl(α•σmin(M)2)\no\nWe also define accordingly the graph fl(G′\nη) whose vertices are the points in Gη ∩\nfl(A′(f)), and with two vertices x, y joined by an edge if and only if fl(B\n′\nf(x))∩fl(B\n′\nf(x)) ̸=\n∅. Its connected components are denoted by fl(U).\nOur algorithm is the following:\nCount Roots 2(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet fl(U1), . . . , fl(Ur) be the connected components of fl(Gη)\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈fl(Ui) and all xj ∈fl(Uj), fl(d(xi, xj)) > fl( 3\n2πη√n + 1)\nand\n(ii) for all x ∈Gη \\ fl(A′(f)), fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D∥f∥)\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\nIn the rest of the section we will see that, when the precision u satisfies u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), this algorithm is correct and halts as soon as η\n≤\nα•\n4D2(n+1)κ(f)2 .\n6.4\nBounding errors for elementary computations\nThe goal of this subsection is to exhibit bounds for the accumulated error in the main\ncomputations of Count Roots 2. We will rely on the basic notations and results described\nin §6.2.\nTo simplify notation, and without loss of generality, in all what follows we assume that\n∥f∥= 1. We denote by S(Hd) the sphere of such systems. Also, we do not discuss in what\nfollows the accumulated error in the computation of φ : Cn →Sn. This is a minor detail\nwhich can be taken care of using Lemma 6.8(i).\nProposition 6.10. Given f ∈S(Hd) and x ∈Sn, we can compute ∥f(x)∥∞with finite\nprecision u such that\nError(∥f(x)∥∞) = [[D + log S]]\nwhere S is a bound on the number of coefficients of each fi.\n18\n\nProof.\nLet f = (f1, . . . , fn). For i ≤n write fi = P cJXJ and let S be the number\nof coefficients of fi. To compute f(x) one computes each monomial cJxJ with fl(cJxJ) =\ncJxJ(1 + θD). Then, one computes fi(x) to get\nfl(fi(x))\n=\nfl(\nX\nfl(cJxJ))\n=\nfl(\nX\ncJxJ(1 + θ(J)\nD ))\n=\nX\ncJxJ(1 + θ(J)\nD ) + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\n=\nfi(x) +\nX\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\nwhere in the third line we reasoned as in the proof of Proposition 6.7. Therefore\nError(∥f(x)∥∞)\n≤\n \nX\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\n \n≤\nX\n|cJ| ∥xJ∥(γD + γlog S + γDγlog S)\n≤\nγD+log S\nwhere we used that for any x ∈Sn, | P |cJ|xJ| ≤∥P |cJ|xJ∥= ∥fi∥≤∥f∥= 1 and\nProposition 6.6 (6). The conclusion follows from Remark 6.9.\nProposition 6.11. Given f ∈S(Hd) and x ∈Sn, let M ∈Rn×n be a matrix representing\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)|TxSn\nin some orthonormal basis of TxSn. Then ∥M∥≤√n. In addition, we can compute such a\nmatrix M with finite precision u such that\n∥Error(M)∥F = [[n(log S + D + log n)]].\nProof.\nStep 1: Let y =\nx−en+1\n∥x−en+1∥. The Householder symmetry\nHy = In+1 −2yyt\nswaps vectors en+1 and x, and fixes y⊥. The first n columns of Hy are therefore an orthonor-\nmal basis of TxSn, while the last column is x. Let H ∈R(n+1)×n denote the submatrix\nobtained from the first n columns of Hy. With that notation, we set\nM =\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)H.\nStep 2: We claim that Pi,x : Hdi →Rn, fi 7→\n1\n√di Dfi(x)|TxSn is an orthogonal projection,\nin the sense that for any fixed x, the map (Pi,x)| ker(Pi,x)⊥is an isometry.\n19\n\nWe use an orthogonal invariance argument. The special orthogonal group SO(n + 1)\nacts on Hdi and on Rn+1 isometrically as follows: to a given Q ∈SO(n + 1), we associate\nrespectively the following isometries:\nx 7→Qx\n,\nfi 7→fi ◦Qt.\nWe set y = Qx and gi = fi ◦Qt. Differentiating the equality gi(Qx) = fi(x), we obtain:\nDgi(y)Q = Dfi(x).\nWhen x is fixed, we can set Q conveniently so that y = en+1. Therefore\nDgi(en+1)Q|TxSn = Dfi(x)|TxSn.\nSince Q(TxSn) = Ten+1Sn we obtain\nDgi(en+1)|Ten+1Sn = Dfi(x)|TxSn.\nThis means that Pi,en+1(fi ◦Qt) = Pi,x(fi). Thus, in order to prove our claim, it is enough\nto show that Pi,en+1 is an orthogonal projection.\nSince for g = P\nJ gJXJ,\n∂g\n∂Xj (en+1) = g(ej+(d−1)en+1) and since Ten+1Sn = ⟨e1, . . . , en⟩,\nwe have that for any gi ∈Hdi,\nPi,en+1(gi) =\n1\n√di\n gi(e1+(di−1)en+1), . . . , gi(en+(di−1)en+1)\n \n.\nHence, for any gi ∈ker(Pi,en+1)⊥, i.e. such that giJ = 0 for all J ̸= ej + (di −1)en+1,\n1 ≤j ≤n, we have\n∥gi∥2 =\nX\nJ\ng2\niJ\n di\nJ\n = ∥Pi,en+1(gi)∥2\n2.\nWe conclude that Pi,x is an orthogonal projection.\nStep 3: From the previous step, for any fi ∈Hdi, using the orthogonal decomposition\nfi = f ◦\ni + f ⊥\ni\nwith f ◦\ni ∈ker Pi,x and f ⊥\ni ∈ker P ⊥\ni,x, we have\n∥Pi,x(fi)∥2\n2 = ∥Pi,x(f ⊥\ni )∥2\n2 = ∥f ⊥\ni ∥2 ≤∥fi∥2.\nIt is now immediate from Step 1 and from the definition of ∥f∥= maxi ∥fi∥that the\nFrobenius norm ∥M∥F of the matrix M satisfies\n∥M∥2\nF =\nn\nX\ni=1\n∥Pi,x(fi)∥2\n2 ≤\nn\nX\ni=1\n∥fi∥2 ≤n∥f∥2 = n\nand hence its spectral norm ∥M∥satisfies ∥M∥≤∥M∥F ≤√n. This bound is independent\nof the choice of the basis for the space TxSn.\nStep 4: We next present the algorithm to compute M, given f and x. This is a non-\noptimal algorithm, and can be significantly improved if more is known on the structure of\nthe polynomial system f.\nWe can compute each entry mij of the matrix M as the scalar product of\n1\n√\ndi Dfi(x)\nand the jth column Hj := (hkj)1≤k≤n+1 of H.\n20\n\nProceeding as in the proof of Proposition 6.10, we can compute\n1\n√\ndi\n∂fi\n∂Xk (x) with\nError\n 1\n√\ndi\n∂fi\n∂Xk\n(x)\n \n= [[D + log S]].\nOn the other hand, the vector y =\nx−en+1\n∥x−en+1∥can be computed using 2n + 4 operations, and\nclearly Error(yj) = [[ log(n)]] for all j. Hence, for all coefficients hkj of H,\nError(hkj) = [[ log(n)]].\nApplying Proposition 6.7 we conclude\nError(mij)\n=\n[[D + log S + log n]]\n\n1\n√di\nDfi(x)\n \n\n ∥Hj∥\n=\n[[D + log S + log n]].\nThe second equality holds because ∥Hj∥= 1 since H is unitary, and because, as in the proof\nof Step 2,\n\n1\n√di\nDfi(x)\n\n2\n=\n\n1\n√di\nDgi(en+1)\n\n2\n= 1\ndi\n∥(gi(e1+(di−1)en+1), . . . , gi(dien+1))∥2 ≤∥gi∥2 ≤1.\nThis implies\n∥Error(M)∥F ≤[[n(D + log S + log n)]].\nLemma 6.12. Let x ∈Sn and M be as in Proposition 6.11. We can compute σmin(M) =\n∥M −1∥−1 satisfying\nError(σmin(M)) = [[n(log S + D + n3/2)]].\nProof.\nLet E′ = M −fl(M). By Proposition 6.11,\n∥E′∥≤∥E′∥F ≤[[n(log S + D + log n)]].\nLet M = fl(M). We compute σmin(M) = ∥M −1∥−1 using a backward stable algorithm\n(e.g., QR factorization). Then the computed fl(σmin(M)) is the exact σmin(M + E′′) for a\nmatrix E′′ with\n∥E′′∥≤cn2u∥M∥\nfor some universal constant c (see, e.g., [11, 17]). Thus,\nfl(σmin(M)) = fl(σmin(M)) = σmin(M + E′′) = σmin(M + E′ + E′′).\nWrite E = E′ + E′′. Then, using ∥M∥≤√n,\n∥E∥\n≤\n∥E′∥+ ∥E′′∥≤∥E′∥+ cn2u∥M∥≤∥E′∥+ cn2u(∥M∥+ ∥E′∥)\n=\n[[n(log S + D + log n)]] + cn2u(√n + [[n(log S + D + log n)]])\n=\n[[n(log S + D + log n)]] + cn2u(√n + c′un(log S + D + n3/2))\n=\n[[n(log S + D + n3/2)]]\nsince the hypothesis on u implies c′un(log S + D + n3/2) is bounded by a constant term.\nTherefore, fl(σmin(M)) = σmin(M + E) which implies by [11, Corollary 8.3.2]:\nError(σmin(M)) ≤∥E∥< [[n(log S + D + n3/2)]].\n21\n\nProposition 6.13. Let f ∈S(Hd)). Assume u ≤\nK\nκ(f)2n2D log S for a small enough constant\nand let x ∈Sn. Then\n(i) If x /∈fl(A′(f)) then α(f, x) ≥1\n3α•.\n(ii) If x ∈fl(A′(f)) then α(f, x) < α•.\nProof.\nFrom Proposition 6.10\nfl(n∥f(x)∥∞D3/2)\n=\n(∥f(x)∥∞+ [[D + log S]])(nD3/2)(1 + θ4)\n≤\nnD3/2∥f(x)∥∞+ [[nD3/2(D + log S)]]\n. Also, from Lemma 6.12, using that σmin(M) ≤√n,\nfl(α•σmin(M)2)\n=\nα•\n \nσmin(M) + [[n(log S + D + n3/2)]]\n 2\n(1 + θ2)\n≥\nα•σmin(M)2 −2α•σmin(M)[[n(log S + D + n3/2)]]\n≥\nα•σmin(M)2 −[[n3/2(log S + D + n3/2)]].\nTherefore,\nn∥f(x)∥∞D3/2 + [[nD3/2(D + log S)]]\n≥\nfl(n∥f(x)∥∞D3/2) ≥fl(α•σ2\nmin)\n≥\nα•σ2\nmin −[[n3/2(log S + D + n3/2)]]\nor yet,\nn∥f(x)∥∞D3/2 −α•σ2\nmin\n≥\n−([[nD3/2(D + log S)]] + [[n3/2(log S + D + n3/2)]])\n≥\n−[[n3D5/2 log S]].\nCase I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, using the hypothesis on u and the inequality\nκ(f) ≥1,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)n2D log S\n≤\nKO(1)n∥f(x)∥∞D3/2 ≤n∥f(x)∥∞D3/2\n2\nthe last by choosing K small enough. Hence, n∥f(x)∥∞D3/2 −α•σ2\nmin ≥−\n \nn∥f(x)∥∞D3/2\n2\n \n,\nwhich implies 3\n2n∥f(x)∥∞D3/2 ≥α•σmin(M)2, i.e., α(f, x) ≥α•\n3 .\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)2n2D log S\n≤\nKO(1)σmin(M)2D3/2 ≤α•σmin(M)2\n3\n22\n\nthe last by choosing K small enough.\nThis implies n∥f(x)∥∞D3/2 −α•σmin(M)2 ≥\n−α•σmin(M)2\n3\nor, equivalently, α(f, x) ≥α•\n3 .\nThis shows part (i). For part (ii), one shows as above that\nn∥f(x)∥∞D3/2 −α•σ2\nmin ≤[[n3D5/2 log S]].\nThen, one proceeds as well by considering the two cases min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞and min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x).\nLemma 6.14. Let y1, y2 ∈Uη and let xi = φ(yi), i = 1, 2. Then d(x1, x2) ≥\nη\n2√n+1.\nProof.\nThe distance d(x1, x2) is minimized at y1 = (1, . . . , 1, 1) and y2 = (1, . . . , 1, 1−η).\nLet N = n + 1. Then\ncos(d(x1, x2))2\n=\n⟨y1, y2⟩2\n∥y1∥2∥y2∥2\n=\n(N −η)2\nN(N −2η + η2)\n=\n1 −\n(N −1)η2\nN 2 −2Nη + Nη2\n≤\n1 −η2 N −1\nN 2\n.\nHence\nd(x1, x2) ≥arccos\n r\n1 −η2 N −1\nN 2\n!\n= arcsin\n η\nN\n√\nN −1\n \n≥\nη\n2\n√\nN\n.\nLemma 6.15. Let u <\nKη2\nn log n for a small enough constant K. For x1, x2 ∈Gη we can\ncompute d(x1, x2) such that\nError(d(x1, x2)) ≤\n √n log n\nη\n \n.\nProof.\nLet yi = φ−1(xi), i = 1, 2, and a = cos(d(x1, x2)), i.e.,\na = ⟨y1, y2⟩\n∥y1∥∥y2∥.\nWe have, using Proposition 6.7,\nfl(⟨y1, y2⟩) = ⟨y1, y2⟩+ θlog n∥y1∥∥y2∥\nand fl(∥y1∥∥y2∥) = ∥y1∥∥y2∥(1+θlog n). Using now Propositions 6.4, 6.5, and 6.6, it follows\nthat fl(a) = a + ε with ε = [[ log n]].\nBy choosing K sufficiently small, ε ≤\nη2n\n12(n+1)2 . Also, from the proof of Lemma 6.14,\na = cos(d(x1, x2)) ≤\ns\n1 −\nη2n\n(n + 1)2\n23\n\nand hence, using that √z + y ≤√z + 3y whenever 0 < z, y ≤1, we obtain\na + ε ≤\ns\n1 −\nη2n\n(n + 1)2 +\nη2n\n12(n + 1)2 ≤\ns\n1 −\n3η2n\n4(n + 1)2 ≤\ns\n1 −\nη2\n3(n + 1).\nUsing Lemma 6.8(ii) it follows that,\narccos(a + ε)\n=\narccos(a) + ε\n \n1\np\n1 −(a + ε)2\n \n=\narccos(a) + [[ log n]]\n \np\n3(n + 1)\nη\n .\nTherefore,\nError(d(x1, x2)) ≤\n √n log n\nη\n \n.\nLemma\n6.16. Let\nf\n∈\nS(Hd).\nAssume that\nη\n≥\nα•\n8D2(n+1)κ(f)2\nand u\n≤\nK\nD2n5/2κ(f)3(log S+n3/2D2κ(f)2) with K small enough, and let x, y ∈Gη. Then\n(i) If y ∈fl(B\n′\nf(x)) then d(x, y) ≤2σβ(f, x).\n(ii) If y /∈fl(B\n′\nf(x)) then d(x, y) > σβ(f, x).\nProof.\nBy Lemmas 6.12 and 6.15 (and using σmin(M) ≤√n and the bound d(x, y) ≤\nπ\n2 η√n + 1 which follows from (7)),\nError(σmin(M)d(x, y))\n=\nO\n d(x, y)Error(σmin(M)) + σmin(M)Error(d(x, y))\n \n=\nη π\n2\n√\nn + 1[[n(log S + D + n3/2)]] + √n\n √n log n\nη\n \n=\nη[[n3/2(log S + D + n3/2)]] +\n n log n\nη\n \n≤\n[[n3/2 log S + n3D2κ(f)2]]\nthe last by the bounds on η. Also, using Proposition 6.10,\nError\n 3\n2σ√n∥f(x)∥∞\n \n≤[[√n(D + log S)]].\nTherefore, for part (i),\nσmin(M)d(x, y) −3\n2σ√n∥f(x)∥∞\n≤fl(σmin(M)d(x, y)) −fl\n 3\n2σ√n∥f(x)∥∞\n \n+ [[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n≤[[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n= [[n3/2 log S + n3D2κ(f)2]].\n24\n\nCase I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, by the hypothesis on u,\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nκ(f)n(log S + n3/2D2κ(f)2)\n≤\nσ√n\n2κ(f) ≤σ√n∥f(x)∥∞\n2\nthe last line by taking K small enough. This implies that σmin(M)d(x, y) ≤2σ√n∥f(x)∥∞,\ni.e., that d(x, y) ≤2σβ(f, x).\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nD2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n≤\n√nα•\n48D2(n + 1)3/2κ(f)3\n≤\n√nη\n8√n + 1κ(f) ≤\n√nd(x, y)\n4κ(f)\n≤σmin(M)d(x, y)\n4\nby taking K small enough and Lemma 6.14.\nThis implies that\n3\n4σmin(M)d(x, y) ≤\n3\n2σ√n∥f(x)∥∞, i.e., that d(x, y) ≤2σβ(f, x).\nThis shows part (i). Part (ii) is shown in a similar way.\nLemma 6.17. Let u ≤Kη2\nlog n with K small enough and x1, x2 ∈Gη.\n(i) If fl(d(x1, x2)) ≤fl( 3\n2πη√n + 1) then d(x1, x2) ≤2πη√n + 1.\n(ii) If fl(d(x1, x2)) > fl( 3\n2πη√n + 1) then d(x1, x2) > πη√n + 1.\nProof.\nBy Lemma 6.15 and the hypothesis on u, we obtain\nError(d(x1, x2)) =\n √n log n\nη\n \n≤O\n √n log n\nη\n Kη2\nlog n ≤π\n2 η\n√\nn + 1,\nthe last by taking K small enough.\nAlso, Error( 3\n2πη√n + 1) ≤\n3\n2πη√n + 1 γ3.\nThe\nstatement easily follows from these two bounds.\nLemma 6.18. Let u ≤\nKη\n√\nnD\nD+log S+η\n√\nnD with K small enough, f ∈S(Hd) and x ∈Sn.\n(i) If fl(∥f(x)∥∞) ≤fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞≤πη\np\n(n + 1)D.\n(ii) If fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞> π\n2 η\np\n(n + 1)D.\n25\n\nProof.\nFor part (i), from Proposition 6.10,\n∥f(x)∥∞≤fl(∥f(x)∥∞) + [[D + log S]].\nAlso,\n√\n2\n2 πη\np\n(n + 1)D ≥fl(\n√\n2\n2 πη\np\n(n + 1)D) −[[η\np\n(n + 1)D]].\nTherefore,\n∥f(x)∥∞−\n√\n2\n2 πη\np\n(n + 1)D\n≤\nfl(∥f(x)∥∞) −fl(\n√\n2\n2 πη\np\n(n + 1)D) + [[D + log S + η\np\n(n + 1)D]]\n≤\n[[D + log S + η\np\n(n + 1)D]]\n=\nO\n D + log S + η\np\n(n + 1)D\n \nKη\n√\nnD\nD + log S + η\n√\nnD\n≤\n(1 −\n√\n2\n2 )η\np\n(n + 1)D,\nthe last by taking K sufficiently small. It follows that ∥f(x)∥∞≤πη\np\n(n + 1)D and hence,\npart (i) of the statement.\nPart (ii) is proved similarly.\n6.5\nProof of Theorem 1.1(4): Correctness\nWe will show that, if u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), and the algorithm halts with\nη ≥\nα•\n8D2(n+1)κ(f)2 , then the value r/2 returned by the algorithm is #R(f).\nThis is a\nconsequence of the floating following versions of Lemmas 5.1 and 5.2.\nLemma 6.19. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each x ∈fl(A′(f)) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each\npoint z ∈fl(B\n′\nf(x)), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈fl(A′(f)). Then ζx = ζy ⇐⇒fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)) ̸= ∅.\nProof.\n(i) Applying Proposition 6.13(ii), x ∈fl(A′(f)) implies that α(f, x) < α•.\nTherefore, by Theorem 6.1, there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover, if\nz ∈fl(B\n′\nf(x)), by Lemma 6.16(i), d(x, z) ≤2σβ(f, x) and the Newton sequence starting at\nz converges to ζx.\n(ii) If ζx = ζy, then Bf(x) ∩Bf(y) ̸= ∅which implies by Lemma 6.16(ii) that there exists\nz ∈fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)).\nThis immediately implies, using that Bf(x) ⊂fl(B\n′\nf(x)) by Lemma 6.16(ii), the follow-\ning corresponding floating version of Lemma 5.2.\nLemma 6.20. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each component fl(U) of fl(G′\nη), there is a unique zero ζU ∈Z(f) such that\nζU ∈Z(fl(U)). Moreover ζU ∈∩x∈fl(U)Bf(x).\n26\n\n(ii) If fl(U) and fl(V ) are different components of fl(G′\nη), then ζU ̸= ζV .\nIn order to show the correctness of Count Roots 2, we only need to prove that Z(f) ⊂\nZ(fl(G′\nη)). This easily follows adapting the proof of Part (1) in Section 5.3 to this situation,\nmaking use of Lemma 6.20 and the facts that Condition (i), fl(d(xi, xj)) > fl( 3\n2πη√n + 1),\nimplies that d(xi, xj) > πη√n + 1 (Lemma 6.17(ii)) and Condition (ii), fl(∥f(x)∥∞) >\nfl(\n√\n2\n2 πη\np\n(n + 1)D), implies that ∥f(x)∥∞> π\n2 η\np\n(n + 1)D (Lemma 6.18(ii)).\n6.6\nProof of Theorem 1.1(4): Complexity\nWe want to show that if η ≤\nα•\n4D2(n+1)κ(f)2 then Count Roots 2(f) halts. Note that this\nmeans that\nα•\n8D2(n + 1)κ(f)2 < η ≤\nα•\n4D2(n + 1)κ(f)2\nand hence, by § 6.5, that it correctly returns #R(f).\nBecause of the hypothesis on η, the hypotheses of Lemmas 6.2, and 6.3 are satisfied.\nLet fl(U) ̸= fl(V ) be different components of fl(G′\nη), and therefore, by Lemma 6.20,\nζU ̸= ζV , and for all x ∈fl(U), y ∈fl(V ), by Lemma 6.2, d(x, y) > 2πη√n + 1 holds. This\nimplies, by Lemma 6.17(i), that Condition (i) in Count Roots 2 is satisfied.\nConsider now x ̸∈fl(A′(f)). By Proposition 6.13(i), α(f, x) ≥α•\n3 . This implies, by\nLemma 6.3, that ∥f(x)∥∞> πη\np\n(n + 1)D, which in turn, by Lemma 6.18(i), ensures that\nCondition (ii) in Count Roots 2 is satisfied. Hence, the algorithm halts.\nAknowledgement. We are grateful to Andr ́e Galligo for a helpful discussion.\nReferences\n[1] B. Bank, M. Giusti, J. Heintz, and L. Pardo.\nGeneralized polar varieties: geometry and\nalgorithms. J. Compl., 21:377–412, 2005.\n[2] L. Blum, F. Cucker, M. Shub, and S. Smale. Complexity and Real Computation. Springer-\nVerlag, 1998.\n[3] P. B ̈urgisser and F. Cucker. Counting complexity classes for numeric computations II: Algebraic\nand semialgebraic sets. J. Compl., 22:147–191, 2006.\n[4] D. Cheung and F. Cucker. Solving linear programs with finite precision: II. Algorithms. J.\nCompl., 22:305–335, 2006.\n[5] G.E. Collins. Quantifier elimination for real closed fields by cylindrical algebraic deccomposi-\ntion, volume 33 of Lect. Notes in Comp. Sci., pages 134–183. Springer-Verlag, 1975.\n[6] F. Cucker. Approximate zeros and condition numbers. J. Compl., 15:214–226, 1999.\n[7] F. Cucker and J. Pe ̃na. A primal-dual algorithm for solving polyhedral conic systems with a\nfinite-precision machine. SIAM J. Optim., 12:522–554, 2002.\n[8] F. Cucker and S. Smale. Complexity estimates depending on condition and round-offerror.\nJournal of the ACM, 46:113–184, 1999.\n[9] F. Cucker and D.X. Zhou. Learning Theory: An Approximation Theory Viewpoint. Cambridge\nUniv. Press, 2007.\n[10] J.-P. Dedieu, P. Priouret, and G. Malajovich.\nNewton method on Riemannian manifolds:\nCovariant alpha-theory. IMA Journal of Numerical Analysis, 23:395–419, 2003.\n27\n\n[11] G. Golub and C. Van Loan. Matrix Computations. John Hopkins Univ. Press, 3rd edition,\n1996.\n[12] D.Yu. Grigoriev. Complexity of deciding Tarski algebra. Journal of Symbolic Computation,\n5:65–108, 1988.\n[13] D.Yu. Grigoriev and N.N. Vorobjov. Solving systems of polynomial inequalities in subexpo-\nnential time. Journal of Symbolic Computation, 5:37–64, 1988.\n[14] D.Yu. Grigoriev and N.N. Vorobjov. Counting connected components of a semialgebraic set\nin subexponential time. Computational Complexity, 2:133–186, 1992.\n[15] Y. Han and R.A. Wagner.\nAn efficient and fast parallel-connected component algorithm.\nJournal of the ACM, 37(3):626–642, 1990.\n[16] J. Heintz, M.-F. Roy, and P. Solerno. Single exponential path finding in semi-algebraic sets\nII: The general case. In C.L. Bajaj, editor, Algebraic Geometry and its Applications, pages\n449–465. Springer-Verlag, 1994.\n[17] N. Higham. Accuracy and Stability of Numerical Algorithms. SIAM, 1996.\n[18] T.Y. Li. Numerical solution of polynomial systems by homotopy continuation methods. In\nP.G. Ciarlet and F. Cucker, editors, Handbook of numerical analysis, volume 11, pages 209–304.\nNorth-Holland, 2003.\n[19] G. Malajovich. On generalized Newton algorithms: Quadratic convergence, path-following and\nerror analysis. Theoret. Comp. Sci., 133:65–84, 1994.\n[20] K. Meer. Counting problems over the reals. Theoret. Comp. Sci., 242:41–58, 2000.\n[21] M. Shub and S. Smale. Complexity of B ́ezout’s theorem I: geometric aspects. Journal of the\nAmer. Math. Soc., 6:459–501, 1993.\n[22] M. Shub and S. Smale. Complexity of B ́ezout’s theorem III: condition number and packing.\nJournal of Complexity, 9:4–14, 1993.\n[23] M. Shub and S. Smale. Complexity of B ́ezout’s theorem IV: probability of success; extensions.\nSIAM J. of Numer. Anal., 33:128–148, 1996.\n[24] S. Smale. Newton’s method estimates from data at one point. In R. Ewing, K. Gross, and\nC. Martin, editors, The Merging of Disciplines: New Directions in Pure, Applied, and Com-\nputational Mathematics. Springer-Verlag, 1986.\n[25] A. Tarski. A Decision Method for Elementary Algebra and Geometry. University of California\nPress, 1951.\n[26] H. Weyl. The Theory of Groups and Quantum Mechanics. Dover, 1932.\n[27] H.R. W ̈uthrich. Ein Entscheidungsverfahren f ̈ur die Theorie der reell-abgeschlossenen K ̈orper.