Numerical Integration with the double exponential method

Size: px

Start display at page:

Download "Numerical Integration with the double exponential method"

Christian Chambers
6 years ago
Views:

1 Numerical Integration with the double exponential method Pascal Molin Université Paris 7 December

2 Numerical integration Number-theory context, want fast quadrature high precision ( digits) rational recognition algebraic dependencies, LLL robust scheme little knowledge of the integrand yet can assume holomorphicity around path rigorous evaluation non-vanishing, inequalities algebraic

3 Digits matter Closed form is not always the best answer maple > int(x*sin(x)/(1+cos(x)^2), x = 0.. Pi); ( ) ( i dilog dilog ) ( i dilog 2 + dilog ) i 2 1 ( ) ( 1 + i + 2 ( + arctan 1 + ) ) 1 2 π 2 1 ( ( ) ) 1 ( arctan 2 1 π + i ln 1 + ) 2 π + 1/2 π 2

4 Digits matter Closed form is not always the best answer maple > int(x*sin(x)/(1+cos(x)^2), x = 0.. Pi); gp > intnum (x=0,pi,x* sin (x) /(1+ cos (x) ^2) ) time = 13 ms. % 1 =

5 Digits matter Closed form is not always the best answer maple > int(x*sin(x)/(1+cos(x)^2), x = 0.. Pi); gp > intnum (x=0,pi,x* sin (x) /(1+ cos (x) ^2) ) time = 13 ms. % 1 = PSLQ algorithm (integer relations with precomputed values) gp > Pi ^2/4 % 1 =

6 Classical quadrature schemes Trapeze sums better than rectangles! error = O(n 2 ) for n evaluations Simpson, Newton-Cotes degree d spline interpolation on each subinterval error = O(n 2d ) for nd evaluations omberg convergence acceleration error = O(n 2k ) for 2 k/2 n evaluations (k steps) Gauss-Legendre interpolation at n chosen points, exact on degree 2n 1 polynomials error = O(2 2n )

7 DE method [Takahasi&Mori,1973] trapezoidal method over for doubly-exponential decay g(x) Me e x h 1 D log D heuristic : D digits with O(D log D) points changes of variable to obtain DE decay g(x) Me e x f (x) Me x, x f (x) M(1 + x ) α, x t = sinh(u) x = sinh(t) x = e t e t t = tanh(u) 0 f (x) Me x, x f (x) M, x [ 1, 1]

8 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast gp > \ p100 gp > 4* intnum (x=0,1,x* log (x +1) ) time = 14 ms. % 2 = gp > 4* intnumgauss (x=0,1,x* log (x +1) ) time = 19 ms. % 3 =

9 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast gp > \p 1000 realprecision = 1001 significant digits (1000 digits displayed ) gp > intnum (x=0,1,x* log (x +1) ); time = 3,246 ms. gp > intnumgauss (x=0,1,x* log (x +1) ); *** Warning : increasing stack size to *** Warning : increasing stack size to *** Warning : increasing stack size to *** Warning : increasing stack size to time = 30,423 ms.

10 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate gp > intnum (x=0,1,x* log (x)) time = 14 ms. % 1 = gp > intnumgauss (x=0,1,x* log (x)) time = 19 ms. % 2 =

11 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate gp > intnum (x=0,1, sqrt (1 -x ^2) ) * 4/ Pi time = 5 ms. % 1 =

12 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate gp > intnum (x=0,1, sqrt (1 -x ^2) ) * 4/ Pi time = 5 ms. % 1 = gp > intnumgauss (x=0,1, sqrt (1 -x ^2) ) * 4/ Pi time = 18 ms. % 1 =

13 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate robust gp > intnum (x=0,1, log (x) ^2) time = 14 ms. % 1 =

14 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate robust gp > intnum (x=0,1, log (x) ^2) time = 14 ms. % 1 = gp > intnumgauss (x=0,1, log (x) ^2) time = 20 ms. % 1 =

15 Empirics [Bailey 2005] best quadrature scheme for number theory (speed + accuracy) fast accurate robust gp > intnum (x = 0, [oo, -I], x ^2* sin (x)) time = 29 ms. % 1 =

