Reachability Problems for Continuous Linear Dynamical Systems
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1 Reachability Problems for Continuous Linear Dynamical Systems James Worrell Department of Computer Science, Oxford University (Joint work with Ventsislav Chonev and Joël Ouaknine) FICS 2015 September 12th, 2015
2 Reachability for Continuous-Time Markov Chains Stagnant market Bull market 0.02 Bear market 0.3
3 Reachability for Continuous-Time Markov Chains Stagnant market Bull market 0.02 Bear market 0.3 Distribution P(t) at time t satisfies P (t) = P(t)Q, where Q = is the rate matrix.
4 Reachability for Continuous-Time Markov Chains Stagnant market Bull market 0.02 Bear market 0.3 Distribution P(t) at time t satisfies P (t) = P(t)Q, where Q = is the rate matrix. Is it ever more likely to be a Bear market than a Bull market? t (P(t) Bear P(t) Bull )
5 Reachability - Some Basic Insights Is it ever more likely to be a Bear market than a Bull market? t (P(t) Bear P(t) Bull )
6 Reachability - Some Basic Insights Is it ever more likely to be a Bear market than a Bull market? t (P(t) Bear P(t) Bull ) Reduce to the time-bounded case by computing the stationary distribution: π = (0.885, 0.071, 0.044)
7 Reachability - Some Basic Insights Is it ever more likely to be a Bear market than a Bull market? t (P(t) Bear P(t) Bull ) Reduce to the time-bounded case by computing the stationary distribution: π = (0.885, 0.071, 0.044) Reduction does not work if π is on boundary of the target set.
8 Reachability - Some Basic Insights Is it ever more likely to be a Bear market than a Bull market? t (P(t) Bear P(t) Bull ) Reduce to the time-bounded case by computing the stationary distribution: π = (0.885, 0.071, 0.044) Reduction does not work if π is on boundary of the target set. Semantic shift consider whether the closure of the orbit {P(t) : t 0} meets the closure of the target set.
9 Cyber-Physical Systems To analyze a cyber-physical system, such as a pacemaker, we need to consider the discrete software controller interacting with the physical world, which is typically modelled by differential equations Rajeev Alur (CACM, 2013)
10 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k
11 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata
12 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata ẋ = c rectangular hybrid automata
13 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata ẋ = c rectangular hybrid automata ẋ = Ax linear hybrid automata
14 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata ẋ = c rectangular hybrid automata ẋ = Ax linear hybrid automata...
15 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata ẋ = c rectangular hybrid automata ẋ = Ax linear hybrid automata... o-minimal flows + strong resets reachability decidable
16 Hybrid Automata: Various Continuous Dynamics Hybrid automaton = states + variables x R k ẋ = 1 timed automata ẋ = c rectangular hybrid automata ẋ = Ax linear hybrid automata... o-minimal flows + strong resets reachability decidable Is this location a trap? x := 2 y := 4 ẋ = 3x y ẏ = x 5y x 10 y 2?
17 Reachability for Continuous Linear Dynamical Systems Is this location a trap? x := 2 y := 4 ẋ = 3x y ẏ = x 5y x 10 y 2? Stagnant market Is ever more likely to be a Bear Bull market Bear market market than a Bull market: t (P(t) Bear P(t) Bull )?
18 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax
19 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
20 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
21 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
22 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
23 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
24 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0)
25 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0) u
26 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0) f (t) = u T x(t)
27 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0) f (t) = u T x(t) f (k) (t) + a k 1 f (k 1) (t) a 1 f (t) + a 0 f (t) = 0
28 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0) f (t) = u T x(t) f (k) (t) + a k 1 f (k 1) (t) a 1 f (t) + a 0 f (t) = 0 m f (t) = P j (t)e λ j t j=1
29 Reachability for Continuous Linear Dynamical Systems x : R 0 R k ẋ = Ax x(t) = exp(at)x(0) f (t) = u T x(t) f (k) (t) + a k 1 f (k 1) (t) a 1 f (t) + a 0 f (t) = 0 m f (t) = P j (t)e λ j t j=1 Note the λ j are complex in general.