\nvolume 43 of Lect. Notes in Comp. Sci., pages 138–162. Springer-Verlag, 1976.\n28","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0710.4508v2 [cs.CC] 19 Mar 2008\nA Numerical Algorithm for Zero Counting.\nI: Complexity and Accuracy\nFelipe Cucker ∗\nDept. of Mathematics\nCity University of Hong Kong\nHONG KONG\ne-mail: macucker@cityu.edu.hk\nTeresa Krick †\nDepartamento de Matem ́atica\nUniv. de Buenos Aires & CONICET\nARGENTINA\ne-mail: krick@dm.uba.ar\nGregorio Malajovich‡\nDepto. de Matem ́atica Aplicada\nUniv. Federal do Rio de Janeiro\nBRASIL\ne-mail: gregorio@ufrj.br\nMario Wschebor\nCentro de Matem ́atica\nUniversidad de la Rep ́ublica\nURUGUAY\ne-mail: wschebor@cmat.edu.uy\nAbstract. We describe an algorithm to count the number of distinct real zeros of\na polynomial (square) system f. The algorithm performs O(log(nDκ(f))) iterations\n(grid refinements) where n is the number of polynomials (as well as the dimension of\nthe ambient space), D is a bound on the polynomials’ degree, and κ(f) is a condition\nnumber for the system. Each iteration uses an exponential number of operations. The\nalgorithm uses finite-precision arithmetic and a major feature in our results is a bound\nfor the precision required to ensure the returned output is correct which is polynomial\nin n and D and logarithmic in κ(f). The algorithm parallelizes well in the sense that\neach iteration can be computed in parallel time polynomial in n, log D and log(κ(f)).\n1\nIntroduction\nIn recent years considerable attention was put on the complexity of counting problems over\nthe reals. The counting complexity class #PR was introduced [20] and completeness results\nfor #PR were established [3] for natural geometric problems notably, for the computation\nof the Euler characteristic of semialgebraic sets. As one could expect, the “basic” #PR-\ncomplete problem consists of counting the real zeros of a system of polynomial equations.\nAlgorithms for counting real zeros have existed since long. One such algorithm follows\nfrom the work of Tarski [25] on quantifier elimination for the theory of the reals. Its complex-\nity is hyperexponential. Algorithms with improved complexity (doubly exponential) were\ndevised in the 70s by Collins [5] and W ̈utrich [27]. A breakthrough was reached a decade\n∗Partially supported by City University SRG grant 7002106.\n†Partially supported by grants UBACyT X112/06-09, CONICET PIP 2461/00 and ANPCyT 33671/05.\n‡Partially supported by CNPq grants 304504/2004-1, 472486/2004-7, 470031/2007-7, 303565/2007-1, and\nby FAPERJ grant E26/170.734/2004.\n1"},{"paragraph_id":"p2","order":2,"text":"later with the introduction of the critical points method by Grigoriev and Vorobjov [13, 12]\nwhich uses exponential time. Algorithms counting connected components (and hence, in\nthe zero-dimensional case, solutions) based on this method can be found in [14, 16], and in\nthe straight-line program model of computation in [1]. These algorithms parallelise well in\nthe sense that one can devise versions of them working in parallel polynomial time when\nan exponential number of processors is available. The #PR-completeness of the problem\nstrongly indicates that this is the best we can hope for.\nAll the algorithms mentioned above are “symbolic algorithms.” They have been devised\nupon the premise that no perturbation or round-offerror is present. Were this not the\ncase, it is not difficult to see that errors would accumulate quite badly. Roughly speaking,\nthese algorithms construct some object of exponential size on which some basic computation\n(e.g., linear algebra) is eventually performed. A question is posed, can one devise “numerical\nalgorithms” (maybe iterative, which need not terminate for ill-posed inputs) with a better\nbehavior viz the accumulation of round-offerrors? For the problem of deciding the existence\nof (or computing) a zero of a polynomial system such algorithms were given in [8, 6, 18].\nThe goal of this article is to describe and analyze a numerical algorithm for zero counting.\nWe will do so by developping appropriate versions of the tools used in [8, 6].\nLet d1, . . . , dn ∈N and d = (d1, . . . , dn). We will denote by Hd the space of polynomial\nsystems f = (f1, . . . , fn) with fi ∈R[X0, . . . , Xn] homogeneous of degree di.\nZero rays of polynomial systems f ∈Hd are associated to pairs of zeros (−ζ, ζ) of the\nrestriction f|Sn of f to the n-dimensional unit sphere Sn ⊂Rn+1. Thus, it will be convenient\nto consider a system f ∈Hd as a (central symmetric, analytic) mapping of Sn into Rn. If\nwe denote by Z(f) = {ζ ∈Sn : f(ζ) = 0} the zero-set of f in Sn then the number #R(f) of\nzero rays of the system f is half the cardinality of Z(f).\nIn this paper we describe a finite-precision algorithm computing #R(f), given f ∈Hd.\nTo analyze its complexity and accuracy, besides the number n of polynomials, we will rely on\ntwo more additional parameters. One is D = maxi≤n di. The other is a condition measure\nκ(f) for the system f. We will describe this measure in detail in Section 2 below. We\nwill also let S = max Si where Si is the number of non-zero coefficients of fi. Note that\nS is bounded by a simple expression in terms of n and D, namely, S =\n n+D\nD"},{"paragraph_id":"p3","order":3,"text":". Yet, we\nwill express dependancy on S since this may be relevant for the case of sparse systems of\npolynomials. Our main result is the following.\nTheorem 1.1. There exists an iterative algorithm which, with input f ∈Hd,\n(1) Returns #R(f).\n(2) Performs O(log(nDκ(f))) iterations and has a total cost (number of arithmetic opera-\ntions) of\nO"},{"paragraph_id":"p4","order":4,"text":"log(nDκ(f))(n + 1)2\n 2(n + 1)D2κ(f)2\nα∗\n 2n!\n,\nwhere α∗≈0.0384629388 . . . is a universal constant.\n(3) Can be well-parallelized in the sense that it admits a parallel version running in time\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2))\nwith a number of processors exponential in this quantity.\n2"},{"paragraph_id":"p5","order":5,"text":"(4) Can be implemented with finite precision (both versions, sequential and parallel). The\nrunning time remains the same (with α∗replaced by α• ≈0.028268 · · ·) and the re-\nturned value is #R(f) as long as the machine precision (i.e., the round-offunit) u\nsatisfies\nu ≤\n1\nO\n D2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n .\n(5) It can be modified to return, in addition and for each real zero ζ ∈Sn of f, an approx-\nimate zero x of f in the sense that Newton’s iteration, starting at x, converges to ζ\nquadratically fast.\nRemark 1.2. (i)\nA system f for which arbitrarily small perturbations may change\nthe value #R(f) is considered ill-posed in our context since for arbitrarily small machine\nprecisions finite precision algorithms may return an incorrect value.\nConsequently, the\ncondition number κ(f) is infinite in these cases (and only then). This happens when f has\nmultiple real zeros and, in particular, when f has infinitely many real zeros. In these cases\nthe algorithm of Theorem 1.1 may not halt.\n(ii)\nNumerical algorithms compute functions φ on real data. Error analysis for algorithms\ncomputing (vectors of) real numbers —i.e. for which the image of φ has non-empty interior—\nare usually expressed in terms of bounds for the relative error of the computed quantities.\nThat is, for data d, bounds in\n∥φ(d) −fl(φ(d)∥\n∥φ(d)∥\nwhere fl(φ(d)) is the vector actually computed with finite precision. This relative error\nvaries continuously with d and depends on the condition of d and on the precision u. Such a\nform of analysis, however, becomes meaningless when computing quantities taking a finite\nnumber of values. Indeed, if Ra denotes the set of input data d for which φ(d) = a the\nfollowing happens. When d is in the interior of Ra we have that the relative error above\nis 0 for sufficiently small u. In contrast, when d is on the boundary of Ra, that error may\nremain constant for all u > 0. Because of this, error analysis for this kind of discrete-valued\nproblems has a different form, as in Theorem 1.1. One bounds how small u needs to be to\nguarantee a correct answer. Such a bound, needless to say, also depends on the condition\nof the data d. Examples of this type of analysis can be found in [4, 6, 7, 8]. In each of these\nreferences a condition number for the problem at hand occurs in the error analysis. We note\nthat the one in [6] is essentially our κ(f).\nThe rest of the paper is organized as follows. In Section 2 we describe the basic objects we\nwill deal with as well as fixing the notation. In Sections 3 and 4 we prove the two technical\nresults our algorithm relies on. In Section5 we describe the algorithm under the assumption\nof infinite precision and we prove parts (1), (2), and (3) of Theorem 1.1. The geometric\nideas making the algorithm work are best seen in this context. Section 6 then describes the\nnecessary modifications to make the algorithm work as well under finite precision. These\nmodifications are simple and can be summarized by saying that we relax a bit the inequalities\ntested in the algorithms to make room for the finite-precision errors to fit in.\n2\nPreliminaries\nDenote by Hd the subspace of R[X0, . . . , Xn] of homogeneous polynomials of degree d. Then,\nHd = Hd1 × · · · × Hdn.\n3"},{"paragraph_id":"p6","order":6,"text":"If g ∈Hd we write\ng(X) =\nX\nJ\ngJXJ\nwhere J = (J0, . . . , Jn) is assumed to range over all multi-indices such that |J| = Pn\nk=0 Jk =\nd, XJ = XJ0\n0 XJ1\n1 . . . XJn\nn\nand gJ ∈R. Multinomial coefficients are defined by:\n d\nJ"},{"paragraph_id":"p7","order":7,"text":"=\nd!\nJ0!J1! · · · Jn!.\nThe space Hd is endowed with the inner product\n⟨g, h⟩=\nX\n|J|=d\ngJhJ\n d\nJ"},{"paragraph_id":"p8","order":8,"text":"which gives rise to the norm ∥g∥=\np\n⟨g, g⟩. These norms, for d1, . . . , dn, induce a norm in\nHd by taking for f = (f1, . . . , fn) ∈Hd:\n∥f∥= ∥(f1, . . . , fn)∥= max\n1≤i≤n ∥fi∥.\nLet O(n + 1) be the orthogonal group. The inner product above is known to be O(n + 1)-\ninvariant: for all Q ∈O(n + 1) and all g, h ∈Hd,\n⟨g ◦Q, h ◦Q⟩= ⟨g, h⟩.\n(This is a direct consequence of [26, III-7] or [2, Theorem 1 p. 218], by considering O(n+ 1)\nas subgroup of U(n + 1)).\nThe associated norm ∥f∥on Hd is therefore also O(n + 1)-\ninvariant. We will use this norm on Hd all along this paper. For x = (x1, . . . , xn) ∈Rn we\nrecall that ∥x∥2 = (x2\n1 + · · · + x2\nn)1/2 and ∥x∥∞= max{|x1|, . . . , |xn|}. We will often denote\n∥x∥2 simply by ∥x∥.\nFor f ∈Hd and x ∈Sn define\nμnorm(f, x) = ∥f∥√n"},{"paragraph_id":"p9","order":9,"text":"Df(x)−1\n|TxSn"},{"paragraph_id":"p10","order":10,"text":"√d1\n√d2\n...\n√dn"},{"paragraph_id":"p11","order":11,"text":"(1)\nwhere Df(x)|TxSn is the restriction to the tangent space of x at Sn of the derivative of f at\nx and the norm is the spectral norm, i.e. the operator norm with respect to ∥∥2. We now\ndefine the condition number κ(f) of f ∈Hd:\nκ(f) = max\nx∈Sn min"},{"paragraph_id":"p12","order":12,"text":"μnorm(f, x),\n∥f∥\n∥f(x)∥∞"},{"paragraph_id":"p13","order":13,"text":".\nRemark 2.1. The quantity κ(f) is closely related to other condition numbers for similar\nproblems.\nA version of the quantity μnorm(f, ζ) was introduced in [21, 22, 23] (see also [2, Chap-\nter 12]) for a complex polynomial system f and a zero ζ of f in the complex unit sphere\nSn\nC ⊂Cn+1. The normalized condition number of such a system f was then defined to be\nμnorm(f) :=\nmax\nζ∈Sn\nC |f(ζ)=0 μnorm(f, ζ).\n(2)\n4"},{"paragraph_id":"p14","order":14,"text":"Actually, the version of μnorm(f, ζ) introduced in [21, 22, 23] differs from (1) in the fact\nthat ∥f∥is defined as (P ∥fi∥2)1/2 (and there is no √n factor). It is bounded above by the\nexpression in (1).\nOver the reals, the right-hand side in (2) may not be well-defined since the zero set\nof f may be empty.\nIn [8] real systems were considered (as in the present paper) and\nan algorithm deciding feasibility of f (i.e., whether f has a real zero) was proposed. Its\ncomplexity was analyzed in terms of a condition number which, using our notation and\nmodulo minor details, is defined as follows"},{"paragraph_id":"p15","order":15,"text":"min\nζ∈Sn|f(ζ)=0 μnorm(f, ζ)\nif f is feasible\nmax\nζ∈Sn\n∥f∥\n∥f(ζ)∥∞\nif f is infeasible.\nNote the use of min (instead of max) in the first line above. This is due to the fact that the\ntime needed for the algorithm in [8] to detect the existence of a zero depends on the best\nconditioned zero of f. The existence of other, poorly conditioned (or even singular), zeros\nof f is irrelevant.\nShortly after, the algorithm in [8] was extended to an algorithm which would, in addition\nand if f is feasible, return a zero of f [6]. The complexity of this extension was studied\nin terms of a condition number (denoted ̺(f) in [6]) which, essentially, coincides with our\nκ(f).\nProposition 2.2. For all f ∈Hd, κ(f) ≥1.\nProof.\nLet x ∈Sn. Because of orthogonal invariance, we may assume without loss of\ngenerality that x = e0 := (1, 0, . . . , 0).\nIt is then immediate that ∥f(x)∥∞≤∥f∥. This shows that the second expression in the\ndefinition of κ is at least 1.\nFor the first expression, i.e., μnorm(f, x), define g = (g1, . . . , gn) ∈Hd by gi(X) = fi(X)−\nfi(e0)Xdi\n0 . Then g(e0) = 0 and [2, Corollary 3 p. 234], μnorm(g, e0) ≥1 (this is shown for\nthe version of μnorm with the 2-norm for ∥f∥, which is bounded above by the expression (1)).\nSince Df(e0) = Dg(e0) and ∥g∥≤∥f∥, we can conclude μnorm(f, e0) ≥μnorm(g, e0) ≥1.\n3\nThe exclusion Lemma\nIn this article, d( , ) denotes the Riemannian (angular) distance in Sn (which satisfies\n0 ≤d(x, y) ≤π, ∀x, y ∈Sn) and for x ∈Sn, r > 0, we set B(x, r) := {y ∈Sn : d(y, x) < r}\nand B(x, r) := {y ∈Sn : d(y, x) ≤r}.\nThe following result can be used to support an exclusion test.\nLemma 3.1. Let f ∈Hd and let x, y ∈Sn such that d(x, y) ≤\n√\n2. Then,\n∥f(x) −f(y)∥∞≤∥f∥\n√\nD d(x, y)\nIn particular, if f(x) ̸= 0, there is no zero of f in B(x, min{∥f(x)∥∞/(∥f∥\n√\nD),\n√\n2}).\n5"},{"paragraph_id":"p16","order":16,"text":"Proof.\nAn immediate consequence of the definition of the O(n + 1)-invariant inner\nproduct is that Hd endowed with this inner product is a reproducing kernel Hilbert space [9,\nProp. 2.21]. This implies that, for all g ∈Hd and x ∈Rn+1,\ng(x) = ⟨g(X), (xT X)deg g⟩.\n(3)\nBecause of orthogonal invariance, we can assume that x = e0 and y = e0 cos θ + e1 sin θ,\nwhere θ = d(x, y). Equation (3) implies that\nfi(x) −fi(y)\n=\n⟨fi(X), (xT X)di⟩−⟨fi(X), (yT X)di⟩= ⟨fi(X), (xT X)di −(yT X)di⟩\n=\n⟨fi(X), Xdi\n0 −(X0 cos θ + X1 sin θ)di⟩.\nHence, Cauchy-Schwarz-Bunyakowsky implies:\n|fi(x) −fi(y)| ≤∥fi∥∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥.\nSince\nXdi\n0 −(X0 cos θ + X1 sin θ)di = Xdi\n0 (1 −(cos θ)di) +\ndi\nX\nk=1\n di\nk"},{"paragraph_id":"p17","order":17,"text":"(cos θ)di−k(sin θ)kXdi−k\n0\nXk\n1 ,\nwe have:\n∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥2\n=\n(1 −(cos θ)di)2 +\ndi\nX\nk=1\n di\nk"},{"paragraph_id":"p18","order":18,"text":"(cos θ)2(di−k)(sin θ)2k\n=\n(1 −(cos θ)di)2 + 1 −(cos θ)2di\n=\n2(1 −(cos θ)di)\n≤\n2(1 −(1 −θ2\n2 )di)\n(4)\n≤\n2(1 −(1 −di\nθ2\n2 ))\n(5)\n≤\ndiθ2,\nwhere the inequality in line (4) is obtained from Taylor expanding cos θ around 0, and the\ninequality in line (5) is due to the fact that (1 −a)d ≥1 −da for a ≤1.\nWe conclude that\n|fi(x) −fi(y)| ≤∥fi∥θ\np\ndi\nand hence\n∥f(x) −f(y)∥∞≤∥f∥θ\nq\nmax\ni\ndi.\nFor the second assertion, we have\n∥f(y)∥∞\n≥\n∥f(x)∥∞−∥f(x) −f(y)∥∞\n≥\n∥f(x)∥∞−∥f∥\n√\nD d(x, y)\nsince d(x, y) ≤\n√\n2\n>\n∥f(x)∥∞−∥f∥\n√\nD ∥f(x)∥∞/(∥f∥\n√\nD)\n=\n0.\n6"},{"paragraph_id":"p19","order":19,"text":"4\nThe proximity Theorem\n4.1\nNewton and Smale\nNewton iteration on the sphere Sn is defined by\nNf :\nSn\n→\nSn\nx\n7→\nNf(x) = expx"},{"paragraph_id":"p20","order":20,"text":"−Df(x)−1\n|TxSnf(x)"},{"paragraph_id":"p21","order":21,"text":"where expx is the exponential map at x,\nexpx h = cos(∥h∥)x + sin(∥h∥)\n∥h∥\nh.\nFurthermore, the standard invariants of α-theory, introduced by Smale in [24], can be\ndefined as:\nβ(f, x)\n="},{"paragraph_id":"p22","order":22,"text":"Df(x)−1\n|TxSnf(x)"},{"paragraph_id":"p23","order":23,"text":",\nγ(f, x)\n=\nsup\nk≥2"},{"paragraph_id":"p24","order":24,"text":"Df(x)−1\n|TxSnDkf(x)|(TxSn)k\nk!"},{"paragraph_id":"p25","order":25,"text":"1/(k−1)\n,\nα(f, x)\n=\nβ(f, x)γ(f, x).\nRemark 4.1.\n(i) It is easy to see that β(f, x) = d(x, Nf(x)).\n(ii) We will not use Newton’s method in our algorithm. We are instead interested in its\nalpha theory which guarantees existence of zeros near points x with α(f, x) small\nenough.\n(iii) The Newton iteration presented above is not the iteration known as ‘projective New-\nton’. There is an alpha theory for that method, available in [19].\nHere we use slight modifications of the quantities α, β and γ, more adapted to our\npurposes. We set\nβ(f, x)\n:=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\nγ(f, x)\n:=\nD3/2\n2\nμnorm(f, x)\nα(f, x)\n:=\nβ(f, x)γ(f, x).\nThe definition of γ is motivated by the estimate of γ [2, Theorem 2 p. 267].\nγ(f, x) ≤γ(f, x).\nwhich yields the lower bound\nκ(f) ≥\nmax\nζ|f(ξ)=0 2D−3/2γ(f, ζ).\n(6)\n7"},{"paragraph_id":"p26","order":26,"text":"We also observe that γ(f, x) ≥D3/2\n2\nsince μnorm(f, x) ≥1 and that β(f, x) ≤β(f, x)\nsince\nβ(f, x) ="},{"paragraph_id":"p27","order":27,"text":"Df(x)−1\n|TxSnf(x)"},{"paragraph_id":"p28","order":28,"text":"≤√n∥f(x)∥∞"},{"paragraph_id":"p29","order":29,"text":"Df(x)−1\n|TxSn"},{"paragraph_id":"p30","order":30,"text":"≤μnorm(f, x)∥f(x)∥∞\n∥f∥\n= β(f, x).\nTherefore α(f, x) ≤α(f, x).\n4.2\nProximity and unicity from data at a point\nDefinition 4.2. We say that x ∈Sn is an approximate zero for f if and only if the Newton\nsequence {xk}k∈N, where x0 := x and xk+1 := Nf(xk), is defined for all k and moreover\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nd(x0, x1).\nThe limit point ζ = limk→∞xk is a fixed point for Newton iteration and a zero of f. It is\ncalled the associated zero to x.\nIn what follows we denote σ := P\nk≥0 2−2k+1 = 1.632843018 . . . and we set\nBf(x) := {y ∈Sn | d(x, y) ≤σβ(f, x)}.\nThe main technical tool in our algorithm is provided by the following result.\nTheorem 4.3. There exists an universal constant α∗:= 0.0384629388 . . . such that for all\nx ∈Sn, if α(f, x) < α∗, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z in Bf(x) the Newton sequence starting at z converges\nto ζ.\n4.3\nBackground material\nTheorem 4.3 is a consequence of the following two results, which are restatements of results\nproved in [10]. While [10] deals with Newton iteration on arbitrary complete real analytic\nRiemannian manifolds, here we reword them in terms of Newton iteration on the unit\nsphere Sn (Example 1 in [10]). The γ-Theorem for mappings [10, Theorem 1.3] becomes\nthe following.\nTheorem 4.4. Let f : Sn →Rn be analytic. Suppose that f(ζ) = 0 and Df(ζ) is an\nisomorphism. Let\nR(f, ζ) := min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nIf d(x, ζ) ≤R(f, ζ), then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and\nd(xk, ζ) ≤\n 1\n2\n 2k−1 d(x, ζ). In particular, {xk} converges to ζ.\n8"},{"paragraph_id":"p31","order":31,"text":"Now let α0 := 0.130716944 . . . denote the smallest positive root of the polynomial ψ(u)2−\n2u, and\ns0 :=\n1\nσ + (1−σα0)2\nψ(σα0)"},{"paragraph_id":"p32","order":32,"text":"1 +\nσ\n1−σα0\n = 0.103621842 . . .\nWe state the α-Theorem for mappings [10, Theorem 1.4] for the sphere Sn.\nTheorem 4.5. Let f : Sn →Rn be analytic. Let x ∈Sn be such that β(f, x) ≤s0π and\nα(f, x) ≤α0. Then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and converges\nto a zero ζ of f. Moreover,\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nβ(f, x)\nand\nd(xk, ζ) ≤σβ(f, x).\nFinally we introduce ψ(u) := 1−4u+2u2, which is positive and decreasing for 0 < u < 1−\n√\n2\n2 ,\nand state [10, Lemma 4.3]:\nLemma 4.6. Let x, y ∈Sn with d(x, y) < π. Suppose that Df(x) is nonsingular and\nν := d(x, y)γ(f, x) < 1 −\n√\n2\n2 .\nThen\nγ(f, y) ≤\nγ(f, x)\n(1 −ν)ψ(ν).\n4.4\nProof of Theorem 4.3\nSet ν∗:= 0.0628039411 . . . for the only real root of the polynomial\nΨ(u) := (3 −\n√\n7)(1 −u)ψ(u) −4u,\nand α∗:= ν∗\nσ = 0.0384629388 . . .. Note that α∗≤min{α0, s0π}.\nSince γ(f, x) ≥D3/2\n2\n, the hypothesis of Theorem 4.5 hold from α(f, x) ≤α(f, x) < α∗≤\nα0 and β(f, x) ≤β(f, x) ≤2α(f,x)\nD3/2\n<\n2α∗\nD3/2 < s0π.\nUsing Remark 4.1(i) it follows that x is an approximate zero of f, and that the associated\nzero ζ satisfies:\nd(x, ζ) ≤σβ(f, x) ≤σβ(f, x).\nThis already proves Parts (i) and (ii) of Theorem 4.3.\nWe show (iii). Since d(x, ζ) ≤σβ(f, x) < σs0π < π,\nν = d(x, ζ)γ(f, x) ≤σβ(f, x)γ(f, x) ≤σα(f, x) ≤σα∗= ν∗< 1 −\n√\n2\n2 ,\nand we can apply Lemma 4.6. Therefore\n4σβ(f, x)γ(f, ζ) ≤4σβ(f, x)γ(f, x)\n1\n(1 −ν)ψ(ν) ≤4ν∗\n1\n(1 −ν∗)ψ(ν∗) = 3 −\n√\n7,\n9"},{"paragraph_id":"p33","order":33,"text":"because (1−u)ψ(u) decreases for 0 < u < 1−\n√\n2\n2 , and ν∗is a zero of (3−\n√\n7)(1−u)ψ(u)−4u.\nThis shows, since 2σβ(f, x) ≤π, that\n2σβ(f, x) ≤R(f, ζ) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nWe conclude applying Theorem 4.4 to z ∈Bf(x), since\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤2σβ(f, x) ≤R(f, ζ).\nIt follows that the Newton sequence {zk}k∈N starting at z converges to ζ.\nRemark 4.7. The hypothesis on the radius of injectivity in [10] was recently found to be\nredundant.\n5\nInfinite precision\n5.1\nGrids and Graphs\nOur algorithm works on a grid on Sn. We easily construct one by projecting onto Sn a\ngrid on the cube Cn = {y | ∥y∥∞= 1}. We make use of the (easy to compute) bijections\nφ : Cn →Sn and φ−1 : Sn →Cn given by φ(y) =\ny\n∥y∥and φ−1(x) =\nx\n∥x∥∞.\nGiven η := 2−k for some k ≥1, we consider the uniform grid Uη of mesh η on Cn. This is\nthe set of points in Cn whose coordinates are of the form i2−k for i ∈{−2k, −2k+1, . . ., 2k},\nwith at least one coordinate equal to 1 or −1. We denote by Gη its image by φ in Sn. Note\nthat, for y1, y2 ∈Cn,\nd(φ(y1), φ(y2)) ≤π\n2 ∥y1 −y2∥2 ≤π\n2\n√\nn + 1 ∥y1 −y2∥∞.\n(7)\nGiven η as above we associate to it a graph Gη as follows. We set A(f) := {x ∈Sn |\nα(f, x) < α∗}. The vertices of the graph are the points in Gη ∩A(f). Two vertices x, y ∈Gη\nare joined by an edge if and only if Bf(x) ∩Bf(y) ̸= ∅.\nNote that as a simple consequence of Theorem 4.3 we obtain the following lemma.\nLemma 5.1.\n(i) For each x ∈A(f) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each point\nz in Bf(x), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈A(f). Then ζx = ζy ⇐⇒Bf(x) ∩Bf(y) ̸= ∅.\nWe define Z(Gη) := S\nx∈Gη Bf(x) ⊂Sn where x ∈Gη has to be understood as x running\nover all the vertices of Gη. Similarly, for a connected component U of Gη, we define\nZ(U) :=\n[\nx∈U\nBf(x).\nLemma 5.2.\n(i) For each component U of Gη, there is a unique zero ζU ∈Z(f) such that ζU ∈Z(U).\nMoreover, ζU ∈∩x∈UBf(x).\n10"},{"paragraph_id":"p34","order":34,"text":"(ii) If U and V are different components of Gη, then ζU ̸= ζV .\nProof.\n(i) Let x ∈U. Since x ∈A(f), by Lemma 5.1 (i) there exists a zero ζx of f\nin Bf(x) ⊆Z(U). This shows the existence. For the second assertion and the uniqueness,\nassume that there exist ζ and ξ zeros of f in Z(U). Let x, y ∈U be such that ζ ∈Bf(x),\nand ξ ∈Bf(y). Since U is connected, there exist x0 = x, x1, . . . , xk−1, xk := y in A(f) such\nthat (xi, xi+1) is an edge of Gη for i = 0, . . . , k −1, that is, Bf(xi)∩Bf(xi+1) ̸= ∅. If ζi and\nζi+1 are the associated zeros of xi and xi+1 in Z(f) respectively, then by Lemma 5.1(ii) we\nhave ζi = ζi+1, and thus ζ = ξ ∈Bf(x) ∩Bf(y).\n(ii) Assume ζU = ζV ∈Bf(x) ∩Bf(y) ⊂Z(U) ∩Z(V ), then x and y are joined by an edge\nand belong to the same connected component.\n5.2\nThe infinite precision algorithm\nCount Roots 1(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet U1, . . . , Ur be the connected components of Gη\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈Ui and all xj ∈Uj, d(xi, xj) > πη√n + 1\nand\n(ii) for all x ∈Gη \\ A(f), ∥f(x)∥∞> π\n2 η\np\n(n + 1)D∥f∥\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\n5.3\nProof of Theorem 1.1(1–3)\nProof of Part (1)\nThis proof requires some arguments of convexity.\nWe can natu-\nrally define spherical convex hulls for sets of points in Hn, an open half-sphere in Sn. If\nx1, . . . , xq ∈Hn we define\nSCH(x1, . . . , xq) := Cone(x1, . . . , xq) ∩Sn\nwhere Cone(x1, . . . , xq) is the smallest convex cone with vertex at the origin and containing\nthe points x1, . . . , xq. Alternatively, we have,\nSCH(x1, . . . , xq) =\n λ1x1 + · · · + λqxq\n∥λ1x1 + · · · + λqxq∥| λ1, . . . , λq ≥0,\nX\nλi = 1"},{"paragraph_id":"p35","order":35,"text":".\nWe will use the following fact.\nLemma 5.3. Let x1, ..., xq ∈Hn ⊂Rn+1. If Tq\ni=1 B(xi, ri) ̸= ∅, then SCH(x1, . . . , xq) ⊂\nSq\ni=1 B(xi, ri).\nProof.\nLet x ∈SCH(x1, . . . , xq) and y ∈Tq\ni=1 B(xi, ri). We will prove that x ∈B(xi, ri)\nfor some i.\nIf x = y, this is obvious.\nIf x ̸= y, let H be the half-space\nH :="},{"paragraph_id":"p36","order":36,"text":"z ∈Rn+1 : ⟨z, y −x⟩< 0"},{"paragraph_id":"p37","order":37,"text":".\n11"},{"paragraph_id":"p38","order":38,"text":"Since ∥x∥= ∥y∥= 1, we have ⟨x + y, y −x⟩= 0, and we note that in this case, x + y\ndetermines the mid-line between x and y. Moreover, since x ̸= y, we have x ∈H since\n⟨x, y −x⟩= ⟨x, y⟩−∥x∥2 < ∥x∥∥y∥−∥x∥2 = 0. Therefore the half-space H is the set of\npoints z in Rn+1 such that the Euclidean distance ∥z −x∥< ∥z −y∥.\nOn the other hand, H must contain at least one point of the set {x1, ..., xq} since if\nthis were not the case, the convex set Cone(CH(x1, . . . , xq)) would be contained in {z :\n⟨z, y −x⟩≥0}, contradicting x ∈SCH(x1, . . . , xq). Let, therefore, xi ∈H. It follows that\n∥x −xi∥< ∥y −xi∥\nwhich implies\nd(x, xi) < d(y, xi) ≤ri.\nWe can now proceed. Assume the algorithm halts, we want to show that if r equals\nthe number of connected components of Gη, then #R(f) = #Z(f)/2 = r/2. We already\nknow by Lemma 5.2 that each connected component U of Gη determines uniquely a zero\nζU ∈Z(f). Thus it is enough to prove that Z(f) ⊂Z(Gη).\nAssume that there is a zero ζ of f in Sn such that ζ is not in Z(Gη). Let B∞(φ−1(ζ), η) :=\n{y ∈Uη | ∥y −φ−1(ζ)∥∞≤η} = {y1, . . . , yq}, the set of all neighbors of φ−1(ζ) in Uη, and\nlet xi = φ(yi), i = 1, . . . , q. Clearly, φ−1(ζ) is in the cone spanned by {y1, . . . , yq} and hence\nζ ∈SCH(x1, . . . , xq).\nWe claim that there exists j ≤q such that xj ̸∈A(f). Indeed, assume this is not the\ncase. We consider two cases.\n(a)\nAll the xi belong to the same connected component U of Gη. By Lemma 5.2 there\nexists a unique zero ζU ∈Sn of f in Z(U) and ζU ∈∩iBf(xi). We may apply Lemma 5.3\nto deduce that\nSCH(x1, . . . , xq) ⊆\n[\nBf(xi).\nIt follows that, for some i ∈{1, . . . , q}, ζ ∈Bf(xi) ⊆Z(U), contradicting that ζ ̸∈Z(Gη).\n(b)\nThere exist l̸= s and 1 ≤i < j ≤r such that xl∈Ui and xs ∈Uj. Since condition\n(i) in the algorithm is satisfied, d(xl, xs) > πη√n + 1. But, by (7),\nd(xl, xs) ≤π\n2\n√\nn + 1∥yl−ys∥∞≤π\n2\n√\nn + 1\n ∥yl−φ−1(ζ)∥∞+ ∥φ−1(ζ) −ys∥∞"},{"paragraph_id":"p39","order":39,"text":"≤πη\n√\nn + 1,\na contradiction.\nWe have thus proved the claim. Let then 1 ≤j ≤q be such that xj ̸∈A(f). Since\ncondition (ii) in the algorithm is satisfied ∥f(xj)∥∞> π\n2 η\np\n(n + 1)D∥f∥. It follows from\nthe inequality d(xj, ζ) ≤π\n2\n√n + 1η and Lemma 3.1 that ∥f(ζ)∥∞> 0, a contradiction.\nProof of Part (2)\nWe need a few lemmas.\nLemma 5.4. If ζ1 ̸= ζ2 ∈Z(f) then\nd(ζ1, ζ2) ≥2(3 −\n√\n7)D−3/2\nκ(f)\n.\n12"},{"paragraph_id":"p40","order":40,"text":"Proof.\nFor i = 1, 2, using (6) and Proposition 2.2,\nR(f, ζi) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζi)\n)\n≥min\n(\nπ, (3 −\n√\n7)D−3/2\nκ(f)\n)\n= (3 −\n√\n7)D−3/2\nκ(f)\n.\nNow suppose that d(ζ1, ζ2) < R(f, ζ1) + R(f, ζ2) and choose x ∈Sn such that d(x, ζ1) <\nR(f, ζ1) and d(x, ζ2) < R(f, ζ2). Then Theorem 4.4 implies that ζ1 = ζ2, a contradiction.\nLemma 5.5. Let x1, x2 ∈Gη with associated zeros ζ1 ̸= ζ2. If η ≤\n2(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen\nd(x1, x2) > πη√n + 1.\nProof.\nAssume d(x1, x2) ≤πη√n + 1.\nSince x2 ̸∈Bf(x1), d(x1, x2) > σβ(f, x1).\nConsequently,\nd(x1, ζ1) ≤σβ(f, x1) < d(x1, x2) ≤πη\n√\nn + 1\nand, similarly, d(x2, ζ2) < πη√n + 1. But then,\nd(ζ1, ζ2) ≤d(ζ1, x1) + d(x1, x2) + d(x2, ζ2) < 3πη\n√\nn + 1 ≤2(3 −\n√\n7)D−3/2\nκ(f)\ncontradicting Lemma 5.4.\nLemma 5.6. Let x ∈Sn such that x ̸∈A(f).\nIf η ≤\nα∗\n(n+1)D2κ(f)2 then ∥f(x)∥∞>\nπ\n2 η\np\n(n + 1)D∥f∥.\nProof.\nSince x ̸∈A(f) we have α(f, x) ≥α∗. We divide the proof in two cases.\nCase I. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n=\n∥f∥\n∥f(x)∥∞\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗∥f(x)∥2\n∞\n(n + 1)D2∥f∥2\nwhich implies, since η ≤1\n2 <\n4D\nπ2α∗,\n∥f(x)∥∞≥\n√η√n + 1D∥f∥\n√α∗\n> π\n2 η\np\n(n + 1)D∥f∥.\nCase II. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗\n(n + 1)D2μnorm(f, x)2\nwhich implies α∗≥η(n + 1)D2μnorm(f, x)2. Also,\nα∗≤α(f, x) = 1\n2β(f, x)μnorm(f, x)D3/2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞.\n13"},{"paragraph_id":"p41","order":41,"text":"Putting both inequalities together we obtain\nη(n + 1)D2μnorm(f, x)2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞\nor yet,\n∥f(x)∥∞≥2η(n + 1)D1/2∥f∥> π\n2 η\np\n(n + 1)D∥f∥.\nWe can now conclude the proof of Part (2).\nAssume η ≤\nα∗\n(n+1)D2κ(f)2 .\nThen the\nhypotheses of Lemmas 5.5 and 5.6 hold. The first of these lemmas ensures that condition (i)\nin the algorithm is satisfied. The second, that condition (ii) is so. Therefore, the algorithm\nhalts as soon as\nα∗\n2(n+1)D2κ(f)2 < η ≤\nα∗\n(n+1)D2κ(f)2 . This gives a bound of O(ln(nDκ(f)))\nfor the number of iterations. Since the number of grid points considered at this iteration\n(η =\nα∗\n(n+1)D2κ(f)2 ) is at most 2(n + 1)"},{"paragraph_id":"p42","order":42,"text":"2(n+1)D2κ(f)2\nα∗\n n\n, the bound for the total complexity\nfollows.\nProof of Parts (3) and (5)\nWe have already seen that the number of iterations is\nbounded by O(ln(nDκ(f))). At each of these iterations, we need to perform a number of\ncomputations on the (at most) 2(n+1)"},{"paragraph_id":"p43","order":43,"text":"2(n+1)D2κ(f)2\nα∗\n n\ngrid points to decide whether they\nare in A(f). These can be done independently. Then, we need to compute the number of\nconnected components of Gη. This can be done (see, e.g., [15]) in parallel time O(ln(|Vη|))2\nwhere |Vη| denotes the number of vertices of Gη and therefore, in parallel time at most\nO(n2(ln(nDκ(f))2 + ln(α∗)2)). Since this is the dominant step in the computation at a\ngiven iteration, it follows that the total parallel time consumed by the algorithm is at most\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2)). This shows part (3). For part (5), just note\nthat, for i = 1, . . . , r, any vertex xi of Ui is an approximate zero of the only zero of f in\nZ(Ui).\n6\nFinite Precision\n6.1\nMaking room to allow errors\nOur finite precision algorithm will be a variation of Algorithm Count Roots 1. But since\nfinite precision computations will be affected by errors, we need to make room in the infinite\nprecision algorithm to allow them.\nFor this aim, we state the corresponding version of\nTheorem 4.3.\nTheorem 6.1. There exist a universal constant α• = 0.028268 · · · such that, for all x ∈Sn,\nif α(f, x) < α•, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z s.t. d(x, z) ≤2σβ(f, x) the Newton sequence starting at\nz converges to ζ.\n14"},{"paragraph_id":"p44","order":44,"text":"Proof.\nParts (i) and (ii) follow from Theorem 4.3 and the fact that α• < α∗. Part (iii)\nis proved by taking ν• = 0.046158 · · · to be the only real root of the polynomial Ψ(u) :=\n(3 −\n√\n7)(1 −u)ψ(u) −6u, and α• = ν•\nσ = 0.028268. Then, one proves as in Theorem 4.3\nthat 3σβ(f, x) ≤R(f, ζ) from which it follows that, for all z s.t. d(x, z) ≤2σβ(f, x),\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤3σβ(f, x) ≤R(f, ζ)\nand hence, that the Newton sequence {zk}k∈N starting at z converges to ζ.\nThe proofs of Lemmas 5.5 and 5.6 yield, mutatis mutandis, the following results.\nLemma 6.2. Let x1, x2 ∈Gη with associated zeros ξ1 and ξ2, ξ1 ̸= ξ2. If η ≤(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen d(x1, x2) > 2πη√n + 1.\nLemma 6.3. Let x ∈Sn such that α(f, x) > α•\n3 . If η ≤\nα•\n4D2(n+1)κ(f)2 then ∥f(x)∥∞>\nπη\np\n(n + 1)D∥f∥.\n6.2\nBasic facts\nWe recall the basics of a floating-point arithmetic which idealizes the usual IEEE standard\narithmetic. This system is defined by a set F ⊂R containing 0 (the floating-point numbers),\na transformation r : R →F (the rounding map), and a constant u ∈R (the round-offunit)\nsatisfying 0 < u < 1. The properties we require for such a system are the following:\n(i) For any x ∈F, r(x) = x. In particular, r(0) = 0.\n(ii) For any x ∈R, r(x) = x(1 + δ) with |δ| ≤u.\nWe also define on F arithmetic operations following the classical scheme\nxe◦y = r(x ◦y)\nfor any x, y ∈F and ◦∈{+, −, ×, /}, so that\ne◦: F × F →F.\nThe following is an immediate consequence of property (ii) above.\nProposition 6.4. For any x, y ∈F we have\nxe◦y = (x ◦y)(1 + δ),\n|δ| ≤u.\nWhen combining many operations in floating-point arithmetic, quantities such as\nQn\ni=1(1 + δi)ρi naturally appear. Our round-offanalysis uses the notations and ideas in\nChapter 3 of [17], from where we quote the following results:\nProposition 6.5. If |δi| ≤u, ρi ∈{−1, 1}, and nu < 1, then\nn\nY\ni=1\n(1 + δi)ρi = 1 + θn,\nwhere\n|θn| ≤γn =\nnu\n1 −nu.\n15"},{"paragraph_id":"p45","order":45,"text":"Proposition 6.6. For any positive integer k such that ku < 1, let θk, θj be any quantities\nsatisfying\n|θk| ≤γk =\nku\n1 −ku\n|θj| ≤γj =\nju\n1 −ju.\nThe following relations hold.\n1. (1 + θk)(1 + θj) = 1 + θk+j for some |θk+j| ≤γk+j.\n2.\n1 + θk\n1 + θj\n=\n 1 + θk+j\nif j ≤k,\n1 + θk+2j\nif j > k.\nfor some |θk+j| ≤γk+j or some |θk+2j| ≤γk+2j.\n3. If ku, ju ≤1/2, then γkγj ≤γmin{k,j}.\n4. iγk ≤γik.\n5. γk + u ≤γk+1.\n6. γk + γj + γkγj ≤γk+j.\nFrom now on, whenever we write an expression containing θk we mean that the same\nexpression is true for some θk, with |θk| ≤γk.\nWhen computing an arithmetic expression q with a round-offalgorithm, errors will\naccumulate and we will obtain another quantity which we will denote by fl(q). We write\nError(q) = |q −fl(q)|.\nAn example of round-offanalysis which will be useful in what follows is given in the next\nproposition, the proof of which can be found in Section 3.1 of [17].\nProposition 6.7. There is a round-offalgorithm which, with input x, y ∈Rn, computes\nthe dot product of x and y. The computed value fl(⟨x, y⟩) satisfies\nfl(⟨x, y⟩) = ⟨x, y⟩+ θ⌈log2 n⌉+1⟨|x|, |y|⟩,\nwhere |x| = (|x1|, . . . , |xn|).\nIn particular, if x = y, the algorithm computes fl(∥x∥2)\nsatisfying\nfl(∥x∥2) = ∥x∥2(1 + θ⌈log2 n⌉+1).\nWe will also have to deal with square roots and arccosinus. The following result will\nhelp us to do so.\nLemma 6.8. (i) Let θ ∈R such that |θ| ≤1/2. Then,\n√\n1 −θ = 1 −θ′ with |θ′| ≤|θ|.\n(ii) Let 0 < a ≤1 and ε ∈R such that 0 < a + ε < 1. Then, arccos(a + ε) = arccos(a) +\nυ\n1\n√\n1−(a+ε)2 with |υ| ≤|ε|.\nProof.\nAssume θ > 0 (if θ < 0 it is done similarly). By the intermediate value theorem\nwe have that 1 −\n√\n1 −θ = θ(√ξ)′ with ξ ∈(1 −θ, 1). But\n(\np\nξ)′ =\n1\n2√ξ ≤\n1\n√\n2,\n16"},{"paragraph_id":"p46","order":46,"text":"the last since ξ ≥1/2. This proves (i).\nPart (ii) is shown similarly. Again, assume for simplicity that ε > 0. Then, for some\nξ ∈(a, a + ε),\narccos(a + ε) −arccos(a) = ε arccos′(ξ) = ε\n1\np\n1 −ξ2 =\nυ\np\n1 −(a + ε)2 .\nWe assume that, besides the four basic arithmetic operations, we are allowed to compute\nsquare roots and arccosinus with finite precision. That is, if op denotes any of these two\noperators, we compute f\nop such that\nf\nop(x) = op(x)(1 + δ),\n|δ| ≤u.\nFrom Lemma 6.8(i) it follows that, for all a > 0,\n^\np\na(1 + θk) = √a(1 + θk+1).\nRemark 6.9. Our choice of the precision u in Theorem 1.1(4) guarantees that ku < 1/2\nholds whenever we encounter θk in what follows, and consequently, |θk| ≤γk ≤2ku. This\nimplies that in all what follows we have γg = O(ug) for all the expressions g we will\nencounter.\nAccording to the previous remark we will introduce a further notation that will consid-\nerably simplify our exposition. For all expression g, we will write\n[[g]] := O(ug).\nThis notation will avoid we burden ourselves with the consideration of multiplicative con-\nstants.\n6.3\nThe finite precision algorithm\nOur finite precision algorithm is a variation of Algorithm Count Roots 1 in Section 5.3.\nGiven x ∈Sn we define below fl(A′(f)) and fl(B\n′\nf(x)), which are convenient floating\nversions of the sets A′(f) ="},{"paragraph_id":"p47","order":47,"text":"x ∈Sn | α(f, x) < 1\n2α•"},{"paragraph_id":"p48","order":48,"text":"and B\n′\nf(y) = {z ∈Sn | d(x, y) ≤\n3\n2σβ(f, x)} respectively.\nGiven f ∈Hd and x ∈Sn, we let M ∈Rn×n be a matrix representing"},{"paragraph_id":"p49","order":49,"text":"1\n√d1\n1\n√d2\n...\n1\n√dn"},{"paragraph_id":"p50","order":50,"text":"Df(x)|TxSn.\nand we set σmin(M) = ∥M −1∥−1. Therefore\nμnorm(f, x)\n=\n∥f∥√n ∥M −1∥= ∥f∥√n σmin(M)−1,\nβ(f, x)\n=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\n= √n σmin(M)−1∥f(x)∥∞,\nα(f, x)\n=\nβ(f, x)μnorm(f, x)D3/2\n2\n= ∥f∥n σmin(M)−2∥f(x)∥∞\nD3/2\n2\n.\n17"},{"paragraph_id":"p51","order":51,"text":"This implies that\ny ∈B\n′\nf(x)\n⇐⇒\nd(x, y) ≤3\n2σβ(f, x)\n⇐⇒\nσmin(M)d(x, y) ≤3\n2σ√n∥f(x)∥∞,\nx ∈A′(f)\n⇐⇒\nα(f, x) < α•\n2\n⇐⇒\n∥f∥n∥f(x)∥∞D3/2 < α•σmin(M)2.\nThese statements are equivalent under infinite precision, but the expressions at the right-\nhand side are more convenient to handle when working with finite precision. This motivates\nour definitions of\nfl(B\n′\nf(x))\n:="},{"paragraph_id":"p52","order":52,"text":"y ∈Sn | fl(σmin(M)d(x, y)) ≤fl(3\n2σ√n∥f(x)∥∞)"},{"paragraph_id":"p53","order":53,"text":"fl(A′(f))\n:=\nn\nx ∈Sn | fl(∥f∥n ∥f(x)∥∞D3/2) < fl(α•σmin(M)2)\no\nWe also define accordingly the graph fl(G′\nη) whose vertices are the points in Gη ∩\nfl(A′(f)), and with two vertices x, y joined by an edge if and only if fl(B\n′\nf(x))∩fl(B\n′\nf(x)) ̸=\n∅. Its connected components are denoted by fl(U).\nOur algorithm is the following:\nCount Roots 2(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet fl(U1), . . . , fl(Ur) be the connected components of fl(Gη)\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈fl(Ui) and all xj ∈fl(Uj), fl(d(xi, xj)) > fl( 3\n2πη√n + 1)\nand\n(ii) for all x ∈Gη \\ fl(A′(f)), fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D∥f∥)\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\nIn the rest of the section we will see that, when the precision u satisfies u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), this algorithm is correct and halts as soon as η\n≤\nα•\n4D2(n+1)κ(f)2 .\n6.4\nBounding errors for elementary computations\nThe goal of this subsection is to exhibit bounds for the accumulated error in the main\ncomputations of Count Roots 2. We will rely on the basic notations and results described\nin §6.2.\nTo simplify notation, and without loss of generality, in all what follows we assume that\n∥f∥= 1. We denote by S(Hd) the sphere of such systems. Also, we do not discuss in what\nfollows the accumulated error in the computation of φ : Cn →Sn. This is a minor detail\nwhich can be taken care of using Lemma 6.8(i).\nProposition 6.10. Given f ∈S(Hd) and x ∈Sn, we can compute ∥f(x)∥∞with finite\nprecision u such that\nError(∥f(x)∥∞) = [[D + log S]]\nwhere S is a bound on the number of coefficients of each fi.