30 Reachability for Continuous Linear Dynamical Systems Let f : R 0 R be given as above, with all coefficients algebraic.
31 Reachability for Continuous Linear Dynamical Systems Let f : R 0 R be given as above, with all coefficients algebraic. BOUNDED-ZERO Problem Instance: f and bounded interval [a, b] Question: Is there t [a, b] such that f (t) = 0?
32 Reachability for Continuous Linear Dynamical Systems Let f : R 0 R be given as above, with all coefficients algebraic. BOUNDED-ZERO Problem Instance: f and bounded interval [a, b] Question: Is there t [a, b] such that f (t) = 0? ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0?
33 Reachability for Continuous Linear Dynamical Systems Let f : R 0 R be given as above, with all coefficients algebraic. BOUNDED-ZERO Problem Instance: f and bounded interval [a, b] Question: Is there t [a, b] such that f (t) = 0? ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0? Decidability open! [Bell, Delvenne, Jungers, Blondel 2010]
34 Related Work A lot of work since 1920s on the zeros of exponential polynomials f (z) = m P j (z)e λ j z j=1 (Polya, Ritt, Tamarkin, Kac, Voorhoeve, van der Poorten,... ) but mostly on distribution of complex zeros.
35 Related Work A lot of work since 1920s on the zeros of exponential polynomials f (z) = m P j (z)e λ j z j=1 (Polya, Ritt, Tamarkin, Kac, Voorhoeve, van der Poorten,... ) but mostly on distribution of complex zeros. CONTINUOUS-ORBIT Problem The problem of whether the trajectory x(t) = e At x(0) reaches a given target point was shown to be decidable by Hainry (2008) and in PTIME by Chen, Han and Yu (2015).
36 Reachability for Continuous Linear Dynamical Systems Theorem (Bell, Delvenne, Jungers, Blondel 2010) In dimension 2, BOUNDED-ZERO and ZERO are decidable.
37 Reachability for Continuous Linear Dynamical Systems Theorem (Bell, Delvenne, Jungers, Blondel 2010) In dimension 2, BOUNDED-ZERO and ZERO are decidable. Theorem (Chonev, Ouaknine, W. 2015) In dimension 3, BOUNDED-ZERO and ZERO are decidable.
38 Reachability for Continuous Linear Dynamical Systems Theorem (Bell, Delvenne, Jungers, Blondel 2010) In dimension 2, BOUNDED-ZERO and ZERO are decidable. Theorem (Chonev, Ouaknine, W. 2015) In dimension 3, BOUNDED-ZERO and ZERO are decidable. Theorem (Chonev, Ouaknine, W. 2015) Assuming Schanuel s Conjecture, BOUNDED-ZERO is decidable in all dimensions.
39 Reachability for Continuous Linear Dynamical Systems Theorem (Bell, Delvenne, Jungers, Blondel 2010) In dimension 2, BOUNDED-ZERO and ZERO are decidable. Theorem (Chonev, Ouaknine, W. 2015) In dimension 3, BOUNDED-ZERO and ZERO are decidable. Theorem (Chonev, Ouaknine, W. 2015) Assuming Schanuel s Conjecture, BOUNDED-ZERO is decidable in all dimensions. It turns out that this result (in fact, a powerful generalisation of it) had already been discovered (but never published) in the early 1990s by Macintyre and Wilkie! [Angus Macintyre, personal communication, July 2015]
40 Reachability for Continuous Linear Dynamical Systems Theorem (Chonev, Ouaknine, W. 2015) In dimension 8 or less, ZERO reduces to BOUNDED-ZERO.
41 Reachability for Continuous Linear Dynamical Systems Theorem (Chonev, Ouaknine, W. 2015) In dimension 8 or less, ZERO reduces to BOUNDED-ZERO. Theorem (Chonev, Ouaknine, W. 2015) In dimension 9 (and above), decidability of ZERO would entail major breakthroughs in Diophantine approximation the Diophantine approximation type of α would be computable to within arbitrary precision.