\n18"},{"paragraph_id":"p54","order":54,"text":"Proof.\nLet f = (f1, . . . , fn). For i ≤n write fi = P cJXJ and let S be the number\nof coefficients of fi. To compute f(x) one computes each monomial cJxJ with fl(cJxJ) =\ncJxJ(1 + θD). Then, one computes fi(x) to get\nfl(fi(x))\n=\nfl(\nX\nfl(cJxJ))\n=\nfl(\nX\ncJxJ(1 + θ(J)\nD ))\n=\nX\ncJxJ(1 + θ(J)\nD ) + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\n=\nfi(x) +\nX\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\nwhere in the third line we reasoned as in the proof of Proposition 6.7. Therefore\nError(∥f(x)∥∞)\n≤"},{"paragraph_id":"p55","order":55,"text":"X\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )"},{"paragraph_id":"p56","order":56,"text":"≤\nX\n|cJ| ∥xJ∥(γD + γlog S + γDγlog S)\n≤\nγD+log S\nwhere we used that for any x ∈Sn, | P |cJ|xJ| ≤∥P |cJ|xJ∥= ∥fi∥≤∥f∥= 1 and\nProposition 6.6 (6). The conclusion follows from Remark 6.9.\nProposition 6.11. Given f ∈S(Hd) and x ∈Sn, let M ∈Rn×n be a matrix representing"},{"paragraph_id":"p57","order":57,"text":"1\n√d1\n1\n√d2\n...\n1\n√dn"},{"paragraph_id":"p58","order":58,"text":"Df(x)|TxSn\nin some orthonormal basis of TxSn. Then ∥M∥≤√n. In addition, we can compute such a\nmatrix M with finite precision u such that\n∥Error(M)∥F = [[n(log S + D + log n)]].\nProof.\nStep 1: Let y =\nx−en+1\n∥x−en+1∥. The Householder symmetry\nHy = In+1 −2yyt\nswaps vectors en+1 and x, and fixes y⊥. The first n columns of Hy are therefore an orthonor-\nmal basis of TxSn, while the last column is x. Let H ∈R(n+1)×n denote the submatrix\nobtained from the first n columns of Hy. With that notation, we set\nM ="},{"paragraph_id":"p59","order":59,"text":"1\n√d1\n1\n√d2\n...\n1\n√dn"},{"paragraph_id":"p60","order":60,"text":"Df(x)H.\nStep 2: We claim that Pi,x : Hdi →Rn, fi 7→\n1\n√di Dfi(x)|TxSn is an orthogonal projection,\nin the sense that for any fixed x, the map (Pi,x)| ker(Pi,x)⊥is an isometry.\n19"},{"paragraph_id":"p61","order":61,"text":"We use an orthogonal invariance argument. The special orthogonal group SO(n + 1)\nacts on Hdi and on Rn+1 isometrically as follows: to a given Q ∈SO(n + 1), we associate\nrespectively the following isometries:\nx 7→Qx\n,\nfi 7→fi ◦Qt.\nWe set y = Qx and gi = fi ◦Qt. Differentiating the equality gi(Qx) = fi(x), we obtain:\nDgi(y)Q = Dfi(x).\nWhen x is fixed, we can set Q conveniently so that y = en+1. Therefore\nDgi(en+1)Q|TxSn = Dfi(x)|TxSn.\nSince Q(TxSn) = Ten+1Sn we obtain\nDgi(en+1)|Ten+1Sn = Dfi(x)|TxSn.\nThis means that Pi,en+1(fi ◦Qt) = Pi,x(fi). Thus, in order to prove our claim, it is enough\nto show that Pi,en+1 is an orthogonal projection.\nSince for g = P\nJ gJXJ,\n∂g\n∂Xj (en+1) = g(ej+(d−1)en+1) and since Ten+1Sn = ⟨e1, . . . , en⟩,\nwe have that for any gi ∈Hdi,\nPi,en+1(gi) =\n1\n√di\n gi(e1+(di−1)en+1), . . . , gi(en+(di−1)en+1)"},{"paragraph_id":"p62","order":62,"text":".\nHence, for any gi ∈ker(Pi,en+1)⊥, i.e. such that giJ = 0 for all J ̸= ej + (di −1)en+1,\n1 ≤j ≤n, we have\n∥gi∥2 =\nX\nJ\ng2\niJ\n di\nJ\n = ∥Pi,en+1(gi)∥2\n2.\nWe conclude that Pi,x is an orthogonal projection.\nStep 3: From the previous step, for any fi ∈Hdi, using the orthogonal decomposition\nfi = f ◦\ni + f ⊥\ni\nwith f ◦\ni ∈ker Pi,x and f ⊥\ni ∈ker P ⊥\ni,x, we have\n∥Pi,x(fi)∥2\n2 = ∥Pi,x(f ⊥\ni )∥2\n2 = ∥f ⊥\ni ∥2 ≤∥fi∥2.\nIt is now immediate from Step 1 and from the definition of ∥f∥= maxi ∥fi∥that the\nFrobenius norm ∥M∥F of the matrix M satisfies\n∥M∥2\nF =\nn\nX\ni=1\n∥Pi,x(fi)∥2\n2 ≤\nn\nX\ni=1\n∥fi∥2 ≤n∥f∥2 = n\nand hence its spectral norm ∥M∥satisfies ∥M∥≤∥M∥F ≤√n. This bound is independent\nof the choice of the basis for the space TxSn.\nStep 4: We next present the algorithm to compute M, given f and x. This is a non-\noptimal algorithm, and can be significantly improved if more is known on the structure of\nthe polynomial system f.\nWe can compute each entry mij of the matrix M as the scalar product of\n1\n√\ndi Dfi(x)\nand the jth column Hj := (hkj)1≤k≤n+1 of H.\n20"},{"paragraph_id":"p63","order":63,"text":"Proceeding as in the proof of Proposition 6.10, we can compute\n1\n√\ndi\n∂fi\n∂Xk (x) with\nError\n 1\n√\ndi\n∂fi\n∂Xk\n(x)"},{"paragraph_id":"p64","order":64,"text":"= [[D + log S]].\nOn the other hand, the vector y =\nx−en+1\n∥x−en+1∥can be computed using 2n + 4 operations, and\nclearly Error(yj) = [[ log(n)]] for all j. Hence, for all coefficients hkj of H,\nError(hkj) = [[ log(n)]].\nApplying Proposition 6.7 we conclude\nError(mij)\n=\n[[D + log S + log n]]"},{"paragraph_id":"p65","order":65,"text":"1\n√di\nDfi(x)"},{"paragraph_id":"p66","order":66,"text":"∥Hj∥\n=\n[[D + log S + log n]].\nThe second equality holds because ∥Hj∥= 1 since H is unitary, and because, as in the proof\nof Step 2,"},{"paragraph_id":"p67","order":67,"text":"1\n√di\nDfi(x)"},{"paragraph_id":"p68","order":68,"text":"2\n="},{"paragraph_id":"p69","order":69,"text":"1\n√di\nDgi(en+1)"},{"paragraph_id":"p70","order":70,"text":"2\n= 1\ndi\n∥(gi(e1+(di−1)en+1), . . . , gi(dien+1))∥2 ≤∥gi∥2 ≤1.\nThis implies\n∥Error(M)∥F ≤[[n(D + log S + log n)]].\nLemma 6.12. Let x ∈Sn and M be as in Proposition 6.11. We can compute σmin(M) =\n∥M −1∥−1 satisfying\nError(σmin(M)) = [[n(log S + D + n3/2)]].\nProof.\nLet E′ = M −fl(M). By Proposition 6.11,\n∥E′∥≤∥E′∥F ≤[[n(log S + D + log n)]].\nLet M = fl(M). We compute σmin(M) = ∥M −1∥−1 using a backward stable algorithm\n(e.g., QR factorization). Then the computed fl(σmin(M)) is the exact σmin(M + E′′) for a\nmatrix E′′ with\n∥E′′∥≤cn2u∥M∥\nfor some universal constant c (see, e.g., [11, 17]). Thus,\nfl(σmin(M)) = fl(σmin(M)) = σmin(M + E′′) = σmin(M + E′ + E′′).\nWrite E = E′ + E′′. Then, using ∥M∥≤√n,\n∥E∥\n≤\n∥E′∥+ ∥E′′∥≤∥E′∥+ cn2u∥M∥≤∥E′∥+ cn2u(∥M∥+ ∥E′∥)\n=\n[[n(log S + D + log n)]] + cn2u(√n + [[n(log S + D + log n)]])\n=\n[[n(log S + D + log n)]] + cn2u(√n + c′un(log S + D + n3/2))\n=\n[[n(log S + D + n3/2)]]\nsince the hypothesis on u implies c′un(log S + D + n3/2) is bounded by a constant term.\nTherefore, fl(σmin(M)) = σmin(M + E) which implies by [11, Corollary 8.3.2]:\nError(σmin(M)) ≤∥E∥< [[n(log S + D + n3/2)]].\n21"},{"paragraph_id":"p71","order":71,"text":"Proposition 6.13. Let f ∈S(Hd)). Assume u ≤\nK\nκ(f)2n2D log S for a small enough constant\nand let x ∈Sn. Then\n(i) If x /∈fl(A′(f)) then α(f, x) ≥1\n3α•.\n(ii) If x ∈fl(A′(f)) then α(f, x) < α•.\nProof.\nFrom Proposition 6.10\nfl(n∥f(x)∥∞D3/2)\n=\n(∥f(x)∥∞+ [[D + log S]])(nD3/2)(1 + θ4)\n≤\nnD3/2∥f(x)∥∞+ [[nD3/2(D + log S)]]\n. Also, from Lemma 6.12, using that σmin(M) ≤√n,\nfl(α•σmin(M)2)\n=\nα•"},{"paragraph_id":"p72","order":72,"text":"σmin(M) + [[n(log S + D + n3/2)]]\n 2\n(1 + θ2)\n≥\nα•σmin(M)2 −2α•σmin(M)[[n(log S + D + n3/2)]]\n≥\nα•σmin(M)2 −[[n3/2(log S + D + n3/2)]].\nTherefore,\nn∥f(x)∥∞D3/2 + [[nD3/2(D + log S)]]\n≥\nfl(n∥f(x)∥∞D3/2) ≥fl(α•σ2\nmin)\n≥\nα•σ2\nmin −[[n3/2(log S + D + n3/2)]]\nor yet,\nn∥f(x)∥∞D3/2 −α•σ2\nmin\n≥\n−([[nD3/2(D + log S)]] + [[n3/2(log S + D + n3/2)]])\n≥\n−[[n3D5/2 log S]].\nCase I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, using the hypothesis on u and the inequality\nκ(f) ≥1,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)n2D log S\n≤\nKO(1)n∥f(x)∥∞D3/2 ≤n∥f(x)∥∞D3/2\n2\nthe last by choosing K small enough. Hence, n∥f(x)∥∞D3/2 −α•σ2\nmin ≥−"},{"paragraph_id":"p73","order":73,"text":"n∥f(x)∥∞D3/2\n2"},{"paragraph_id":"p74","order":74,"text":",\nwhich implies 3\n2n∥f(x)∥∞D3/2 ≥α•σmin(M)2, i.e., α(f, x) ≥α•\n3 .\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)2n2D log S\n≤\nKO(1)σmin(M)2D3/2 ≤α•σmin(M)2\n3\n22"},{"paragraph_id":"p75","order":75,"text":"the last by choosing K small enough.\nThis implies n∥f(x)∥∞D3/2 −α•σmin(M)2 ≥\n−α•σmin(M)2\n3\nor, equivalently, α(f, x) ≥α•\n3 .\nThis shows part (i). For part (ii), one shows as above that\nn∥f(x)∥∞D3/2 −α•σ2\nmin ≤[[n3D5/2 log S]].\nThen, one proceeds as well by considering the two cases min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞and min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x).\nLemma 6.14. Let y1, y2 ∈Uη and let xi = φ(yi), i = 1, 2. Then d(x1, x2) ≥\nη\n2√n+1.\nProof.\nThe distance d(x1, x2) is minimized at y1 = (1, . . . , 1, 1) and y2 = (1, . . . , 1, 1−η).\nLet N = n + 1. Then\ncos(d(x1, x2))2\n=\n⟨y1, y2⟩2\n∥y1∥2∥y2∥2\n=\n(N −η)2\nN(N −2η + η2)\n=\n1 −\n(N −1)η2\nN 2 −2Nη + Nη2\n≤\n1 −η2 N −1\nN 2\n.\nHence\nd(x1, x2) ≥arccos\n r\n1 −η2 N −1\nN 2\n!\n= arcsin\n η\nN\n√\nN −1"},{"paragraph_id":"p76","order":76,"text":"≥\nη\n2\n√\nN\n.\nLemma 6.15. Let u <\nKη2\nn log n for a small enough constant K. For x1, x2 ∈Gη we can\ncompute d(x1, x2) such that\nError(d(x1, x2)) ≤\n √n log n\nη"},{"paragraph_id":"p77","order":77,"text":".\nProof.\nLet yi = φ−1(xi), i = 1, 2, and a = cos(d(x1, x2)), i.e.,\na = ⟨y1, y2⟩\n∥y1∥∥y2∥.\nWe have, using Proposition 6.7,\nfl(⟨y1, y2⟩) = ⟨y1, y2⟩+ θlog n∥y1∥∥y2∥\nand fl(∥y1∥∥y2∥) = ∥y1∥∥y2∥(1+θlog n). Using now Propositions 6.4, 6.5, and 6.6, it follows\nthat fl(a) = a + ε with ε = [[ log n]].\nBy choosing K sufficiently small, ε ≤\nη2n\n12(n+1)2 . Also, from the proof of Lemma 6.14,\na = cos(d(x1, x2)) ≤\ns\n1 −\nη2n\n(n + 1)2\n23"},{"paragraph_id":"p78","order":78,"text":"and hence, using that √z + y ≤√z + 3y whenever 0 < z, y ≤1, we obtain\na + ε ≤\ns\n1 −\nη2n\n(n + 1)2 +\nη2n\n12(n + 1)2 ≤\ns\n1 −\n3η2n\n4(n + 1)2 ≤\ns\n1 −\nη2\n3(n + 1).\nUsing Lemma 6.8(ii) it follows that,\narccos(a + ε)\n=\narccos(a) + ε"},{"paragraph_id":"p79","order":79,"text":"1\np\n1 −(a + ε)2"},{"paragraph_id":"p80","order":80,"text":"=\narccos(a) + [[ log n]]"},{"paragraph_id":"p81","order":81,"text":"p\n3(n + 1)\nη\n .\nTherefore,\nError(d(x1, x2)) ≤\n √n log n\nη"},{"paragraph_id":"p82","order":82,"text":".\nLemma\n6.16. Let\nf\n∈\nS(Hd).\nAssume that\nη\n≥\nα•\n8D2(n+1)κ(f)2\nand u\n≤\nK\nD2n5/2κ(f)3(log S+n3/2D2κ(f)2) with K small enough, and let x, y ∈Gη. Then\n(i) If y ∈fl(B\n′\nf(x)) then d(x, y) ≤2σβ(f, x).\n(ii) If y /∈fl(B\n′\nf(x)) then d(x, y) > σβ(f, x).\nProof.\nBy Lemmas 6.12 and 6.15 (and using σmin(M) ≤√n and the bound d(x, y) ≤\nπ\n2 η√n + 1 which follows from (7)),\nError(σmin(M)d(x, y))\n=\nO\n d(x, y)Error(σmin(M)) + σmin(M)Error(d(x, y))"},{"paragraph_id":"p83","order":83,"text":"=\nη π\n2\n√\nn + 1[[n(log S + D + n3/2)]] + √n\n √n log n\nη"},{"paragraph_id":"p84","order":84,"text":"=\nη[[n3/2(log S + D + n3/2)]] +\n n log n\nη"},{"paragraph_id":"p85","order":85,"text":"≤\n[[n3/2 log S + n3D2κ(f)2]]\nthe last by the bounds on η. Also, using Proposition 6.10,\nError\n 3\n2σ√n∥f(x)∥∞"},{"paragraph_id":"p86","order":86,"text":"≤[[√n(D + log S)]].\nTherefore, for part (i),\nσmin(M)d(x, y) −3\n2σ√n∥f(x)∥∞\n≤fl(σmin(M)d(x, y)) −fl\n 3\n2σ√n∥f(x)∥∞"},{"paragraph_id":"p87","order":87,"text":"+ [[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n≤[[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n= [[n3/2 log S + n3D2κ(f)2]].\n24"},{"paragraph_id":"p88","order":88,"text":"Case I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, by the hypothesis on u,\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nκ(f)n(log S + n3/2D2κ(f)2)\n≤\nσ√n\n2κ(f) ≤σ√n∥f(x)∥∞\n2\nthe last line by taking K small enough. This implies that σmin(M)d(x, y) ≤2σ√n∥f(x)∥∞,\ni.e., that d(x, y) ≤2σβ(f, x).\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nD2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n≤\n√nα•\n48D2(n + 1)3/2κ(f)3\n≤\n√nη\n8√n + 1κ(f) ≤\n√nd(x, y)\n4κ(f)\n≤σmin(M)d(x, y)\n4\nby taking K small enough and Lemma 6.14.\nThis implies that\n3\n4σmin(M)d(x, y) ≤\n3\n2σ√n∥f(x)∥∞, i.e., that d(x, y) ≤2σβ(f, x).\nThis shows part (i). Part (ii) is shown in a similar way.\nLemma 6.17. Let u ≤Kη2\nlog n with K small enough and x1, x2 ∈Gη.\n(i) If fl(d(x1, x2)) ≤fl( 3\n2πη√n + 1) then d(x1, x2) ≤2πη√n + 1.\n(ii) If fl(d(x1, x2)) > fl( 3\n2πη√n + 1) then d(x1, x2) > πη√n + 1.\nProof.\nBy Lemma 6.15 and the hypothesis on u, we obtain\nError(d(x1, x2)) =\n √n log n\nη"},{"paragraph_id":"p89","order":89,"text":"≤O\n √n log n\nη\n Kη2\nlog n ≤π\n2 η\n√\nn + 1,\nthe last by taking K small enough.\nAlso, Error( 3\n2πη√n + 1) ≤\n3\n2πη√n + 1 γ3.\nThe\nstatement easily follows from these two bounds.\nLemma 6.18. Let u ≤\nKη\n√\nnD\nD+log S+η\n√\nnD with K small enough, f ∈S(Hd) and x ∈Sn.\n(i) If fl(∥f(x)∥∞) ≤fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞≤πη\np\n(n + 1)D.\n(ii) If fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞> π\n2 η\np\n(n + 1)D.\n25"},{"paragraph_id":"p90","order":90,"text":"Proof.\nFor part (i), from Proposition 6.10,\n∥f(x)∥∞≤fl(∥f(x)∥∞) + [[D + log S]].\nAlso,\n√\n2\n2 πη\np\n(n + 1)D ≥fl(\n√\n2\n2 πη\np\n(n + 1)D) −[[η\np\n(n + 1)D]].\nTherefore,\n∥f(x)∥∞−\n√\n2\n2 πη\np\n(n + 1)D\n≤\nfl(∥f(x)∥∞) −fl(\n√\n2\n2 πη\np\n(n + 1)D) + [[D + log S + η\np\n(n + 1)D]]\n≤\n[[D + log S + η\np\n(n + 1)D]]\n=\nO\n D + log S + η\np\n(n + 1)D"},{"paragraph_id":"p91","order":91,"text":"Kη\n√\nnD\nD + log S + η\n√\nnD\n≤\n(1 −\n√\n2\n2 )η\np\n(n + 1)D,\nthe last by taking K sufficiently small. It follows that ∥f(x)∥∞≤πη\np\n(n + 1)D and hence,\npart (i) of the statement.\nPart (ii) is proved similarly.\n6.5\nProof of Theorem 1.1(4): Correctness\nWe will show that, if u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), and the algorithm halts with\nη ≥\nα•\n8D2(n+1)κ(f)2 , then the value r/2 returned by the algorithm is #R(f).\nThis is a\nconsequence of the floating following versions of Lemmas 5.1 and 5.2.\nLemma 6.19. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each x ∈fl(A′(f)) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each\npoint z ∈fl(B\n′\nf(x)), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈fl(A′(f)). Then ζx = ζy ⇐⇒fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)) ̸= ∅.\nProof.\n(i) Applying Proposition 6.13(ii), x ∈fl(A′(f)) implies that α(f, x) < α•.\nTherefore, by Theorem 6.1, there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover, if\nz ∈fl(B\n′\nf(x)), by Lemma 6.16(i), d(x, z) ≤2σβ(f, x) and the Newton sequence starting at\nz converges to ζx.\n(ii) If ζx = ζy, then Bf(x) ∩Bf(y) ̸= ∅which implies by Lemma 6.16(ii) that there exists\nz ∈fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)).\nThis immediately implies, using that Bf(x) ⊂fl(B\n′\nf(x)) by Lemma 6.16(ii), the follow-\ning corresponding floating version of Lemma 5.2.\nLemma 6.20. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each component fl(U) of fl(G′\nη), there is a unique zero ζU ∈Z(f) such that\nζU ∈Z(fl(U)). Moreover ζU ∈∩x∈fl(U)Bf(x).\n26"},{"paragraph_id":"p92","order":92,"text":"(ii) If fl(U) and fl(V ) are different components of fl(G′\nη), then ζU ̸= ζV .\nIn order to show the correctness of Count Roots 2, we only need to prove that Z(f) ⊂\nZ(fl(G′\nη)). This easily follows adapting the proof of Part (1) in Section 5.3 to this situation,\nmaking use of Lemma 6.20 and the facts that Condition (i), fl(d(xi, xj)) > fl( 3\n2πη√n + 1),\nimplies that d(xi, xj) > πη√n + 1 (Lemma 6.17(ii)) and Condition (ii), fl(∥f(x)∥∞) >\nfl(\n√\n2\n2 πη\np\n(n + 1)D), implies that ∥f(x)∥∞> π\n2 η\np\n(n + 1)D (Lemma 6.18(ii)).\n6.6\nProof of Theorem 1.1(4): Complexity\nWe want to show that if η ≤\nα•\n4D2(n+1)κ(f)2 then Count Roots 2(f) halts. Note that this\nmeans that\nα•\n8D2(n + 1)κ(f)2 < η ≤\nα•\n4D2(n + 1)κ(f)2\nand hence, by § 6.5, that it correctly returns #R(f).\nBecause of the hypothesis on η, the hypotheses of Lemmas 6.2, and 6.3 are satisfied.\nLet fl(U) ̸= fl(V ) be different components of fl(G′\nη), and therefore, by Lemma 6.20,\nζU ̸= ζV , and for all x ∈fl(U), y ∈fl(V ), by Lemma 6.2, d(x, y) > 2πη√n + 1 holds. This\nimplies, by Lemma 6.17(i), that Condition (i) in Count Roots 2 is satisfied.\nConsider now x ̸∈fl(A′(f)). By Proposition 6.13(i), α(f, x) ≥α•\n3 . This implies, by\nLemma 6.3, that ∥f(x)∥∞> πη\np\n(n + 1)D, which in turn, by Lemma 6.18(i), ensures that\nCondition (ii) in Count Roots 2 is satisfied. Hence, the algorithm halts.\nAknowledgement. We are grateful to Andr ́e Galligo for a helpful discussion.\nReferences\n[1] B. Bank, M. Giusti, J. Heintz, and L. Pardo.\nGeneralized polar varieties: geometry and\nalgorithms. J. Compl., 21:377–412, 2005.\n[2] L. Blum, F. Cucker, M. Shub, and S. Smale. Complexity and Real Computation. Springer-\nVerlag, 1998.\n[3] P. B ̈urgisser and F. Cucker. Counting complexity classes for numeric computations II: Algebraic\nand semialgebraic sets. J. Compl., 22:147–191, 2006.\n[4] D. Cheung and F. Cucker. Solving linear programs with finite precision: II. Algorithms. J.\nCompl., 22:305–335, 2006.\n[5] G.E. Collins. Quantifier elimination for real closed fields by cylindrical algebraic deccomposi-\ntion, volume 33 of Lect. Notes in Comp. Sci., pages 134–183. 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On generalized Newton algorithms: Quadratic convergence, path-following and\nerror analysis. Theoret. Comp. Sci., 133:65–84, 1994.\n[20] K. Meer. Counting problems over the reals. Theoret. Comp. Sci., 242:41–58, 2000.\n[21] M. Shub and S. Smale. Complexity of B ́ezout’s theorem I: geometric aspects. Journal of the\nAmer. Math. Soc., 6:459–501, 1993.\n[22] M. Shub and S. Smale. Complexity of B ́ezout’s theorem III: condition number and packing.\nJournal of Complexity, 9:4–14, 1993.\n[23] M. Shub and S. Smale. Complexity of B ́ezout’s theorem IV: probability of success; extensions.\nSIAM J. of Numer. Anal., 33:128–148, 1996.\n[24] S. Smale. Newton’s method estimates from data at one point. In R. Ewing, K. Gross, and\nC. Martin, editors, The Merging of Disciplines: New Directions in Pure, Applied, and Com-\nputational Mathematics. Springer-Verlag, 1986.\n[25] A. Tarski. A Decision Method for Elementary Algebra and Geometry. University of California\nPress, 1951.\n[26] H. Weyl. The Theory of Groups and Quantum Mechanics. Dover, 1932.\n[27] H.R. W ̈uthrich. Ein Entscheidungsverfahren f ̈ur die Theorie der reell-abgeschlossenen K ̈orper.\nvolume 43 of Lect. Notes in Comp. Sci., pages 138–162. Springer-Verlag, 1976.\n28"}],"pages":[{"page":1,"text":"arXiv:0710.4508v2 [cs.CC] 19 Mar 2008\nA Numerical Algorithm for Zero Counting.\nI: Complexity and Accuracy\nFelipe Cucker ∗\nDept. of Mathematics\nCity University of Hong Kong\nHONG KONG\ne-mail: macucker@cityu.edu.hk\nTeresa Krick †\nDepartamento de Matem ́atica\nUniv. de Buenos Aires & CONICET\nARGENTINA\ne-mail: krick@dm.uba.ar\nGregorio Malajovich‡\nDepto. de Matem ́atica Aplicada\nUniv. Federal do Rio de Janeiro\nBRASIL\ne-mail: gregorio@ufrj.br\nMario Wschebor\nCentro de Matem ́atica\nUniversidad de la Rep ́ublica\nURUGUAY\ne-mail: wschebor@cmat.edu.uy\nAbstract. We describe an algorithm to count the number of distinct real zeros of\na polynomial (square) system f. The algorithm performs O(log(nDκ(f))) iterations\n(grid refinements) where n is the number of polynomials (as well as the dimension of\nthe ambient space), D is a bound on the polynomials’ degree, and κ(f) is a condition\nnumber for the system. Each iteration uses an exponential number of operations. The\nalgorithm uses finite-precision arithmetic and a major feature in our results is a bound\nfor the precision required to ensure the returned output is correct which is polynomial\nin n and D and logarithmic in κ(f). The algorithm parallelizes well in the sense that\neach iteration can be computed in parallel time polynomial in n, log D and log(κ(f)).\n1\nIntroduction\nIn recent years considerable attention was put on the complexity of counting problems over\nthe reals. The counting complexity class #PR was introduced [20] and completeness results\nfor #PR were established [3] for natural geometric problems notably, for the computation\nof the Euler characteristic of semialgebraic sets. As one could expect, the “basic” #PR-\ncomplete problem consists of counting the real zeros of a system of polynomial equations.\nAlgorithms for counting real zeros have existed since long. One such algorithm follows\nfrom the work of Tarski [25] on quantifier elimination for the theory of the reals. Its complex-\nity is hyperexponential. Algorithms with improved complexity (doubly exponential) were\ndevised in the 70s by Collins [5] and W ̈utrich [27]. A breakthrough was reached a decade\n∗Partially supported by City University SRG grant 7002106.\n†Partially supported by grants UBACyT X112/06-09, CONICET PIP 2461/00 and ANPCyT 33671/05.\n‡Partially supported by CNPq grants 304504/2004-1, 472486/2004-7, 470031/2007-7, 303565/2007-1, and\nby FAPERJ grant E26/170.734/2004.\n1"},{"page":2,"text":"later with the introduction of the critical points method by Grigoriev and Vorobjov [13, 12]\nwhich uses exponential time. Algorithms counting connected components (and hence, in\nthe zero-dimensional case, solutions) based on this method can be found in [14, 16], and in\nthe straight-line program model of computation in [1]. These algorithms parallelise well in\nthe sense that one can devise versions of them working in parallel polynomial time when\nan exponential number of processors is available. The #PR-completeness of the problem\nstrongly indicates that this is the best we can hope for.\nAll the algorithms mentioned above are “symbolic algorithms.” They have been devised\nupon the premise that no perturbation or round-offerror is present. Were this not the\ncase, it is not difficult to see that errors would accumulate quite badly. Roughly speaking,\nthese algorithms construct some object of exponential size on which some basic computation\n(e.g., linear algebra) is eventually performed. A question is posed, can one devise “numerical\nalgorithms” (maybe iterative, which need not terminate for ill-posed inputs) with a better\nbehavior viz the accumulation of round-offerrors? For the problem of deciding the existence\nof (or computing) a zero of a polynomial system such algorithms were given in [8, 6, 18].\nThe goal of this article is to describe and analyze a numerical algorithm for zero counting.\nWe will do so by developping appropriate versions of the tools used in [8, 6].\nLet d1, . . . , dn ∈N and d = (d1, . . . , dn). We will denote by Hd the space of polynomial\nsystems f = (f1, . . . , fn) with fi ∈R[X0, . . . , Xn] homogeneous of degree di.\nZero rays of polynomial systems f ∈Hd are associated to pairs of zeros (−ζ, ζ) of the\nrestriction f|Sn of f to the n-dimensional unit sphere Sn ⊂Rn+1. Thus, it will be convenient\nto consider a system f ∈Hd as a (central symmetric, analytic) mapping of Sn into Rn. If\nwe denote by Z(f) = {ζ ∈Sn : f(ζ) = 0} the zero-set of f in Sn then the number #R(f) of\nzero rays of the system f is half the cardinality of Z(f).\nIn this paper we describe a finite-precision algorithm computing #R(f), given f ∈Hd.\nTo analyze its complexity and accuracy, besides the number n of polynomials, we will rely on\ntwo more additional parameters. One is D = maxi≤n di. The other is a condition measure\nκ(f) for the system f. We will describe this measure in detail in Section 2 below. We\nwill also let S = max Si where Si is the number of non-zero coefficients of fi. Note that\nS is bounded by a simple expression in terms of n and D, namely, S =\n n+D\nD\n \n. Yet, we\nwill express dependancy on S since this may be relevant for the case of sparse systems of\npolynomials. Our main result is the following.\nTheorem 1.1. There exists an iterative algorithm which, with input f ∈Hd,\n(1) Returns #R(f).\n(2) Performs O(log(nDκ(f))) iterations and has a total cost (number of arithmetic opera-\ntions) of\nO\n \nlog(nDκ(f))(n + 1)2\n 2(n + 1)D2κ(f)2\nα∗\n 2n!\n,\nwhere α∗≈0.0384629388 . . . is a universal constant.\n(3) Can be well-parallelized in the sense that it admits a parallel version running in time\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2))\nwith a number of processors exponential in this quantity.\n2"},{"page":3,"text":"(4) Can be implemented with finite precision (both versions, sequential and parallel). The\nrunning time remains the same (with α∗replaced by α• ≈0.028268 · · ·) and the re-\nturned value is #R(f) as long as the machine precision (i.e., the round-offunit) u\nsatisfies\nu ≤\n1\nO\n D2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n .\n(5) It can be modified to return, in addition and for each real zero ζ ∈Sn of f, an approx-\nimate zero x of f in the sense that Newton’s iteration, starting at x, converges to ζ\nquadratically fast.\nRemark 1.2. (i)\nA system f for which arbitrarily small perturbations may change\nthe value #R(f) is considered ill-posed in our context since for arbitrarily small machine\nprecisions finite precision algorithms may return an incorrect value.\nConsequently, the\ncondition number κ(f) is infinite in these cases (and only then). This happens when f has\nmultiple real zeros and, in particular, when f has infinitely many real zeros. In these cases\nthe algorithm of Theorem 1.1 may not halt.\n(ii)\nNumerical algorithms compute functions φ on real data. Error analysis for algorithms\ncomputing (vectors of) real numbers —i.e. for which the image of φ has non-empty interior—\nare usually expressed in terms of bounds for the relative error of the computed quantities.\nThat is, for data d, bounds in\n∥φ(d) −fl(φ(d)∥\n∥φ(d)∥\nwhere fl(φ(d)) is the vector actually computed with finite precision. This relative error\nvaries continuously with d and depends on the condition of d and on the precision u. Such a\nform of analysis, however, becomes meaningless when computing quantities taking a finite\nnumber of values. Indeed, if Ra denotes the set of input data d for which φ(d) = a the\nfollowing happens. When d is in the interior of Ra we have that the relative error above\nis 0 for sufficiently small u. In contrast, when d is on the boundary of Ra, that error may\nremain constant for all u > 0. Because of this, error analysis for this kind of discrete-valued\nproblems has a different form, as in Theorem 1.1. One bounds how small u needs to be to\nguarantee a correct answer. Such a bound, needless to say, also depends on the condition\nof the data d. Examples of this type of analysis can be found in [4, 6, 7, 8]. In each of these\nreferences a condition number for the problem at hand occurs in the error analysis. We note\nthat the one in [6] is essentially our κ(f).\nThe rest of the paper is organized as follows. In Section 2 we describe the basic objects we\nwill deal with as well as fixing the notation. In Sections 3 and 4 we prove the two technical\nresults our algorithm relies on. In Section5 we describe the algorithm under the assumption\nof infinite precision and we prove parts (1), (2), and (3) of Theorem 1.1. The geometric\nideas making the algorithm work are best seen in this context. Section 6 then describes the\nnecessary modifications to make the algorithm work as well under finite precision. These\nmodifications are simple and can be summarized by saying that we relax a bit the inequalities\ntested in the algorithms to make room for the finite-precision errors to fit in.\n2\nPreliminaries\nDenote by Hd the subspace of R[X0, . . . , Xn] of homogeneous polynomials of degree d. Then,\nHd = Hd1 × · · · × Hdn.\n3"},{"page":4,"text":"If g ∈Hd we write\ng(X) =\nX\nJ\ngJXJ\nwhere J = (J0, . . . , Jn) is assumed to range over all multi-indices such that |J| = Pn\nk=0 Jk =\nd, XJ = XJ0\n0 XJ1\n1 . . . XJn\nn\nand gJ ∈R. Multinomial coefficients are defined by:\n d\nJ\n \n=\nd!\nJ0!J1! · · · Jn!.\nThe space Hd is endowed with the inner product\n⟨g, h⟩=\nX\n|J|=d\ngJhJ\n d\nJ\n \nwhich gives rise to the norm ∥g∥=\np\n⟨g, g⟩. These norms, for d1, . . . , dn, induce a norm in\nHd by taking for f = (f1, . . . , fn) ∈Hd:\n∥f∥= ∥(f1, . . . , fn)∥= max\n1≤i≤n ∥fi∥.\nLet O(n + 1) be the orthogonal group. The inner product above is known to be O(n + 1)-\ninvariant: for all Q ∈O(n + 1) and all g, h ∈Hd,\n⟨g ◦Q, h ◦Q⟩= ⟨g, h⟩.\n(This is a direct consequence of [26, III-7] or [2, Theorem 1 p. 218], by considering O(n+ 1)\nas subgroup of U(n + 1)).\nThe associated norm ∥f∥on Hd is therefore also O(n + 1)-\ninvariant. We will use this norm on Hd all along this paper. For x = (x1, . . . , xn) ∈Rn we\nrecall that ∥x∥2 = (x2\n1 + · · · + x2\nn)1/2 and ∥x∥∞= max{|x1|, . . . , |xn|}. We will often denote\n∥x∥2 simply by ∥x∥.\nFor f ∈Hd and x ∈Sn define\nμnorm(f, x) = ∥f∥√n\n\nDf(x)−1\n|TxSn\n \n \n√d1\n√d2\n...\n√dn\n \n \n\n(1)\nwhere Df(x)|TxSn is the restriction to the tangent space of x at Sn of the derivative of f at\nx and the norm is the spectral norm, i.e. the operator norm with respect to ∥∥2. We now\ndefine the condition number κ(f) of f ∈Hd:\nκ(f) = max\nx∈Sn min\n \nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\n \n.\nRemark 2.1. The quantity κ(f) is closely related to other condition numbers for similar\nproblems.\nA version of the quantity μnorm(f, ζ) was introduced in [21, 22, 23] (see also [2, Chap-\nter 12]) for a complex polynomial system f and a zero ζ of f in the complex unit sphere\nSn\nC ⊂Cn+1. The normalized condition number of such a system f was then defined to be\nμnorm(f) :=\nmax\nζ∈Sn\nC |f(ζ)=0 μnorm(f, ζ).\n(2)\n4"},{"page":5,"text":"Actually, the version of μnorm(f, ζ) introduced in [21, 22, 23] differs from (1) in the fact\nthat ∥f∥is defined as (P ∥fi∥2)1/2 (and there is no √n factor). It is bounded above by the\nexpression in (1).\nOver the reals, the right-hand side in (2) may not be well-defined since the zero set\nof f may be empty.\nIn [8] real systems were considered (as in the present paper) and\nan algorithm deciding feasibility of f (i.e., whether f has a real zero) was proposed. Its\ncomplexity was analyzed in terms of a condition number which, using our notation and\nmodulo minor details, is defined as follows\n \n \n \n \n \nmin\nζ∈Sn|f(ζ)=0 μnorm(f, ζ)\nif f is feasible\nmax\nζ∈Sn\n∥f∥\n∥f(ζ)∥∞\nif f is infeasible.\nNote the use of min (instead of max) in the first line above. This is due to the fact that the\ntime needed for the algorithm in [8] to detect the existence of a zero depends on the best\nconditioned zero of f. The existence of other, poorly conditioned (or even singular), zeros\nof f is irrelevant.\nShortly after, the algorithm in [8] was extended to an algorithm which would, in addition\nand if f is feasible, return a zero of f [6]. The complexity of this extension was studied\nin terms of a condition number (denoted ̺(f) in [6]) which, essentially, coincides with our\nκ(f).\nProposition 2.2. For all f ∈Hd, κ(f) ≥1.\nProof.\nLet x ∈Sn. Because of orthogonal invariance, we may assume without loss of\ngenerality that x = e0 := (1, 0, . . . , 0).\nIt is then immediate that ∥f(x)∥∞≤∥f∥. This shows that the second expression in the\ndefinition of κ is at least 1.\nFor the first expression, i.e., μnorm(f, x), define g = (g1, . . . , gn) ∈Hd by gi(X) = fi(X)−\nfi(e0)Xdi\n0 . Then g(e0) = 0 and [2, Corollary 3 p. 234], μnorm(g, e0) ≥1 (this is shown for\nthe version of μnorm with the 2-norm for ∥f∥, which is bounded above by the expression (1)).\nSince Df(e0) = Dg(e0) and ∥g∥≤∥f∥, we can conclude μnorm(f, e0) ≥μnorm(g, e0) ≥1.\n3\nThe exclusion Lemma\nIn this article, d( , ) denotes the Riemannian (angular) distance in Sn (which satisfies\n0 ≤d(x, y) ≤π, ∀x, y ∈Sn) and for x ∈Sn, r > 0, we set B(x, r) := {y ∈Sn : d(y, x) < r}\nand B(x, r) := {y ∈Sn : d(y, x) ≤r}.\nThe following result can be used to support an exclusion test.\nLemma 3.1. Let f ∈Hd and let x, y ∈Sn such that d(x, y) ≤\n√\n2. Then,\n∥f(x) −f(y)∥∞≤∥f∥\n√\nD d(x, y)\nIn particular, if f(x) ̸= 0, there is no zero of f in B(x, min{∥f(x)∥∞/(∥f∥\n√\nD),\n√\n2}).\n5"},{"page":6,"text":"Proof.\nAn immediate consequence of the definition of the O(n + 1)-invariant inner\nproduct is that Hd endowed with this inner product is a reproducing kernel Hilbert space [9,\nProp. 2.21]. This implies that, for all g ∈Hd and x ∈Rn+1,\ng(x) = ⟨g(X), (xT X)deg g⟩.\n(3)\nBecause of orthogonal invariance, we can assume that x = e0 and y = e0 cos θ + e1 sin θ,\nwhere θ = d(x, y). Equation (3) implies that\nfi(x) −fi(y)\n=\n⟨fi(X), (xT X)di⟩−⟨fi(X), (yT X)di⟩= ⟨fi(X), (xT X)di −(yT X)di⟩\n=\n⟨fi(X), Xdi\n0 −(X0 cos θ + X1 sin θ)di⟩.\nHence, Cauchy-Schwarz-Bunyakowsky implies:\n|fi(x) −fi(y)| ≤∥fi∥∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥.\nSince\nXdi\n0 −(X0 cos θ + X1 sin θ)di = Xdi\n0 (1 −(cos θ)di) +\ndi\nX\nk=1\n di\nk\n \n(cos θ)di−k(sin θ)kXdi−k\n0\nXk\n1 ,\nwe have:\n∥Xdi\n0 −(X0 cos θ + X1 sin θ)di∥2\n=\n(1 −(cos θ)di)2 +\ndi\nX\nk=1\n di\nk\n \n(cos θ)2(di−k)(sin θ)2k\n=\n(1 −(cos θ)di)2 + 1 −(cos θ)2di\n=\n2(1 −(cos θ)di)\n≤\n2(1 −(1 −θ2\n2 )di)\n(4)\n≤\n2(1 −(1 −di\nθ2\n2 ))\n(5)\n≤\ndiθ2,\nwhere the inequality in line (4) is obtained from Taylor expanding cos θ around 0, and the\ninequality in line (5) is due to the fact that (1 −a)d ≥1 −da for a ≤1.\nWe conclude that\n|fi(x) −fi(y)| ≤∥fi∥θ\np\ndi\nand hence\n∥f(x) −f(y)∥∞≤∥f∥θ\nq\nmax\ni\ndi.\nFor the second assertion, we have\n∥f(y)∥∞\n≥\n∥f(x)∥∞−∥f(x) −f(y)∥∞\n≥\n∥f(x)∥∞−∥f∥\n√\nD d(x, y)\nsince d(x, y) ≤\n√\n2\n>\n∥f(x)∥∞−∥f∥\n√\nD ∥f(x)∥∞/(∥f∥\n√\nD)\n=\n0.\n6"},{"page":7,"text":"4\nThe proximity Theorem\n4.1\nNewton and Smale\nNewton iteration on the sphere Sn is defined by\nNf :\nSn\n→\nSn\nx\n7→\nNf(x) = expx\n \n−Df(x)−1\n|TxSnf(x)\n \nwhere expx is the exponential map at x,\nexpx h = cos(∥h∥)x + sin(∥h∥)\n∥h∥\nh.\nFurthermore, the standard invariants of α-theory, introduced by Smale in [24], can be\ndefined as:\nβ(f, x)\n=\n\nDf(x)−1\n|TxSnf(x)\n\n ,\nγ(f, x)\n=\nsup\nk≥2\n\nDf(x)−1\n|TxSnDkf(x)|(TxSn)k\nk!\n\n1/(k−1)\n,\nα(f, x)\n=\nβ(f, x)γ(f, x).\nRemark 4.1.\n(i) It is easy to see that β(f, x) = d(x, Nf(x)).\n(ii) We will not use Newton’s method in our algorithm. We are instead interested in its\nalpha theory which guarantees existence of zeros near points x with α(f, x) small\nenough.\n(iii) The Newton iteration presented above is not the iteration known as ‘projective New-\nton’. There is an alpha theory for that method, available in [19].\nHere we use slight modifications of the quantities α, β and γ, more adapted to our\npurposes. We set\nβ(f, x)\n:=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\nγ(f, x)\n:=\nD3/2\n2\nμnorm(f, x)\nα(f, x)\n:=\nβ(f, x)γ(f, x).\nThe definition of γ is motivated by the estimate of γ [2, Theorem 2 p. 267].\nγ(f, x) ≤γ(f, x).\nwhich yields the lower bound\nκ(f) ≥\nmax\nζ|f(ξ)=0 2D−3/2γ(f, ζ).\n(6)\n7"},{"page":8,"text":"We also observe that γ(f, x) ≥D3/2\n2\nsince μnorm(f, x) ≥1 and that β(f, x) ≤β(f, x)\nsince\nβ(f, x) =\n\nDf(x)−1\n|TxSnf(x)\n\n ≤√n∥f(x)∥∞\n\nDf(x)−1\n|TxSn\n\n ≤μnorm(f, x)∥f(x)∥∞\n∥f∥\n= β(f, x).\nTherefore α(f, x) ≤α(f, x).\n4.2\nProximity and unicity from data at a point\nDefinition 4.2. We say that x ∈Sn is an approximate zero for f if and only if the Newton\nsequence {xk}k∈N, where x0 := x and xk+1 := Nf(xk), is defined for all k and moreover\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nd(x0, x1).\nThe limit point ζ = limk→∞xk is a fixed point for Newton iteration and a zero of f. It is\ncalled the associated zero to x.\nIn what follows we denote σ := P\nk≥0 2−2k+1 = 1.632843018 . . . and we set\nBf(x) := {y ∈Sn | d(x, y) ≤σβ(f, x)}.\nThe main technical tool in our algorithm is provided by the following result.\nTheorem 4.3. There exists an universal constant α∗:= 0.0384629388 . . . such that for all\nx ∈Sn, if α(f, x) < α∗, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z in Bf(x) the Newton sequence starting at z converges\nto ζ.\n4.3\nBackground material\nTheorem 4.3 is a consequence of the following two results, which are restatements of results\nproved in [10]. While [10] deals with Newton iteration on arbitrary complete real analytic\nRiemannian manifolds, here we reword them in terms of Newton iteration on the unit\nsphere Sn (Example 1 in [10]). The γ-Theorem for mappings [10, Theorem 1.3] becomes\nthe following.\nTheorem 4.4. Let f : Sn →Rn be analytic. Suppose that f(ζ) = 0 and Df(ζ) is an\nisomorphism. Let\nR(f, ζ) := min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nIf d(x, ζ) ≤R(f, ζ), then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and\nd(xk, ζ) ≤\n 1\n2\n 2k−1 d(x, ζ). In particular, {xk} converges to ζ.\n8"},{"page":9,"text":"Now let α0 := 0.130716944 . . . denote the smallest positive root of the polynomial ψ(u)2−\n2u, and\ns0 :=\n1\nσ + (1−σα0)2\nψ(σα0)\n \n1 +\nσ\n1−σα0\n = 0.103621842 . . .\nWe state the α-Theorem for mappings [10, Theorem 1.4] for the sphere Sn.\nTheorem 4.5. Let f : Sn →Rn be analytic. Let x ∈Sn be such that β(f, x) ≤s0π and\nα(f, x) ≤α0. Then the Newton sequence xk = N k\nf (x) is defined for all k ≥0 and converges\nto a zero ζ of f. Moreover,\nd(xk, xk+1) ≤\n 1\n2\n 2k−1\nβ(f, x)\nand\nd(xk, ζ) ≤σβ(f, x).\nFinally we introduce ψ(u) := 1−4u+2u2, which is positive and decreasing for 0 < u < 1−\n√\n2\n2 ,\nand state [10, Lemma 4.3]:\nLemma 4.6. Let x, y ∈Sn with d(x, y) < π. Suppose that Df(x) is nonsingular and\nν := d(x, y)γ(f, x) < 1 −\n√\n2\n2 .\nThen\nγ(f, y) ≤\nγ(f, x)\n(1 −ν)ψ(ν).\n4.4\nProof of Theorem 4.3\nSet ν∗:= 0.0628039411 . . . for the only real root of the polynomial\nΨ(u) := (3 −\n√\n7)(1 −u)ψ(u) −4u,\nand α∗:= ν∗\nσ = 0.0384629388 . . .. Note that α∗≤min{α0, s0π}.\nSince γ(f, x) ≥D3/2\n2\n, the hypothesis of Theorem 4.5 hold from α(f, x) ≤α(f, x) < α∗≤\nα0 and β(f, x) ≤β(f, x) ≤2α(f,x)\nD3/2\n<\n2α∗\nD3/2 < s0π.\nUsing Remark 4.1(i) it follows that x is an approximate zero of f, and that the associated\nzero ζ satisfies:\nd(x, ζ) ≤σβ(f, x) ≤σβ(f, x).\nThis already proves Parts (i) and (ii) of Theorem 4.3.\nWe show (iii). Since d(x, ζ) ≤σβ(f, x) < σs0π < π,\nν = d(x, ζ)γ(f, x) ≤σβ(f, x)γ(f, x) ≤σα(f, x) ≤σα∗= ν∗< 1 −\n√\n2\n2 ,\nand we can apply Lemma 4.6. Therefore\n4σβ(f, x)γ(f, ζ) ≤4σβ(f, x)γ(f, x)\n1\n(1 −ν)ψ(ν) ≤4ν∗\n1\n(1 −ν∗)ψ(ν∗) = 3 −\n√\n7,\n9"},{"page":10,"text":"because (1−u)ψ(u) decreases for 0 < u < 1−\n√\n2\n2 , and ν∗is a zero of (3−\n√\n7)(1−u)ψ(u)−4u.\nThis shows, since 2σβ(f, x) ≤π, that\n2σβ(f, x) ≤R(f, ζ) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζ)\n)\n.\nWe conclude applying Theorem 4.4 to z ∈Bf(x), since\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤2σβ(f, x) ≤R(f, ζ).\nIt follows that the Newton sequence {zk}k∈N starting at z converges to ζ.\nRemark 4.7. The hypothesis on the radius of injectivity in [10] was recently found to be\nredundant.\n5\nInfinite precision\n5.1\nGrids and Graphs\nOur algorithm works on a grid on Sn. We easily construct one by projecting onto Sn a\ngrid on the cube Cn = {y | ∥y∥∞= 1}. We make use of the (easy to compute) bijections\nφ : Cn →Sn and φ−1 : Sn →Cn given by φ(y) =\ny\n∥y∥and φ−1(x) =\nx\n∥x∥∞.\nGiven η := 2−k for some k ≥1, we consider the uniform grid Uη of mesh η on Cn. This is\nthe set of points in Cn whose coordinates are of the form i2−k for i ∈{−2k, −2k+1, . . ., 2k},\nwith at least one coordinate equal to 1 or −1. We denote by Gη its image by φ in Sn. Note\nthat, for y1, y2 ∈Cn,\nd(φ(y1), φ(y2)) ≤π\n2 ∥y1 −y2∥2 ≤π\n2\n√\nn + 1 ∥y1 −y2∥∞.\n(7)\nGiven η as above we associate to it a graph Gη as follows. We set A(f) := {x ∈Sn |\nα(f, x) < α∗}. The vertices of the graph are the points in Gη ∩A(f). Two vertices x, y ∈Gη\nare joined by an edge if and only if Bf(x) ∩Bf(y) ̸= ∅.\nNote that as a simple consequence of Theorem 4.3 we obtain the following lemma.\nLemma 5.1.\n(i) For each x ∈A(f) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each point\nz in Bf(x), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈A(f). Then ζx = ζy ⇐⇒Bf(x) ∩Bf(y) ̸= ∅.\nWe define Z(Gη) := S\nx∈Gη Bf(x) ⊂Sn where x ∈Gη has to be understood as x running\nover all the vertices of Gη. Similarly, for a connected component U of Gη, we define\nZ(U) :=\n[\nx∈U\nBf(x).\nLemma 5.2.