42 Schanuel s Conjecture Theorem (Lindemann-Weierstrass) If a 1,..., a n are algebraic numbers linearly independent over Q, then e a 1,..., e an are algebraically independent.
43 Schanuel s Conjecture Theorem (Lindemann-Weierstrass) If a 1,..., a n are algebraic numbers linearly independent over Q, then e a 1,..., e an are algebraically independent. Schanuel s Conjecture If z 1,..., z n are complex numbers linearly independent over Q then some n-element subset of {z 1,..., z n, e z 1,..., e zn } is algebraically independent.
44 Schanuel s Conjecture Theorem (Lindemann-Weierstrass) If a 1,..., a n are algebraic numbers linearly independent over Q, then e a 1,..., e an are algebraically independent. Schanuel s Conjecture If z 1,..., z n are complex numbers linearly independent over Q then some n-element subset of {z 1,..., z n, e z 1,..., e zn } is algebraically independent. Example By Schanuel s conjecture some two-element subset of {1, πi, e 1, e πi } is algebraically independent.
45 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1
46 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) t
47 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a b t
48 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a b t
49 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t non-trivial zero t transcendental
50 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t non-trivial zero t transcendental
51 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t non-trivial zero t transcendental
52 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a b t
53 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a b t
54 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t
55 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t Can this situation arise?
56 The BOUNDED-ZERO Problem m Real-valued exponential polynomial f (t) = P j (t)e λ j t j=1 f(t) a t* b t Easily! For example, f (t) = 2 + e it + e it.
57 Laurent Polynomials and Factorisation Example Write f (t) = 2 + e it + e it in the form f (t) = P(e it ) for the Laurent polynomial P(z) = 2 + z + z 1.
58 Laurent Polynomials and Factorisation Example Write f (t) = 2 + e it + e it in the form f (t) = P(e it ) for the Laurent polynomial P(z) = 2 + z + z 1. Factorisation P(z) = (1 + z)(1 + z 1 ) induces a factorisation f (t) = (1 + e it ) (1 e it ) }{{}}{{} f 1 (t) f 2 (t)
59 Laurent Polynomials and Factorisation Example Write f (t) = 2 + e it + e it in the form f (t) = P(e it ) for the Laurent polynomial P(z) = 2 + z + z 1. Factorisation P(z) = (1 + z)(1 + z 1 ) induces a factorisation f (t) = (1 + e it ) (1 e it ) }{{}}{{} f 1 (t) f 2 (t) Common zeros of f 1 and f 2 are tangential zeros of f
60 Laurent Polynomials and Factorisation Example Write f (t) = 2 + e it + e it in the form f (t) = P(e it ) for the Laurent polynomial P(z) = 2 + z + z 1. Factorisation P(z) = (1 + z)(1 + z 1 ) induces a factorisation f (t) = (1 + e it ) (1 e it ) }{{}}{{} f 1 (t) f 2 (t) Common zeros of f 1 and f 2 are tangential zeros of f Idea: factorise f.
61 Laurent Polynomials and Factorisation Example Write f (t) = 2 + e it + e it in the form f (t) = P(e it ) for the Laurent polynomial P(z) = 2 + z + z 1. Factorisation P(z) = (1 + z)(1 + z 1 ) induces a factorisation f (t) = (1 + e it ) (1 e it ) }{{}}{{} f 1 (t) f 2 (t) Common zeros of f 1 and f 2 are tangential zeros of f Idea: factorise f. Noting that factors may be complex-valued!
62 The Real Case Any exponential polynomial f (t) can be written with f (t) = P(t, e a 1t,..., e amt ) P C[x, x ±1 1,..., x ±1 m ] and {a 1,..., a m } a set of real and imaginary algebraic numbers that is linearly independent over Q.
63 The Real Case Any exponential polynomial f (t) can be written with f (t) = P(t, e a 1t,..., e amt ) P C[x, x ±1 1,..., x ±1 m ] and {a 1,..., a m } a set of real and imaginary algebraic numbers that is linearly independent over Q. Lemma Assuming Schanuel s conjecture, if f is real valued and P is irreducible then f has no tangential zeros.