\n(i) For each component U of Gη, there is a unique zero ζU ∈Z(f) such that ζU ∈Z(U).\nMoreover, ζU ∈∩x∈UBf(x).\n10"},{"page":11,"text":"(ii) If U and V are different components of Gη, then ζU ̸= ζV .\nProof.\n(i) Let x ∈U. Since x ∈A(f), by Lemma 5.1 (i) there exists a zero ζx of f\nin Bf(x) ⊆Z(U). This shows the existence. For the second assertion and the uniqueness,\nassume that there exist ζ and ξ zeros of f in Z(U). Let x, y ∈U be such that ζ ∈Bf(x),\nand ξ ∈Bf(y). Since U is connected, there exist x0 = x, x1, . . . , xk−1, xk := y in A(f) such\nthat (xi, xi+1) is an edge of Gη for i = 0, . . . , k −1, that is, Bf(xi)∩Bf(xi+1) ̸= ∅. If ζi and\nζi+1 are the associated zeros of xi and xi+1 in Z(f) respectively, then by Lemma 5.1(ii) we\nhave ζi = ζi+1, and thus ζ = ξ ∈Bf(x) ∩Bf(y).\n(ii) Assume ζU = ζV ∈Bf(x) ∩Bf(y) ⊂Z(U) ∩Z(V ), then x and y are joined by an edge\nand belong to the same connected component.\n5.2\nThe infinite precision algorithm\nCount Roots 1(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet U1, . . . , Ur be the connected components of Gη\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈Ui and all xj ∈Uj, d(xi, xj) > πη√n + 1\nand\n(ii) for all x ∈Gη \\ A(f), ∥f(x)∥∞> π\n2 η\np\n(n + 1)D∥f∥\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\n5.3\nProof of Theorem 1.1(1–3)\nProof of Part (1)\nThis proof requires some arguments of convexity.\nWe can natu-\nrally define spherical convex hulls for sets of points in Hn, an open half-sphere in Sn. If\nx1, . . . , xq ∈Hn we define\nSCH(x1, . . . , xq) := Cone(x1, . . . , xq) ∩Sn\nwhere Cone(x1, . . . , xq) is the smallest convex cone with vertex at the origin and containing\nthe points x1, . . . , xq. Alternatively, we have,\nSCH(x1, . . . , xq) =\n λ1x1 + · · · + λqxq\n∥λ1x1 + · · · + λqxq∥| λ1, . . . , λq ≥0,\nX\nλi = 1\n \n.\nWe will use the following fact.\nLemma 5.3. Let x1, ..., xq ∈Hn ⊂Rn+1. If Tq\ni=1 B(xi, ri) ̸= ∅, then SCH(x1, . . . , xq) ⊂\nSq\ni=1 B(xi, ri).\nProof.\nLet x ∈SCH(x1, . . . , xq) and y ∈Tq\ni=1 B(xi, ri). We will prove that x ∈B(xi, ri)\nfor some i.\nIf x = y, this is obvious.\nIf x ̸= y, let H be the half-space\nH :=\n \nz ∈Rn+1 : ⟨z, y −x⟩< 0\n \n.\n11"},{"page":12,"text":"Since ∥x∥= ∥y∥= 1, we have ⟨x + y, y −x⟩= 0, and we note that in this case, x + y\ndetermines the mid-line between x and y. Moreover, since x ̸= y, we have x ∈H since\n⟨x, y −x⟩= ⟨x, y⟩−∥x∥2 < ∥x∥∥y∥−∥x∥2 = 0. Therefore the half-space H is the set of\npoints z in Rn+1 such that the Euclidean distance ∥z −x∥< ∥z −y∥.\nOn the other hand, H must contain at least one point of the set {x1, ..., xq} since if\nthis were not the case, the convex set Cone(CH(x1, . . . , xq)) would be contained in {z :\n⟨z, y −x⟩≥0}, contradicting x ∈SCH(x1, . . . , xq). Let, therefore, xi ∈H. It follows that\n∥x −xi∥< ∥y −xi∥\nwhich implies\nd(x, xi) < d(y, xi) ≤ri.\nWe can now proceed. Assume the algorithm halts, we want to show that if r equals\nthe number of connected components of Gη, then #R(f) = #Z(f)/2 = r/2. We already\nknow by Lemma 5.2 that each connected component U of Gη determines uniquely a zero\nζU ∈Z(f). Thus it is enough to prove that Z(f) ⊂Z(Gη).\nAssume that there is a zero ζ of f in Sn such that ζ is not in Z(Gη). Let B∞(φ−1(ζ), η) :=\n{y ∈Uη | ∥y −φ−1(ζ)∥∞≤η} = {y1, . . . , yq}, the set of all neighbors of φ−1(ζ) in Uη, and\nlet xi = φ(yi), i = 1, . . . , q. Clearly, φ−1(ζ) is in the cone spanned by {y1, . . . , yq} and hence\nζ ∈SCH(x1, . . . , xq).\nWe claim that there exists j ≤q such that xj ̸∈A(f). Indeed, assume this is not the\ncase. We consider two cases.\n(a)\nAll the xi belong to the same connected component U of Gη. By Lemma 5.2 there\nexists a unique zero ζU ∈Sn of f in Z(U) and ζU ∈∩iBf(xi). We may apply Lemma 5.3\nto deduce that\nSCH(x1, . . . , xq) ⊆\n[\nBf(xi).\nIt follows that, for some i ∈{1, . . . , q}, ζ ∈Bf(xi) ⊆Z(U), contradicting that ζ ̸∈Z(Gη).\n(b)\nThere exist l̸= s and 1 ≤i < j ≤r such that xl∈Ui and xs ∈Uj. Since condition\n(i) in the algorithm is satisfied, d(xl, xs) > πη√n + 1. But, by (7),\nd(xl, xs) ≤π\n2\n√\nn + 1∥yl−ys∥∞≤π\n2\n√\nn + 1\n ∥yl−φ−1(ζ)∥∞+ ∥φ−1(ζ) −ys∥∞\n \n≤πη\n√\nn + 1,\na contradiction.\nWe have thus proved the claim. Let then 1 ≤j ≤q be such that xj ̸∈A(f). Since\ncondition (ii) in the algorithm is satisfied ∥f(xj)∥∞> π\n2 η\np\n(n + 1)D∥f∥. It follows from\nthe inequality d(xj, ζ) ≤π\n2\n√n + 1η and Lemma 3.1 that ∥f(ζ)∥∞> 0, a contradiction.\nProof of Part (2)\nWe need a few lemmas.\nLemma 5.4. If ζ1 ̸= ζ2 ∈Z(f) then\nd(ζ1, ζ2) ≥2(3 −\n√\n7)D−3/2\nκ(f)\n.\n12"},{"page":13,"text":"Proof.\nFor i = 1, 2, using (6) and Proposition 2.2,\nR(f, ζi) = min\n(\nπ, 3 −\n√\n7\n2γ(f, ζi)\n)\n≥min\n(\nπ, (3 −\n√\n7)D−3/2\nκ(f)\n)\n= (3 −\n√\n7)D−3/2\nκ(f)\n.\nNow suppose that d(ζ1, ζ2) < R(f, ζ1) + R(f, ζ2) and choose x ∈Sn such that d(x, ζ1) <\nR(f, ζ1) and d(x, ζ2) < R(f, ζ2). Then Theorem 4.4 implies that ζ1 = ζ2, a contradiction.\nLemma 5.5. Let x1, x2 ∈Gη with associated zeros ζ1 ̸= ζ2. If η ≤\n2(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen\nd(x1, x2) > πη√n + 1.\nProof.\nAssume d(x1, x2) ≤πη√n + 1.\nSince x2 ̸∈Bf(x1), d(x1, x2) > σβ(f, x1).\nConsequently,\nd(x1, ζ1) ≤σβ(f, x1) < d(x1, x2) ≤πη\n√\nn + 1\nand, similarly, d(x2, ζ2) < πη√n + 1. But then,\nd(ζ1, ζ2) ≤d(ζ1, x1) + d(x1, x2) + d(x2, ζ2) < 3πη\n√\nn + 1 ≤2(3 −\n√\n7)D−3/2\nκ(f)\ncontradicting Lemma 5.4.\nLemma 5.6. Let x ∈Sn such that x ̸∈A(f).\nIf η ≤\nα∗\n(n+1)D2κ(f)2 then ∥f(x)∥∞>\nπ\n2 η\np\n(n + 1)D∥f∥.\nProof.\nSince x ̸∈A(f) we have α(f, x) ≥α∗. We divide the proof in two cases.\nCase I. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n=\n∥f∥\n∥f(x)∥∞\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗∥f(x)∥2\n∞\n(n + 1)D2∥f∥2\nwhich implies, since η ≤1\n2 <\n4D\nπ2α∗,\n∥f(x)∥∞≥\n√η√n + 1D∥f∥\n√α∗\n> π\n2 η\np\n(n + 1)D∥f∥.\nCase II. min\nn\nμnorm(f, x),\n∥f∥\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case\nη ≤\nα∗\n(n + 1)D2κ(f)2 ≤\nα∗\n(n + 1)D2μnorm(f, x)2\nwhich implies α∗≥η(n + 1)D2μnorm(f, x)2. Also,\nα∗≤α(f, x) = 1\n2β(f, x)μnorm(f, x)D3/2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞.\n13"},{"page":14,"text":"Putting both inequalities together we obtain\nη(n + 1)D2μnorm(f, x)2 ≤\n1\n2∥f∥μnorm(f, x)2D3/2∥f(x)∥∞\nor yet,\n∥f(x)∥∞≥2η(n + 1)D1/2∥f∥> π\n2 η\np\n(n + 1)D∥f∥.\nWe can now conclude the proof of Part (2).\nAssume η ≤\nα∗\n(n+1)D2κ(f)2 .\nThen the\nhypotheses of Lemmas 5.5 and 5.6 hold. The first of these lemmas ensures that condition (i)\nin the algorithm is satisfied. The second, that condition (ii) is so. Therefore, the algorithm\nhalts as soon as\nα∗\n2(n+1)D2κ(f)2 < η ≤\nα∗\n(n+1)D2κ(f)2 . This gives a bound of O(ln(nDκ(f)))\nfor the number of iterations. Since the number of grid points considered at this iteration\n(η =\nα∗\n(n+1)D2κ(f)2 ) is at most 2(n + 1)\n \n2(n+1)D2κ(f)2\nα∗\n n\n, the bound for the total complexity\nfollows.\nProof of Parts (3) and (5)\nWe have already seen that the number of iterations is\nbounded by O(ln(nDκ(f))). At each of these iterations, we need to perform a number of\ncomputations on the (at most) 2(n+1)\n \n2(n+1)D2κ(f)2\nα∗\n n\ngrid points to decide whether they\nare in A(f). These can be done independently. Then, we need to compute the number of\nconnected components of Gη. This can be done (see, e.g., [15]) in parallel time O(ln(|Vη|))2\nwhere |Vη| denotes the number of vertices of Gη and therefore, in parallel time at most\nO(n2(ln(nDκ(f))2 + ln(α∗)2)). Since this is the dominant step in the computation at a\ngiven iteration, it follows that the total parallel time consumed by the algorithm is at most\nO(n2 ln(nDκ(f))(ln(nDκ(f))2 + ln(α∗)2)). This shows part (3). For part (5), just note\nthat, for i = 1, . . . , r, any vertex xi of Ui is an approximate zero of the only zero of f in\nZ(Ui).\n6\nFinite Precision\n6.1\nMaking room to allow errors\nOur finite precision algorithm will be a variation of Algorithm Count Roots 1. But since\nfinite precision computations will be affected by errors, we need to make room in the infinite\nprecision algorithm to allow them.\nFor this aim, we state the corresponding version of\nTheorem 4.3.\nTheorem 6.1. There exist a universal constant α• = 0.028268 · · · such that, for all x ∈Sn,\nif α(f, x) < α•, then\n(i) x is an approximate zero of f.\n(ii) If ζ denotes its associated zero then ζ ∈Bf(x).\n(iii) Furthermore, for each point z s.t. d(x, z) ≤2σβ(f, x) the Newton sequence starting at\nz converges to ζ.\n14"},{"page":15,"text":"Proof.\nParts (i) and (ii) follow from Theorem 4.3 and the fact that α• < α∗. Part (iii)\nis proved by taking ν• = 0.046158 · · · to be the only real root of the polynomial Ψ(u) :=\n(3 −\n√\n7)(1 −u)ψ(u) −6u, and α• = ν•\nσ = 0.028268. Then, one proves as in Theorem 4.3\nthat 3σβ(f, x) ≤R(f, ζ) from which it follows that, for all z s.t. d(x, z) ≤2σβ(f, x),\nd(z, ζ) ≤d(z, x) + d(x, ζ) ≤3σβ(f, x) ≤R(f, ζ)\nand hence, that the Newton sequence {zk}k∈N starting at z converges to ζ.\nThe proofs of Lemmas 5.5 and 5.6 yield, mutatis mutandis, the following results.\nLemma 6.2. Let x1, x2 ∈Gη with associated zeros ξ1 and ξ2, ξ1 ̸= ξ2. If η ≤(3−\n√\n7)D−3/2\n3πκ(f)√n+1\nthen d(x1, x2) > 2πη√n + 1.\nLemma 6.3. Let x ∈Sn such that α(f, x) > α•\n3 . If η ≤\nα•\n4D2(n+1)κ(f)2 then ∥f(x)∥∞>\nπη\np\n(n + 1)D∥f∥.\n6.2\nBasic facts\nWe recall the basics of a floating-point arithmetic which idealizes the usual IEEE standard\narithmetic. This system is defined by a set F ⊂R containing 0 (the floating-point numbers),\na transformation r : R →F (the rounding map), and a constant u ∈R (the round-offunit)\nsatisfying 0 < u < 1. The properties we require for such a system are the following:\n(i) For any x ∈F, r(x) = x. In particular, r(0) = 0.\n(ii) For any x ∈R, r(x) = x(1 + δ) with |δ| ≤u.\nWe also define on F arithmetic operations following the classical scheme\nxe◦y = r(x ◦y)\nfor any x, y ∈F and ◦∈{+, −, ×, /}, so that\ne◦: F × F →F.\nThe following is an immediate consequence of property (ii) above.\nProposition 6.4. For any x, y ∈F we have\nxe◦y = (x ◦y)(1 + δ),\n|δ| ≤u.\nWhen combining many operations in floating-point arithmetic, quantities such as\nQn\ni=1(1 + δi)ρi naturally appear. Our round-offanalysis uses the notations and ideas in\nChapter 3 of [17], from where we quote the following results:\nProposition 6.5. If |δi| ≤u, ρi ∈{−1, 1}, and nu < 1, then\nn\nY\ni=1\n(1 + δi)ρi = 1 + θn,\nwhere\n|θn| ≤γn =\nnu\n1 −nu.\n15"},{"page":16,"text":"Proposition 6.6. For any positive integer k such that ku < 1, let θk, θj be any quantities\nsatisfying\n|θk| ≤γk =\nku\n1 −ku\n|θj| ≤γj =\nju\n1 −ju.\nThe following relations hold.\n1. (1 + θk)(1 + θj) = 1 + θk+j for some |θk+j| ≤γk+j.\n2.\n1 + θk\n1 + θj\n=\n 1 + θk+j\nif j ≤k,\n1 + θk+2j\nif j > k.\nfor some |θk+j| ≤γk+j or some |θk+2j| ≤γk+2j.\n3. If ku, ju ≤1/2, then γkγj ≤γmin{k,j}.\n4. iγk ≤γik.\n5. γk + u ≤γk+1.\n6. γk + γj + γkγj ≤γk+j.\nFrom now on, whenever we write an expression containing θk we mean that the same\nexpression is true for some θk, with |θk| ≤γk.\nWhen computing an arithmetic expression q with a round-offalgorithm, errors will\naccumulate and we will obtain another quantity which we will denote by fl(q). We write\nError(q) = |q −fl(q)|.\nAn example of round-offanalysis which will be useful in what follows is given in the next\nproposition, the proof of which can be found in Section 3.1 of [17].\nProposition 6.7. There is a round-offalgorithm which, with input x, y ∈Rn, computes\nthe dot product of x and y. The computed value fl(⟨x, y⟩) satisfies\nfl(⟨x, y⟩) = ⟨x, y⟩+ θ⌈log2 n⌉+1⟨|x|, |y|⟩,\nwhere |x| = (|x1|, . . . , |xn|).\nIn particular, if x = y, the algorithm computes fl(∥x∥2)\nsatisfying\nfl(∥x∥2) = ∥x∥2(1 + θ⌈log2 n⌉+1).\nWe will also have to deal with square roots and arccosinus. The following result will\nhelp us to do so.\nLemma 6.8. (i) Let θ ∈R such that |θ| ≤1/2. Then,\n√\n1 −θ = 1 −θ′ with |θ′| ≤|θ|.\n(ii) Let 0 < a ≤1 and ε ∈R such that 0 < a + ε < 1. Then, arccos(a + ε) = arccos(a) +\nυ\n1\n√\n1−(a+ε)2 with |υ| ≤|ε|.\nProof.\nAssume θ > 0 (if θ < 0 it is done similarly). By the intermediate value theorem\nwe have that 1 −\n√\n1 −θ = θ(√ξ)′ with ξ ∈(1 −θ, 1). But\n(\np\nξ)′ =\n1\n2√ξ ≤\n1\n√\n2,\n16"},{"page":17,"text":"the last since ξ ≥1/2. This proves (i).\nPart (ii) is shown similarly. Again, assume for simplicity that ε > 0. Then, for some\nξ ∈(a, a + ε),\narccos(a + ε) −arccos(a) = ε arccos′(ξ) = ε\n1\np\n1 −ξ2 =\nυ\np\n1 −(a + ε)2 .\nWe assume that, besides the four basic arithmetic operations, we are allowed to compute\nsquare roots and arccosinus with finite precision. That is, if op denotes any of these two\noperators, we compute f\nop such that\nf\nop(x) = op(x)(1 + δ),\n|δ| ≤u.\nFrom Lemma 6.8(i) it follows that, for all a > 0,\n^\np\na(1 + θk) = √a(1 + θk+1).\nRemark 6.9. Our choice of the precision u in Theorem 1.1(4) guarantees that ku < 1/2\nholds whenever we encounter θk in what follows, and consequently, |θk| ≤γk ≤2ku. This\nimplies that in all what follows we have γg = O(ug) for all the expressions g we will\nencounter.\nAccording to the previous remark we will introduce a further notation that will consid-\nerably simplify our exposition. For all expression g, we will write\n[[g]] := O(ug).\nThis notation will avoid we burden ourselves with the consideration of multiplicative con-\nstants.\n6.3\nThe finite precision algorithm\nOur finite precision algorithm is a variation of Algorithm Count Roots 1 in Section 5.3.\nGiven x ∈Sn we define below fl(A′(f)) and fl(B\n′\nf(x)), which are convenient floating\nversions of the sets A′(f) =\n \nx ∈Sn | α(f, x) < 1\n2α•\n \nand B\n′\nf(y) = {z ∈Sn | d(x, y) ≤\n3\n2σβ(f, x)} respectively.\nGiven f ∈Hd and x ∈Sn, we let M ∈Rn×n be a matrix representing\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)|TxSn.\nand we set σmin(M) = ∥M −1∥−1. Therefore\nμnorm(f, x)\n=\n∥f∥√n ∥M −1∥= ∥f∥√n σmin(M)−1,\nβ(f, x)\n=\nμnorm(f, x)∥f(x)∥∞\n∥f∥\n= √n σmin(M)−1∥f(x)∥∞,\nα(f, x)\n=\nβ(f, x)μnorm(f, x)D3/2\n2\n= ∥f∥n σmin(M)−2∥f(x)∥∞\nD3/2\n2\n.\n17"},{"page":18,"text":"This implies that\ny ∈B\n′\nf(x)\n⇐⇒\nd(x, y) ≤3\n2σβ(f, x)\n⇐⇒\nσmin(M)d(x, y) ≤3\n2σ√n∥f(x)∥∞,\nx ∈A′(f)\n⇐⇒\nα(f, x) < α•\n2\n⇐⇒\n∥f∥n∥f(x)∥∞D3/2 < α•σmin(M)2.\nThese statements are equivalent under infinite precision, but the expressions at the right-\nhand side are more convenient to handle when working with finite precision. This motivates\nour definitions of\nfl(B\n′\nf(x))\n:=\n \ny ∈Sn | fl(σmin(M)d(x, y)) ≤fl(3\n2σ√n∥f(x)∥∞)\n \nfl(A′(f))\n:=\nn\nx ∈Sn | fl(∥f∥n ∥f(x)∥∞D3/2) < fl(α•σmin(M)2)\no\nWe also define accordingly the graph fl(G′\nη) whose vertices are the points in Gη ∩\nfl(A′(f)), and with two vertices x, y joined by an edge if and only if fl(B\n′\nf(x))∩fl(B\n′\nf(x)) ̸=\n∅. Its connected components are denoted by fl(U).\nOur algorithm is the following:\nCount Roots 2(f)\nlet η :=\n2\n√\n2\nπ√n+1\n(1)\nlet fl(U1), . . . , fl(Ur) be the connected components of fl(Gη)\nif\n(i) for 1 ≤i < j ≤r\nfor all xi ∈fl(Ui) and all xj ∈fl(Uj), fl(d(xi, xj)) > fl( 3\n2πη√n + 1)\nand\n(ii) for all x ∈Gη \\ fl(A′(f)), fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D∥f∥)\nthen HALT and return r/2\nelse η := η/2\ngo to (1)\nIn the rest of the section we will see that, when the precision u satisfies u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), this algorithm is correct and halts as soon as η\n≤\nα•\n4D2(n+1)κ(f)2 .\n6.4\nBounding errors for elementary computations\nThe goal of this subsection is to exhibit bounds for the accumulated error in the main\ncomputations of Count Roots 2. We will rely on the basic notations and results described\nin §6.2.\nTo simplify notation, and without loss of generality, in all what follows we assume that\n∥f∥= 1. We denote by S(Hd) the sphere of such systems. Also, we do not discuss in what\nfollows the accumulated error in the computation of φ : Cn →Sn. This is a minor detail\nwhich can be taken care of using Lemma 6.8(i).\nProposition 6.10. Given f ∈S(Hd) and x ∈Sn, we can compute ∥f(x)∥∞with finite\nprecision u such that\nError(∥f(x)∥∞) = [[D + log S]]\nwhere S is a bound on the number of coefficients of each fi.\n18"},{"page":19,"text":"Proof.\nLet f = (f1, . . . , fn). For i ≤n write fi = P cJXJ and let S be the number\nof coefficients of fi. To compute f(x) one computes each monomial cJxJ with fl(cJxJ) =\ncJxJ(1 + θD). Then, one computes fi(x) to get\nfl(fi(x))\n=\nfl(\nX\nfl(cJxJ))\n=\nfl(\nX\ncJxJ(1 + θ(J)\nD ))\n=\nX\ncJxJ(1 + θ(J)\nD ) + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\n=\nfi(x) +\nX\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\nwhere in the third line we reasoned as in the proof of Proposition 6.7. Therefore\nError(∥f(x)∥∞)\n≤\n \nX\ncJxJθ(J)\nD + θlog S\nX\n|cJxJ|(1 + θ(J)\nD )\n \n≤\nX\n|cJ| ∥xJ∥(γD + γlog S + γDγlog S)\n≤\nγD+log S\nwhere we used that for any x ∈Sn, | P |cJ|xJ| ≤∥P |cJ|xJ∥= ∥fi∥≤∥f∥= 1 and\nProposition 6.6 (6). The conclusion follows from Remark 6.9.\nProposition 6.11. Given f ∈S(Hd) and x ∈Sn, let M ∈Rn×n be a matrix representing\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)|TxSn\nin some orthonormal basis of TxSn. Then ∥M∥≤√n. In addition, we can compute such a\nmatrix M with finite precision u such that\n∥Error(M)∥F = [[n(log S + D + log n)]].\nProof.\nStep 1: Let y =\nx−en+1\n∥x−en+1∥. The Householder symmetry\nHy = In+1 −2yyt\nswaps vectors en+1 and x, and fixes y⊥. The first n columns of Hy are therefore an orthonor-\nmal basis of TxSn, while the last column is x. Let H ∈R(n+1)×n denote the submatrix\nobtained from the first n columns of Hy. With that notation, we set\nM =\n \n \n1\n√d1\n1\n√d2\n...\n1\n√dn\n \n \nDf(x)H.\nStep 2: We claim that Pi,x : Hdi →Rn, fi 7→\n1\n√di Dfi(x)|TxSn is an orthogonal projection,\nin the sense that for any fixed x, the map (Pi,x)| ker(Pi,x)⊥is an isometry.\n19"},{"page":20,"text":"We use an orthogonal invariance argument. The special orthogonal group SO(n + 1)\nacts on Hdi and on Rn+1 isometrically as follows: to a given Q ∈SO(n + 1), we associate\nrespectively the following isometries:\nx 7→Qx\n,\nfi 7→fi ◦Qt.\nWe set y = Qx and gi = fi ◦Qt. Differentiating the equality gi(Qx) = fi(x), we obtain:\nDgi(y)Q = Dfi(x).\nWhen x is fixed, we can set Q conveniently so that y = en+1. Therefore\nDgi(en+1)Q|TxSn = Dfi(x)|TxSn.\nSince Q(TxSn) = Ten+1Sn we obtain\nDgi(en+1)|Ten+1Sn = Dfi(x)|TxSn.\nThis means that Pi,en+1(fi ◦Qt) = Pi,x(fi). Thus, in order to prove our claim, it is enough\nto show that Pi,en+1 is an orthogonal projection.\nSince for g = P\nJ gJXJ,\n∂g\n∂Xj (en+1) = g(ej+(d−1)en+1) and since Ten+1Sn = ⟨e1, . . . , en⟩,\nwe have that for any gi ∈Hdi,\nPi,en+1(gi) =\n1\n√di\n gi(e1+(di−1)en+1), . . . , gi(en+(di−1)en+1)\n \n.\nHence, for any gi ∈ker(Pi,en+1)⊥, i.e. such that giJ = 0 for all J ̸= ej + (di −1)en+1,\n1 ≤j ≤n, we have\n∥gi∥2 =\nX\nJ\ng2\niJ\n di\nJ\n = ∥Pi,en+1(gi)∥2\n2.\nWe conclude that Pi,x is an orthogonal projection.\nStep 3: From the previous step, for any fi ∈Hdi, using the orthogonal decomposition\nfi = f ◦\ni + f ⊥\ni\nwith f ◦\ni ∈ker Pi,x and f ⊥\ni ∈ker P ⊥\ni,x, we have\n∥Pi,x(fi)∥2\n2 = ∥Pi,x(f ⊥\ni )∥2\n2 = ∥f ⊥\ni ∥2 ≤∥fi∥2.\nIt is now immediate from Step 1 and from the definition of ∥f∥= maxi ∥fi∥that the\nFrobenius norm ∥M∥F of the matrix M satisfies\n∥M∥2\nF =\nn\nX\ni=1\n∥Pi,x(fi)∥2\n2 ≤\nn\nX\ni=1\n∥fi∥2 ≤n∥f∥2 = n\nand hence its spectral norm ∥M∥satisfies ∥M∥≤∥M∥F ≤√n. This bound is independent\nof the choice of the basis for the space TxSn.\nStep 4: We next present the algorithm to compute M, given f and x. This is a non-\noptimal algorithm, and can be significantly improved if more is known on the structure of\nthe polynomial system f.\nWe can compute each entry mij of the matrix M as the scalar product of\n1\n√\ndi Dfi(x)\nand the jth column Hj := (hkj)1≤k≤n+1 of H.\n20"},{"page":21,"text":"Proceeding as in the proof of Proposition 6.10, we can compute\n1\n√\ndi\n∂fi\n∂Xk (x) with\nError\n 1\n√\ndi\n∂fi\n∂Xk\n(x)\n \n= [[D + log S]].\nOn the other hand, the vector y =\nx−en+1\n∥x−en+1∥can be computed using 2n + 4 operations, and\nclearly Error(yj) = [[ log(n)]] for all j. Hence, for all coefficients hkj of H,\nError(hkj) = [[ log(n)]].\nApplying Proposition 6.7 we conclude\nError(mij)\n=\n[[D + log S + log n]]\n\n1\n√di\nDfi(x)\n \n\n ∥Hj∥\n=\n[[D + log S + log n]].\nThe second equality holds because ∥Hj∥= 1 since H is unitary, and because, as in the proof\nof Step 2,\n\n1\n√di\nDfi(x)\n\n2\n=\n\n1\n√di\nDgi(en+1)\n\n2\n= 1\ndi\n∥(gi(e1+(di−1)en+1), . . . , gi(dien+1))∥2 ≤∥gi∥2 ≤1.\nThis implies\n∥Error(M)∥F ≤[[n(D + log S + log n)]].\nLemma 6.12. Let x ∈Sn and M be as in Proposition 6.11. We can compute σmin(M) =\n∥M −1∥−1 satisfying\nError(σmin(M)) = [[n(log S + D + n3/2)]].\nProof.\nLet E′ = M −fl(M). By Proposition 6.11,\n∥E′∥≤∥E′∥F ≤[[n(log S + D + log n)]].\nLet M = fl(M). We compute σmin(M) = ∥M −1∥−1 using a backward stable algorithm\n(e.g., QR factorization). Then the computed fl(σmin(M)) is the exact σmin(M + E′′) for a\nmatrix E′′ with\n∥E′′∥≤cn2u∥M∥\nfor some universal constant c (see, e.g., [11, 17]). Thus,\nfl(σmin(M)) = fl(σmin(M)) = σmin(M + E′′) = σmin(M + E′ + E′′).\nWrite E = E′ + E′′. Then, using ∥M∥≤√n,\n∥E∥\n≤\n∥E′∥+ ∥E′′∥≤∥E′∥+ cn2u∥M∥≤∥E′∥+ cn2u(∥M∥+ ∥E′∥)\n=\n[[n(log S + D + log n)]] + cn2u(√n + [[n(log S + D + log n)]])\n=\n[[n(log S + D + log n)]] + cn2u(√n + c′un(log S + D + n3/2))\n=\n[[n(log S + D + n3/2)]]\nsince the hypothesis on u implies c′un(log S + D + n3/2) is bounded by a constant term.\nTherefore, fl(σmin(M)) = σmin(M + E) which implies by [11, Corollary 8.3.2]:\nError(σmin(M)) ≤∥E∥< [[n(log S + D + n3/2)]].\n21"},{"page":22,"text":"Proposition 6.13. Let f ∈S(Hd)). Assume u ≤\nK\nκ(f)2n2D log S for a small enough constant\nand let x ∈Sn. Then\n(i) If x /∈fl(A′(f)) then α(f, x) ≥1\n3α•.\n(ii) If x ∈fl(A′(f)) then α(f, x) < α•.\nProof.\nFrom Proposition 6.10\nfl(n∥f(x)∥∞D3/2)\n=\n(∥f(x)∥∞+ [[D + log S]])(nD3/2)(1 + θ4)\n≤\nnD3/2∥f(x)∥∞+ [[nD3/2(D + log S)]]\n. Also, from Lemma 6.12, using that σmin(M) ≤√n,\nfl(α•σmin(M)2)\n=\nα•\n \nσmin(M) + [[n(log S + D + n3/2)]]\n 2\n(1 + θ2)\n≥\nα•σmin(M)2 −2α•σmin(M)[[n(log S + D + n3/2)]]\n≥\nα•σmin(M)2 −[[n3/2(log S + D + n3/2)]].\nTherefore,\nn∥f(x)∥∞D3/2 + [[nD3/2(D + log S)]]\n≥\nfl(n∥f(x)∥∞D3/2) ≥fl(α•σ2\nmin)\n≥\nα•σ2\nmin −[[n3/2(log S + D + n3/2)]]\nor yet,\nn∥f(x)∥∞D3/2 −α•σ2\nmin\n≥\n−([[nD3/2(D + log S)]] + [[n3/2(log S + D + n3/2)]])\n≥\n−[[n3D5/2 log S]].\nCase I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, using the hypothesis on u and the inequality\nκ(f) ≥1,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)n2D log S\n≤\nKO(1)n∥f(x)∥∞D3/2 ≤n∥f(x)∥∞D3/2\n2\nthe last by choosing K small enough. Hence, n∥f(x)∥∞D3/2 −α•σ2\nmin ≥−\n \nn∥f(x)∥∞D3/2\n2\n \n,\nwhich implies 3\n2n∥f(x)∥∞D3/2 ≥α•σmin(M)2, i.e., α(f, x) ≥α•\n3 .\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u,\n[[n3D5/2 log S]]\n=\nuO(n3D5/2 log S) ≤K O(n3D5/2 log S)\nκ(f)2n2D log S\n≤\nKO(1)σmin(M)2D3/2 ≤α•σmin(M)2\n3\n22"},{"page":23,"text":"the last by choosing K small enough.\nThis implies n∥f(x)∥∞D3/2 −α•σmin(M)2 ≥\n−α•σmin(M)2\n3\nor, equivalently, α(f, x) ≥α•\n3 .\nThis shows part (i). For part (ii), one shows as above that\nn∥f(x)∥∞D3/2 −α•σ2\nmin ≤[[n3D5/2 log S]].\nThen, one proceeds as well by considering the two cases min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞and min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x).\nLemma 6.14. Let y1, y2 ∈Uη and let xi = φ(yi), i = 1, 2. Then d(x1, x2) ≥\nη\n2√n+1.\nProof.\nThe distance d(x1, x2) is minimized at y1 = (1, . . . , 1, 1) and y2 = (1, . . . , 1, 1−η).\nLet N = n + 1. Then\ncos(d(x1, x2))2\n=\n⟨y1, y2⟩2\n∥y1∥2∥y2∥2\n=\n(N −η)2\nN(N −2η + η2)\n=\n1 −\n(N −1)η2\nN 2 −2Nη + Nη2\n≤\n1 −η2 N −1\nN 2\n.\nHence\nd(x1, x2) ≥arccos\n r\n1 −η2 N −1\nN 2\n!\n= arcsin\n η\nN\n√\nN −1\n \n≥\nη\n2\n√\nN\n.\nLemma 6.15. Let u <\nKη2\nn log n for a small enough constant K. For x1, x2 ∈Gη we can\ncompute d(x1, x2) such that\nError(d(x1, x2)) ≤\n √n log n\nη\n \n.\nProof.\nLet yi = φ−1(xi), i = 1, 2, and a = cos(d(x1, x2)), i.e.,\na = ⟨y1, y2⟩\n∥y1∥∥y2∥.\nWe have, using Proposition 6.7,\nfl(⟨y1, y2⟩) = ⟨y1, y2⟩+ θlog n∥y1∥∥y2∥\nand fl(∥y1∥∥y2∥) = ∥y1∥∥y2∥(1+θlog n). Using now Propositions 6.4, 6.5, and 6.6, it follows\nthat fl(a) = a + ε with ε = [[ log n]].\nBy choosing K sufficiently small, ε ≤\nη2n\n12(n+1)2 . Also, from the proof of Lemma 6.14,\na = cos(d(x1, x2)) ≤\ns\n1 −\nη2n\n(n + 1)2\n23"},{"page":24,"text":"and hence, using that √z + y ≤√z + 3y whenever 0 < z, y ≤1, we obtain\na + ε ≤\ns\n1 −\nη2n\n(n + 1)2 +\nη2n\n12(n + 1)2 ≤\ns\n1 −\n3η2n\n4(n + 1)2 ≤\ns\n1 −\nη2\n3(n + 1).\nUsing Lemma 6.8(ii) it follows that,\narccos(a + ε)\n=\narccos(a) + ε\n \n1\np\n1 −(a + ε)2\n \n=\narccos(a) + [[ log n]]\n \np\n3(n + 1)\nη\n .\nTherefore,\nError(d(x1, x2)) ≤\n √n log n\nη\n \n.\nLemma\n6.16. Let\nf\n∈\nS(Hd).\nAssume that\nη\n≥\nα•\n8D2(n+1)κ(f)2\nand u\n≤\nK\nD2n5/2κ(f)3(log S+n3/2D2κ(f)2) with K small enough, and let x, y ∈Gη. Then\n(i) If y ∈fl(B\n′\nf(x)) then d(x, y) ≤2σβ(f, x).\n(ii) If y /∈fl(B\n′\nf(x)) then d(x, y) > σβ(f, x).\nProof.\nBy Lemmas 6.12 and 6.15 (and using σmin(M) ≤√n and the bound d(x, y) ≤\nπ\n2 η√n + 1 which follows from (7)),\nError(σmin(M)d(x, y))\n=\nO\n d(x, y)Error(σmin(M)) + σmin(M)Error(d(x, y))\n \n=\nη π\n2\n√\nn + 1[[n(log S + D + n3/2)]] + √n\n √n log n\nη\n \n=\nη[[n3/2(log S + D + n3/2)]] +\n n log n\nη\n \n≤\n[[n3/2 log S + n3D2κ(f)2]]\nthe last by the bounds on η. Also, using Proposition 6.10,\nError\n 3\n2σ√n∥f(x)∥∞\n \n≤[[√n(D + log S)]].\nTherefore, for part (i),\nσmin(M)d(x, y) −3\n2σ√n∥f(x)∥∞\n≤fl(σmin(M)d(x, y)) −fl\n 3\n2σ√n∥f(x)∥∞\n \n+ [[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n≤[[n3/2 log S + n3D2κ(f)2]] + [[√n(D + log S)]]\n= [[n3/2 log S + n3D2κ(f)2]].\n24"},{"page":25,"text":"Case I. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n=\n1\n∥f(x)∥∞\nIn this case κ(f) ≥\n1\n∥f(x)∥∞and, therefore, by the hypothesis on u,\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nκ(f)n(log S + n3/2D2κ(f)2)\n≤\nσ√n\n2κ(f) ≤σ√n∥f(x)∥∞\n2\nthe last line by taking K small enough. This implies that σmin(M)d(x, y) ≤2σ√n∥f(x)∥∞,\ni.e., that d(x, y) ≤2σβ(f, x).\nCase II. min\nn\nμnorm(f, x),\n1\n∥f(x)∥∞\no\n= μnorm(f, x)\nIn this case κ(f) ≥μnorm(f, x) =\n√n\nσmin(M). By the hypothesis on u\n[[n3/2 log S + n3D2κ(f)2]]\n=\nO(n3/2 log S + n3D2κ(f)2)\nK\nD2n5/2κ(f)3(log S + n3/2D2κ(f)2)\n≤\n√nα•\n48D2(n + 1)3/2κ(f)3\n≤\n√nη\n8√n + 1κ(f) ≤\n√nd(x, y)\n4κ(f)\n≤σmin(M)d(x, y)\n4\nby taking K small enough and Lemma 6.14.\nThis implies that\n3\n4σmin(M)d(x, y) ≤\n3\n2σ√n∥f(x)∥∞, i.e., that d(x, y) ≤2σβ(f, x).\nThis shows part (i). Part (ii) is shown in a similar way.\nLemma 6.17. Let u ≤Kη2\nlog n with K small enough and x1, x2 ∈Gη.\n(i) If fl(d(x1, x2)) ≤fl( 3\n2πη√n + 1) then d(x1, x2) ≤2πη√n + 1.\n(ii) If fl(d(x1, x2)) > fl( 3\n2πη√n + 1) then d(x1, x2) > πη√n + 1.\nProof.\nBy Lemma 6.15 and the hypothesis on u, we obtain\nError(d(x1, x2)) =\n √n log n\nη\n \n≤O\n √n log n\nη\n Kη2\nlog n ≤π\n2 η\n√\nn + 1,\nthe last by taking K small enough.\nAlso, Error( 3\n2πη√n + 1) ≤\n3\n2πη√n + 1 γ3.\nThe\nstatement easily follows from these two bounds.\nLemma 6.18. Let u ≤\nKη\n√\nnD\nD+log S+η\n√\nnD with K small enough, f ∈S(Hd) and x ∈Sn.\n(i) If fl(∥f(x)∥∞) ≤fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞≤πη\np\n(n + 1)D.\n(ii) If fl(∥f(x)∥∞) > fl(\n√\n2\n2 πη\np\n(n + 1)D) then ∥f(x)∥∞> π\n2 η\np\n(n + 1)D.\n25"},{"page":26,"text":"Proof.\nFor part (i), from Proposition 6.10,\n∥f(x)∥∞≤fl(∥f(x)∥∞) + [[D + log S]].\nAlso,\n√\n2\n2 πη\np\n(n + 1)D ≥fl(\n√\n2\n2 πη\np\n(n + 1)D) −[[η\np\n(n + 1)D]].\nTherefore,\n∥f(x)∥∞−\n√\n2\n2 πη\np\n(n + 1)D\n≤\nfl(∥f(x)∥∞) −fl(\n√\n2\n2 πη\np\n(n + 1)D) + [[D + log S + η\np\n(n + 1)D]]\n≤\n[[D + log S + η\np\n(n + 1)D]]\n=\nO\n D + log S + η\np\n(n + 1)D\n \nKη\n√\nnD\nD + log S + η\n√\nnD\n≤\n(1 −\n√\n2\n2 )η\np\n(n + 1)D,\nthe last by taking K sufficiently small. It follows that ∥f(x)∥∞≤πη\np\n(n + 1)D and hence,\npart (i) of the statement.\nPart (ii) is proved similarly.\n6.5\nProof of Theorem 1.1(4): Correctness\nWe will show that, if u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)), and the algorithm halts with\nη ≥\nα•\n8D2(n+1)κ(f)2 , then the value r/2 returned by the algorithm is #R(f).\nThis is a\nconsequence of the floating following versions of Lemmas 5.1 and 5.2.\nLemma 6.19. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each x ∈fl(A′(f)) there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover for each\npoint z ∈fl(B\n′\nf(x)), the Newton sequence starting at z converges to ζx.\n(ii) Let x, y ∈fl(A′(f)). Then ζx = ζy ⇐⇒fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)) ̸= ∅.\nProof.\n(i) Applying Proposition 6.13(ii), x ∈fl(A′(f)) implies that α(f, x) < α•.\nTherefore, by Theorem 6.1, there exists ζx ∈Z(f) such that ζx ∈Bf(x). Moreover, if\nz ∈fl(B\n′\nf(x)), by Lemma 6.16(i), d(x, z) ≤2σβ(f, x) and the Newton sequence starting at\nz converges to ζx.\n(ii) If ζx = ζy, then Bf(x) ∩Bf(y) ̸= ∅which implies by Lemma 6.16(ii) that there exists\nz ∈fl(B\n′\nf(x)) ∩fl(B\n′\nf(y)).\nThis immediately implies, using that Bf(x) ⊂fl(B\n′\nf(x)) by Lemma 6.16(ii), the follow-\ning corresponding floating version of Lemma 5.2.\nLemma 6.20. Let f ∈S(Hd), η ≥\nα•\n8D2(n+1)κ(f)2 and u ≤\n1\nO(D2n5/2κ(f)3(log S+n3/2D2κ(f)2)).\n(i) For each component fl(U) of fl(G′\nη), there is a unique zero ζU ∈Z(f) such that\nζU ∈Z(fl(U)). Moreover ζU ∈∩x∈fl(U)Bf(x).\n26"},{"page":27,"text":"(ii) If fl(U) and fl(V ) are different components of fl(G′\nη), then ζU ̸= ζV .\nIn order to show the correctness of Count Roots 2, we only need to prove that Z(f) ⊂\nZ(fl(G′\nη)). This easily follows adapting the proof of Part (1) in Section 5.3 to this situation,\nmaking use of Lemma 6.20 and the facts that Condition (i), fl(d(xi, xj)) > fl( 3\n2πη√n + 1),\nimplies that d(xi, xj) > πη√n + 1 (Lemma 6.17(ii)) and Condition (ii), fl(∥f(x)∥∞) >\nfl(\n√\n2\n2 πη\np\n(n + 1)D), implies that ∥f(x)∥∞> π\n2 η\np\n(n + 1)D (Lemma 6.18(ii)).\n6.6\nProof of Theorem 1.1(4): Complexity\nWe want to show that if η ≤\nα•\n4D2(n+1)κ(f)2 then Count Roots 2(f) halts. Note that this\nmeans that\nα•\n8D2(n + 1)κ(f)2 < η ≤\nα•\n4D2(n + 1)κ(f)2\nand hence, by § 6.5, that it correctly returns #R(f).\nBecause of the hypothesis on η, the hypotheses of Lemmas 6.2, and 6.3 are satisfied.\nLet fl(U) ̸= fl(V ) be different components of fl(G′\nη), and therefore, by Lemma 6.20,\nζU ̸= ζV , and for all x ∈fl(U), y ∈fl(V ), by Lemma 6.2, d(x, y) > 2πη√n + 1 holds. This\nimplies, by Lemma 6.17(i), that Condition (i) in Count Roots 2 is satisfied.\nConsider now x ̸∈fl(A′(f)). By Proposition 6.13(i), α(f, x) ≥α•\n3 . This implies, by\nLemma 6.3, that ∥f(x)∥∞> πη\np\n(n + 1)D, which in turn, by Lemma 6.18(i), ensures that\nCondition (ii) in Count Roots 2 is satisfied. Hence, the algorithm halts.\nAknowledgement. We are grateful to Andr ́e Galligo for a helpful discussion.\nReferences\n[1] B. Bank, M. Giusti, J. Heintz, and L. Pardo.\nGeneralized polar varieties: geometry and\nalgorithms. J. Compl., 21:377–412, 2005.\n[2] L. Blum, F. Cucker, M. Shub, and S. Smale. Complexity and Real Computation. Springer-\nVerlag, 1998.\n[3] P. B ̈urgisser and F. Cucker. Counting complexity classes for numeric computations II: Algebraic\nand semialgebraic sets. J. Compl., 22:147–191, 2006.\n[4] D. Cheung and F. Cucker. Solving linear programs with finite precision: II. Algorithms. J.\nCompl., 22:305–335, 2006.\n[5] G.E. Collins. Quantifier elimination for real closed fields by cylindrical algebraic deccomposi-\ntion, volume 33 of Lect. Notes in Comp. Sci., pages 134–183. Springer-Verlag, 1975.\n[6] F. Cucker. Approximate zeros and condition numbers. J. Compl., 15:214–226, 1999.\n[7] F. Cucker and J. Pe ̃na. A primal-dual algorithm for solving polyhedral conic systems with a\nfinite-precision machine. SIAM J. Optim., 12:522–554, 2002.\n[8] F. Cucker and S. Smale. Complexity estimates depending on condition and round-offerror.\nJournal of the ACM, 46:113–184, 1999.\n[9] F. Cucker and D.X. Zhou. Learning Theory: An Approximation Theory Viewpoint. Cambridge\nUniv. Press, 2007.\n[10] J.-P. Dedieu, P. Priouret, and G. Malajovich.\nNewton method on Riemannian manifolds:\nCovariant alpha-theory. IMA Journal of Numerical Analysis, 23:395–419, 2003.\n27"},{"page":28,"text":"[11] G. Golub and C. Van Loan. Matrix Computations. John Hopkins Univ. Press, 3rd edition,\n1996.\n[12] D.Yu. Grigoriev. Complexity of deciding Tarski algebra. Journal of Symbolic Computation,\n5:65–108, 1988.\n[13] D.Yu. Grigoriev and N.N. Vorobjov. Solving systems of polynomial inequalities in subexpo-\nnential time. Journal of Symbolic Computation, 5:37–64, 1988.\n[14] D.Yu. Grigoriev and N.N. Vorobjov. Counting connected components of a semialgebraic set\nin subexponential time. Computational Complexity, 2:133–186, 1992.\n[15] Y. Han and R.A. Wagner.\nAn efficient and fast parallel-connected component algorithm.\nJournal of the ACM, 37(3):626–642, 1990.\n[16] J. Heintz, M.-F. Roy, and P. Solerno. Single exponential path finding in semi-algebraic sets\nII: The general case. In C.L. Bajaj, editor, Algebraic Geometry and its Applications, pages\n449–465. Springer-Verlag, 1994.\n[17] N. Higham. Accuracy and Stability of Numerical Algorithms. SIAM, 1996.\n[18] T.Y. Li. Numerical solution of polynomial systems by homotopy continuation methods. In\nP.G. Ciarlet and F. Cucker, editors, Handbook of numerical analysis, volume 11, pages 209–304.\nNorth-Holland, 2003.\n[19] G. Malajovich. On generalized Newton algorithms: Quadratic convergence, path-following and\nerror analysis. Theoret. Comp. Sci., 133:65–84, 1994.\n[20] K. Meer. Counting problems over the reals. Theoret. Comp. Sci., 242:41–58, 2000.\n[21] M. Shub and S. Smale. Complexity of B ́ezout’s theorem I: geometric aspects. Journal of the\nAmer. Math. Soc., 6:459–501, 1993.\n[22] M. Shub and S. Smale. Complexity of B ́ezout’s theorem III: condition number and packing.\nJournal of Complexity, 9:4–14, 1993.\n[23] M. Shub and S. Smale. Complexity of B ́ezout’s theorem IV: probability of success; extensions.\nSIAM J. of Numer. Anal., 33:128–148, 1996.\n[24] S. Smale. Newton’s method estimates from data at one point. In R. Ewing, K. Gross, and\nC. Martin, editors, The Merging of Disciplines: New Directions in Pure, Applied, and Com-\nputational Mathematics. Springer-Verlag, 1986.\n[25] A. Tarski. A Decision Method for Elementary Algebra and Geometry. University of California\nPress, 1951.\n[26] H. Weyl. The Theory of Groups and Quantum Mechanics. Dover, 1932.\n[27] H.R. W ̈uthrich. Ein Entscheidungsverfahren f ̈ur die Theorie der reell-abgeschlossenen K ̈orper.\nvolume 43 of Lect. Notes in Comp. Sci., pages 138–162. Springer-Verlag, 1976.\n28"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"Let d1, . . . , dn ∈N and d = (d1, . . . , dn). We will denote by Hd the space of polynomial","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"systems f = (f1, . . . , fn) with fi ∈R[X0, . . . , Xn] homogeneous of degree di.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"we denote by Z(f) = {ζ ∈Sn : f(ζ) = 0} the zero-set of f in Sn then the number #R(f) of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"two more additional parameters. One is D = maxi≤n di. The other is a condition measure","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"will also let S = max Si where Si is the number of non-zero coefficients of fi. Note that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"S is bounded by a simple expression in terms of n and D, namely, S =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"number of values. Indeed, if Ra denotes the set of input data d for which φ(d) = a the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"Hd = Hd1 × · · · × Hdn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"g(X) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"where J = (J0, . . . , Jn) is assumed to range over all multi-indices such that |J| = Pn","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"k=0 Jk =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"d, XJ = XJ0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"⟨g, h⟩=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"|J|=d","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"which gives rise to the norm ∥g∥=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"Hd by taking for f = (f1, . . . , fn) ∈Hd:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"∥f∥= ∥(f1, . . . , fn)∥= max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"⟨g ◦Q, h ◦Q⟩= ⟨g, h⟩.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"invariant. We will use this norm on Hd all along this paper. For x = (x1, . . . , xn) ∈Rn we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"recall that ∥x∥2 = (x2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"n)1/2 and ∥x∥∞= max{|x1|, . . . , |xn|}. We will often denote","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"μnorm(f, x) = ∥f∥√n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"κ(f) = max","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"μnorm(f) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"C |f(ζ)=0 μnorm(f, ζ).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"ζ∈Sn|f(ζ)=0 μnorm(f, ζ)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"generality that x = e0 := (1, 0, . . . , 0).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"For the first expression, i.e., μnorm(f, x), define g = (g1, . . . , gn) ∈Hd by gi(X) = fi(X)−","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"0 . Then g(e0) = 0 and [2, Corollary 3 p. 234], μnorm(g, e0) ≥1 (this is shown for","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"Since Df(e0) = Dg(e0) and ∥g∥≤∥f∥, we can conclude μnorm(f, e0) ≥μnorm(g, e0) ≥1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"0 ≤d(x, y) ≤π, ∀x, y ∈Sn) and for x ∈Sn, r > 0, we set B(x, r) := {y ∈Sn : d(y, x) < r}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"and B(x, r) := {y ∈Sn : d(y, x) ≤r}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"In particular, if f(x) ̸= 0, there is no zero of f in B(x, min{∥f(x)∥∞/(∥f∥","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"g(x) = ⟨g(X), (xT X)deg g⟩.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"Because of orthogonal invariance, we can assume that x = e0 and y = e0 cos θ + e1 sin θ,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"where θ = d(x, y). Equation (3) implies that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"⟨fi(X), (xT X)di⟩−⟨fi(X), (yT X)di⟩= ⟨fi(X), (xT X)di −(yT X)di⟩","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"0 −(X0 cos θ + X1 sin θ)di = Xdi","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"k=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"Nf(x) = expx","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"expx h = cos(∥h∥)x + sin(∥h∥)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"(i) It is easy to see that β(f, x) = d(x, Nf(x)).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"ζ|f(ξ)=0 2D−3/2γ(f, ζ).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"β(f, x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"= β(f, x).