64 The Real Case Any exponential polynomial f (t) can be written with f (t) = P(t, e a 1t,..., e amt ) P C[x, x ±1 1,..., x ±1 m ] and {a 1,..., a m } a set of real and imaginary algebraic numbers that is linearly independent over Q. Lemma Assuming Schanuel s conjecture, if f is real valued and P is irreducible then f has no tangential zeros. Complex case requires some new ideas...
65 The Unbounded Case there are known unknowns; that is to say we know there are some things we do not know.
66 Continued Fractions Finite continued fractions: [3, 7, 15, 1, 292] =
67 Continued Fractions Finite continued fractions: [3, 7, 15, 1, 292] = =
68 Continued Fractions Finite continued fractions: [3, 7, 15, 1, 292] = 3 + Infinite continued fractions: = [a 0, a 1, a 2, a 3,...] = a a a 2 + a 3 +
69 Real Algebraic Numbers Theorem The continued fraction expansion of a real quadratic irrational number is periodic.
70 Real Algebraic Numbers Theorem The continued fraction expansion of a real quadratic irrational number is periodic. 389 = [19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38,...]
71 Real Algebraic Numbers Theorem The continued fraction expansion of a real quadratic irrational number is periodic. 389 = [19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38,...] What about numbers of degree 3?
72 Real Algebraic Numbers Theorem The continued fraction expansion of a real quadratic irrational number is periodic. 389 = [19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38,...] What about numbers of degree 3? 3 2 = [1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1 3, 4, 1, 1, 2, 14, 3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1,...]
73 Real Algebraic Numbers Theorem The continued fraction expansion of a real quadratic irrational number is periodic. 389 = [19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38,...] What about numbers of degree 3? 3 2 = [1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1 3, 4, 1, 1, 2, 14, 3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1,...] Lang and Trotter: no significant departure from random behaviour
74 An Open Problem [... ] no continued fraction development of an algebraic number of higher degree than the second is known. It is not even known if such a development has bounded elements. A. Khinchin
75 An Open Problem [... ] no continued fraction development of an algebraic number of higher degree than the second is known. It is not even known if such a development has bounded elements. A. Khinchin Is there an algebraic number of degree higher than two whose simple continued fraction has unbounded partial quotients? Does every such number have unbounded partial quotients? R. K. Guy, 2004
76 A Mathematical Obstacle at Dimension 9 Given x = [a 0, a 1, a 2,...], define S(x) = sup n N a n.
77 A Mathematical Obstacle at Dimension 9 Given x = [a 0, a 1, a 2,...], define S(x) = sup n N a n. Theorem (Chonev, Ouaknine, W., 2015) If the ZERO PROBLEM is decidable at dimension 9 then {x R A : S(x) < } is recursively enumerable.
78 A Mathematical Obstacle at Dimension 9 Given x = [a 0, a 1, a 2,...], define S(x) = sup n N a n. Theorem (Chonev, Ouaknine, W., 2015) If the ZERO PROBLEM is decidable at dimension 9 then {x R A : S(x) < } is recursively enumerable. Remark Perhaps this set is recursive it may even be or R A. However proving recursive enumerability would be a significant achievement.
79 Diophantine Approximation How well can one approximate a real number x with rationals? x m n
80 Diophantine Approximation How well can one approximate a real number x with rationals? x m n Theorem (Dirichlet 1842) There are infinitely many integers m, n such that x m < 1 n n 2.
81 Diophantine Approximation How well can one approximate a real number x with rationals? x m n Theorem (Dirichlet 1842) There are infinitely many integers m, n such that x m < 1 n n 2. S(x) < if and only if there exists ε > 0 such that x m < ε n n 2 has no solutions.
82 Diophantine Approximation How well can one approximate a real number x with rationals? x m n Theorem (Dirichlet 1842) There are infinitely many integers m, n such that x m < 1 n n 2. S(x) < if and only if there exists ε > 0 such that x m < ε n n 2 has no solutions. Relate this to the existence of zeros of order-9 exponential polynomial f (t) with terms e ixt and e it.
83 The ZERO Problem ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0?
84 The ZERO Problem ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0? Theorem In dimension 8 or less, ZERO reduces to BOUNDED-ZERO.
85 The ZERO Problem ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0? Theorem In dimension 8 or less, ZERO reduces to BOUNDED-ZERO. In the limit f is either never zero or infinitely often zero, and we decide which is the case.
86 The ZERO Problem ZERO Problem Instance: f Question: Is there t R 0 such that f (t) = 0? Theorem In dimension 8 or less, ZERO reduces to BOUNDED-ZERO. In the limit f is either never zero or infinitely often zero, and we decide which is the case. Diophantine-approximation bounds play a key role in the proof.
87 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t
88 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
89 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
90 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
91 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
92 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
93 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
94 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
95 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
96 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
97 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
98 Illustrative Example Consider the exponential polynomial f (t) = cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
99 Illustrative Example Consider the exponential polynomial f (t) = 2 + cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2
100 Illustrative Example Consider the exponential polynomial f (t) = 2 + cos(t + ϕ 1 ) + cos( 2t + ϕ 2 ) e t Kronecker s Theorem: (t, 2t) mod 2π is dense in [0, 2π] 2 Baker s Theorem: ( t, ) 2t (π ϕ 1, π ϕ 2 ) 1 poly(t)
101 Conclusion and Perspectives
102 The Discrete Case A linear recurrence sequence is a sequence u 0, u 1, u 2,... of integers such that there exist constants a 1,..., a k, such that for all n 0. u n+k = a 1 u n+k 1 + a 2 u n+k a k u n
103 The Discrete Case A linear recurrence sequence is a sequence u 0, u 1, u 2,... of integers such that there exist constants a 1,..., a k, such that for all n 0. u n+k = a 1 u n+k 1 + a 2 u n+k a k u n Theorem (Skolem 1934; Mahler 1935, 1956; Lech 1953) The set of zeros of a linear recurrence sequence is semi-linear: {n : u n = 0} = F A 1... A l where F is finite and each A i is a full arithmetic progression.
104 The Discrete Case A linear recurrence sequence is a sequence u 0, u 1, u 2,... of integers such that there exist constants a 1,..., a k, such that for all n 0. u n+k = a 1 u n+k 1 + a 2 u n+k a k u n Theorem (Skolem 1934; Mahler 1935, 1956; Lech 1953) The set of zeros of a linear recurrence sequence is semi-linear: {n : u n = 0} = F A 1... A l where F is finite and each A i is a full arithmetic progression. Theorem (Berstel and Mignotte 1976) In Skolem-Mahler-Lech, the infinite part (arithmetic progressions A 1,..., A l ) is fully constructive.
105 The Skolem Problem Skolem Problem Does n such that u n = 0?
106 The Skolem Problem Skolem Problem Does n such that u n = 0? It is faintly outrageous that this problem is still open; it is saying that we do not know how to decide the Halting Problem even for linear automata! Terence Tao
107 The Skolem Problem Skolem Problem Does n such that u n = 0? It is faintly outrageous that this problem is still open; it is saying that we do not know how to decide the Halting Problem even for linear automata! Terence Tao... a mathematical embarrassment... Richard Lipton
108 Wrapping Things Up Continuous Skolem Problem Does t such that f (t) = 0?
109 Wrapping Things Up Continuous Skolem Problem Does t such that f (t) = 0? Even the bounded problem is hard (apparently).
110 Wrapping Things Up Continuous Skolem Problem Does t such that f (t) = 0? Even the bounded problem is hard (apparently). Formidable mathematical obstacle at dimension 9 in the unbounded case.
111 Wrapping Things Up Continuous Skolem Problem Does t such that f (t) = 0? Even the bounded problem is hard (apparently). Formidable mathematical obstacle at dimension 9 in the unbounded case. The infinite-zeros problem is also hard.
112 Wrapping Things Up Continuous Skolem Problem Does t such that f (t) = 0? Even the bounded problem is hard (apparently). Formidable mathematical obstacle at dimension 9 in the unbounded case. The infinite-zeros problem is also hard. Diophantine-approximation techniques unavoidable.
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