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"sequence {xk}k∈N, where x0 := x and xk+1 := Nf(xk), is defined for all k and moreover","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"The limit point ζ = limk→∞xk is a fixed point for Newton iteration and a zero of f. It is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"In what follows we denote σ := P","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"k≥0 2−2k+1 = 1.632843018 . . . and we set","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"Bf(x) := {y ∈Sn | d(x, y) ≤σβ(f, x)}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"Theorem 4.3. There exists an universal constant α∗:= 0.0384629388 . . . such that for all","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"Theorem 4.4. Let f : Sn →Rn be analytic. Suppose that f(ζ) = 0 and Df(ζ) is an","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"R(f, ζ) := min","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"If d(x, ζ) ≤R(f, ζ), then the Newton sequence xk = N k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"Now let α0 := 0.130716944 . . . denote the smallest positive root of the polynomial ψ(u)2−","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"s0 :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"α(f, x) ≤α0. Then the Newton sequence xk = N k","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"Finally we introduce ψ(u) := 1−4u+2u2, which is positive and decreasing for 0 < u < 1−","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"ν := d(x, y)γ(f, x) < 1 −","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"Set ν∗:= 0.0628039411 . . . for the only real root of the polynomial","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"Ψ(u) := (3 −","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"and α∗:= ν∗","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"σ = 0.0384629388 . . .. Note that α∗≤min{α0, s0π}.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"ν = d(x, ζ)γ(f, x) ≤σβ(f, x)γ(f, x) ≤σα(f, x) ≤σα∗= ν∗< 1 −","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"2σβ(f, x) ≤R(f, ζ) = min","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"grid on the cube Cn = {y | ∥y∥∞= 1}. We make use of the (easy to compute) bijections","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"φ : Cn →Sn and φ−1 : Sn →Cn given by φ(y) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"∥y∥and φ−1(x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"Given η := 2−k for some k ≥1, we consider the uniform grid Uη of mesh η on Cn. This is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"Given η as above we associate to it a graph Gη as follows. We set A(f) := {x ∈Sn |","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"are joined by an edge if and only if Bf(x) ∩Bf(y) ̸= ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"(ii) Let x, y ∈A(f). Then ζx = ζy ⇐⇒Bf(x) ∩Bf(y) ̸= ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"We define Z(Gη) := S","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"Z(U) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"(ii) If U and V are different components of Gη, then ζU ̸= ζV .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"and ξ ∈Bf(y). Since U is connected, there exist x0 = x, x1, . . . , xk−1, xk := y in A(f) such","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"that (xi, xi+1) is an edge of Gη for i = 0, . . . , k −1, that is, Bf(xi)∩Bf(xi+1) ̸= ∅. If ζi and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"have ζi = ζi+1, and thus ζ = ξ ∈Bf(x) ∩Bf(y).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"(ii) Assume ζU = ζV ∈Bf(x) ∩Bf(y) ⊂Z(U) ∩Z(V ), then x and y are joined by an edge","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"let η :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"else η := η/2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"SCH(x1, . . . , xq) := Cone(x1, . . . , xq) ∩Sn","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"SCH(x1, . . . , xq) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"λi = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"i=1 B(xi, ri) ̸= ∅, then SCH(x1, . . . , xq) ⊂","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"i=1 B(xi, ri).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"i=1 B(xi, ri). We will prove that x ∈B(xi, ri)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"If x = y, this is obvious.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"If x ̸= y, let H be the half-space","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"H :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"Since ∥x∥= ∥y∥= 1, we have ⟨x + y, y −x⟩= 0, and we note that in this case, x + y","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"determines the mid-line between x and y. Moreover, since x ̸= y, we have x ∈H since","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"⟨x, y −x⟩= ⟨x, y⟩−∥x∥2 < ∥x∥∥y∥−∥x∥2 = 0. Therefore the half-space H is the set of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"the number of connected components of Gη, then #R(f) = #Z(f)/2 = r/2. We already","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"Assume that there is a zero ζ of f in Sn such that ζ is not in Z(Gη). Let B∞(φ−1(ζ), η) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"{y ∈Uη | ∥y −φ−1(ζ)∥∞≤η} = {y1, . . . , yq}, the set of all neighbors of φ−1(ζ) in Uη, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"let xi = φ(yi), i = 1, . . . , q. Clearly, φ−1(ζ) is in the cone spanned by {y1, . . . , yq} and hence","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"There exist l̸= s and 1 ≤i < j ≤r such that xl∈Ui and xs ∈Uj. Since condition","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"Lemma 5.4. If ζ1 ̸= ζ2 ∈Z(f) then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"For i = 1, 2, using (6) and Proposition 2.2,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"R(f, ζi) = min","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"R(f, ζ1) and d(x, ζ2) < R(f, ζ2). Then Theorem 4.4 implies that ζ1 = ζ2, a contradiction.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"Lemma 5.5. Let x1, x2 ∈Gη with associated zeros ζ1 ̸= ζ2. If η ≤","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"= μnorm(f, x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"α∗≤α(f, x) = 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"that, for i = 1, . . . , r, any vertex xi of Ui is an approximate zero of the only zero of f in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"Theorem 6.1. There exist a universal constant α• = 0.028268 · · · such that, for all x ∈Sn,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"is proved by taking ν• = 0.046158 · · · to be the only real root of the polynomial Ψ(u) :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"7)(1 −u)ψ(u) −6u, and α• = ν•","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"σ = 0.028268. Then, one proves as in Theorem 4.3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"Lemma 6.2. Let x1, x2 ∈Gη with associated zeros ξ1 and ξ2, ξ1 ̸= ξ2. If η ≤(3−","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"(i) For any x ∈F, r(x) = x. In particular, r(0) = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"(ii) For any x ∈R, r(x) = x(1 + δ) with |δ| ≤u.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"xe◦y = r(x ◦y)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq116","equation_number":null,"raw_text":"xe◦y = (x ◦y)(1 + δ),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq117","equation_number":null,"raw_text":"i=1(1 + δi)ρi naturally appear. Our round-offanalysis uses the notations and ideas in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq118","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq119","equation_number":null,"raw_text":"(1 + δi)ρi = 1 + θn,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq120","equation_number":null,"raw_text":"|θn| ≤γn =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq121","equation_number":null,"raw_text":"|θk| ≤γk =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq122","equation_number":null,"raw_text":"|θj| ≤γj =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq123","equation_number":null,"raw_text":"1. (1 + θk)(1 + θj) = 1 + θk+j for some |θk+j| ≤γk+j.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq124","equation_number":null,"raw_text":"Error(q) = |q −fl(q)|.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq125","equation_number":null,"raw_text":"fl(⟨x, y⟩) = ⟨x, y⟩+ θ⌈log2 n⌉+1⟨|x|, |y|⟩,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq126","equation_number":null,"raw_text":"where |x| = (|x1|, . . . , |xn|).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq127","equation_number":null,"raw_text":"In particular, if x = y, the algorithm computes fl(∥x∥2)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq128","equation_number":null,"raw_text":"fl(∥x∥2) = ∥x∥2(1 + θ⌈log2 n⌉+1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq129","equation_number":null,"raw_text":"1 −θ = 1 −θ′ with |θ′| ≤|θ|.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq130","equation_number":null,"raw_text":"(ii) Let 0 < a ≤1 and ε ∈R such that 0 < a + ε < 1. Then, arccos(a + ε) = arccos(a) +","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq131","equation_number":null,"raw_text":"1 −θ = θ(√ξ)′ with ξ ∈(1 −θ, 1). But","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq132","equation_number":null,"raw_text":"arccos(a + ε) −arccos(a) = ε arccos′(ξ) = ε","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq133","equation_number":null,"raw_text":"op(x) = op(x)(1 + δ),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq134","equation_number":null,"raw_text":"a(1 + θk) = √a(1 + θk+1).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq135","equation_number":null,"raw_text":"implies that in all what follows we have γg = O(ug) for all the expressions g we will","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq136","equation_number":null,"raw_text":"[[g]] := O(ug).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq137","equation_number":null,"raw_text":"versions of the sets A′(f) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq138","equation_number":null,"raw_text":"f(y) = {z ∈Sn | d(x, y) ≤","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq139","equation_number":null,"raw_text":"and we set σmin(M) = ∥M −1∥−1. Therefore","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq140","equation_number":null,"raw_text":"∥f∥√n ∥M −1∥= ∥f∥√n σmin(M)−1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq141","equation_number":null,"raw_text":"= √n σmin(M)−1∥f(x)∥∞,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq142","equation_number":null,"raw_text":"= ∥f∥n σmin(M)−2∥f(x)∥∞","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq143","equation_number":null,"raw_text":"f(x)) ̸=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq144","equation_number":null,"raw_text":"let η :=","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq145","equation_number":null,"raw_text":"else η := η/2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq146","equation_number":null,"raw_text":"∥f∥= 1. We denote by S(Hd) the sphere of such systems. Also, we do not discuss in what","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq147","equation_number":null,"raw_text":"Error(∥f(x)∥∞) = [[D + log S]]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq148","equation_number":null,"raw_text":"Let f = (f1, . . . , fn). For i ≤n write fi = P cJXJ and let S be the number","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq149","equation_number":null,"raw_text":"of coefficients of fi. To compute f(x) one computes each monomial cJxJ with fl(cJxJ) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq150","equation_number":null,"raw_text":"where we used that for any x ∈Sn, | P |cJ|xJ| ≤∥P |cJ|xJ∥= ∥fi∥≤∥f∥= 1 and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq151","equation_number":null,"raw_text":"∥Error(M)∥F = [[n(log S + D + log n)]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq152","equation_number":null,"raw_text":"Step 1: Let y =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq153","equation_number":null,"raw_text":"Hy = In+1 −2yyt","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq154","equation_number":null,"raw_text":"M =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq155","equation_number":null,"raw_text":"We set y = Qx and gi = fi ◦Qt. Differentiating the equality gi(Qx) = fi(x), we obtain:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq156","equation_number":null,"raw_text":"Dgi(y)Q = Dfi(x).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq157","equation_number":null,"raw_text":"When x is fixed, we can set Q conveniently so that y = en+1. Therefore","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq158","equation_number":null,"raw_text":"Dgi(en+1)Q|TxSn = Dfi(x)|TxSn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq159","equation_number":null,"raw_text":"Since Q(TxSn) = Ten+1Sn we obtain","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq160","equation_number":null,"raw_text":"Dgi(en+1)|Ten+1Sn = Dfi(x)|TxSn.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq161","equation_number":null,"raw_text":"This means that Pi,en+1(fi ◦Qt) = Pi,x(fi). Thus, in order to prove our claim, it is enough","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq162","equation_number":null,"raw_text":"Since for g = P","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq163","equation_number":null,"raw_text":"∂Xj (en+1) = g(ej+(d−1)en+1) and since Ten+1Sn = ⟨e1, . . . , en⟩,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq164","equation_number":null,"raw_text":"Pi,en+1(gi) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq165","equation_number":null,"raw_text":"Hence, for any gi ∈ker(Pi,en+1)⊥, i.e. such that giJ = 0 for all J ̸= ej + (di −1)en+1,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq166","equation_number":null,"raw_text":"∥gi∥2 =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq167","equation_number":null,"raw_text":"= ∥Pi,en+1(gi)∥2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq168","equation_number":null,"raw_text":"fi = f ◦","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq169","equation_number":null,"raw_text":"2 = ∥Pi,x(f ⊥","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq170","equation_number":null,"raw_text":"2 = ∥f ⊥","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq171","equation_number":null,"raw_text":"It is now immediate from Step 1 and from the definition of ∥f∥= maxi ∥fi∥that the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq172","equation_number":null,"raw_text":"F =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq173","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq174","equation_number":null,"raw_text":"i=1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq175","equation_number":null,"raw_text":"∥fi∥2 ≤n∥f∥2 = n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq176","equation_number":null,"raw_text":"and the jth column Hj := (hkj)1≤k≤n+1 of H.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq177","equation_number":null,"raw_text":"= [[D + log S]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq178","equation_number":null,"raw_text":"On the other hand, the vector y =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq179","equation_number":null,"raw_text":"clearly Error(yj) = [[ log(n)]] for all j. Hence, for all coefficients hkj of H,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq180","equation_number":null,"raw_text":"Error(hkj) = [[ log(n)]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq181","equation_number":null,"raw_text":"The second equality holds because ∥Hj∥= 1 since H is unitary, and because, as in the proof","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq182","equation_number":null,"raw_text":"Lemma 6.12. Let x ∈Sn and M be as in Proposition 6.11. We can compute σmin(M) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq183","equation_number":null,"raw_text":"Error(σmin(M)) = [[n(log S + D + n3/2)]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq184","equation_number":null,"raw_text":"Let E′ = M −fl(M). By Proposition 6.11,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq185","equation_number":null,"raw_text":"Let M = fl(M). We compute σmin(M) = ∥M −1∥−1 using a backward stable algorithm","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq186","equation_number":null,"raw_text":"fl(σmin(M)) = fl(σmin(M)) = σmin(M + E′′) = σmin(M + E′ + E′′).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq187","equation_number":null,"raw_text":"Write E = E′ + E′′. Then, using ∥M∥≤√n,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq188","equation_number":null,"raw_text":"Therefore, fl(σmin(M)) = σmin(M + E) which implies by [11, Corollary 8.3.2]:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq189","equation_number":null,"raw_text":"= μnorm(f, x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq190","equation_number":null,"raw_text":"In this case κ(f) ≥μnorm(f, x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq191","equation_number":null,"raw_text":"= μnorm(f, x).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq192","equation_number":null,"raw_text":"Lemma 6.14. Let y1, y2 ∈Uη and let xi = φ(yi), i = 1, 2. Then d(x1, x2) ≥","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq193","equation_number":null,"raw_text":"The distance d(x1, x2) is minimized at y1 = (1, . . . , 1, 1) and y2 = (1, . . . , 1, 1−η).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq194","equation_number":null,"raw_text":"Let N = n + 1. Then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq195","equation_number":null,"raw_text":"= arcsin","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq196","equation_number":null,"raw_text":"Let yi = φ−1(xi), i = 1, 2, and a = cos(d(x1, x2)), i.e.,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq197","equation_number":null,"raw_text":"a = ⟨y1, y2⟩","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq198","equation_number":null,"raw_text":"fl(⟨y1, y2⟩) = ⟨y1, y2⟩+ θlog n∥y1∥∥y2∥","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq199","equation_number":null,"raw_text":"and fl(∥y1∥∥y2∥) = ∥y1∥∥y2∥(1+θlog n). Using now Propositions 6.4, 6.5, and 6.6, it follows","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq200","equation_number":null,"raw_text":"that fl(a) = a + ε with ε = [[ log n]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq201","equation_number":null,"raw_text":"a = cos(d(x1, x2)) ≤","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq202","equation_number":null,"raw_text":"= [[n3/2 log S + n3D2κ(f)2]].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq203","equation_number":null,"raw_text":"= μnorm(f, x)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq204","equation_number":null,"raw_text":"In this case κ(f) ≥μnorm(f, x) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq205","equation_number":null,"raw_text":"Error(d(x1, x2)) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq206","equation_number":null,"raw_text":"(ii) Let x, y ∈fl(A′(f)). Then ζx = ζy ⇐⇒fl(B","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq207","equation_number":null,"raw_text":"f(y)) ̸= ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq208","equation_number":null,"raw_text":"(ii) If ζx = ζy, then Bf(x) ∩Bf(y) ̸= ∅which implies by Lemma 6.16(ii) that there exists","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq209","equation_number":null,"raw_text":"η), then ζU ̸= ζV .","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq210","equation_number":null,"raw_text":"Let fl(U) ̸= fl(V ) be different components of fl(G′","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq211","equation_number":null,"raw_text":"ζU ̸= ζV , and for all x ∈fl(U), y ∈fl(V ), by Lemma 6.2, d(x, y) > 2πη√n + 1 holds. This","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":53804,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}