16 Bugs? singularities near path gp > \ p100 gp > intnum (x=-oo,oo,1/(1+ x ^2) ) / Pi time = 16 ms. % 2 =

17 Bugs? singularities near path gp > \ p100 gp > intnum (x=-oo,oo,1/(1+ x ^2) ) / Pi time = 16 ms. % 2 = gp > intnum (t=-oo,oo,1/(1+( t +10) ^2) ) / Pi time = 16 ms. % 1 =

18 Bugs? singularities near path + = lim X X X gp > intnum (x=[ -oo,2],[ oo,2], exp (-x ^2) ) / sqrt (Pi) time = 14 ms. % 1 =

19 Bugs? singularities near path + = lim X X X gp > intnum (x=[ -oo,2],[ oo,2], exp (-x ^2) ) / sqrt (Pi) time = 14 ms. % 1 = gp > exp ( -20^2) % 1 = E -174

20 Bugs? singularities near path + = lim X X X gp > intnum (x=[ -oo,2],[ oo,2], exp (-x ^2) ) / sqrt (Pi) time = 14 ms. % 1 = gp > intnum (x= -20,20, exp (-x ^2) ) / sqrt (Pi) time = 12 ms. % 1 =

21 Bugs? singularities near path + = lim X X X gp > intnum (x=[ -oo,2],[ oo,2], exp (-x ^2) ) / sqrt (Pi) time = 14 ms. % 1 = gp > intnum (x= -50,50, exp (-x ^2) ) / sqrt (Pi) time = 12 ms. % 1 =

22 Goal : rigorous convergence Theorem Let f : [ 1, 1] C such that f has holomorphic continuation to B(0, 2) = { z < 2}. Then 1 1 f (x) dx for x k, w k given by where h = n k= n w k f (x k ) x k = tanh(sinh(kh)) and w k = log(n+2) n+2+2/7 log(n). 8 sup f 2 10n log(n+2) (1) B(0,2) h cosh(kh) cosh(sinh(kh)) 2 (2)

23 Comparison with Gauss Theorem Let f : [ 1, 1] C such that f has holomorphic continuation to B(0, 2) = { z < 2}. Then 1 1 f (x) dx for x k, w k given by n k= n ( w k f (x k ) π sup f D(0,2) ) 2 2n (3) P n (x k ) = 0 and w k = where P n are Legendre polynomials { P 0 (x) = 0, P 1 (x) = 1 2 (n + 1)P n(x k )P n+1 (x k ) (k + 1)P k+1 (x) = (2k + 1)xP k (x) kp k 1 (x) (4) (5)

24 Comparison with Gauss Integral of usual functions to D digits. method Gauss DE # points D D log D cost points D 3 log D D 2 log D cost integral D 2 log 2 D D 2 log 3 D many nice integrals single nice integral unknown behaviour

25 Theory

26 Poisson formula g(x) = O(x 2 ), g C 2 (), ĝ(x ) = e 2iπXt g(t) dt h k >n g(kh) + h } {{ } 1 g(k) = ĝ(k) k Z k Z n k= n g(kh) = g + k Z ĝ( k h ) }{{} 2 nh h nh

27 Poisson formula g(x) = O(x 2 ), g C 2 (), ĝ(x ) = e 2iπXt g(t) dt h k >n g(kh) + h } {{ } 1 g(k) = ĝ(k) k Z k Z n k= n g(kh) = g + k Z ĝ( k h ) }{{} 2 Errors 1(nh) and 2( 1 h ). Summation interesting when both g and ĝ vanish quickly (exponentially).

28 g(x) = sin x x, ĝ(x) = 1 [ 1 2π, 1 2π ] k sin(k) k g(x) = e x2, ĝ(x) = e πx2 = Poisson summation sin(x) x dt 1(nh) + 2(h) e D n e D 1(nh) + 2(h) e D n D π g(x) = e a cosh(x), ĝ(x) = 2K i2πx (a) e π2 x x 1(nh) + 2(h) e D n D log(d) π 2

29 g(x) = sin x x, ĝ(x) = 1 [ 1 2π, 1 2π ] k sin(mk) k = g(x) = e x2, ĝ(x) = e πx2 sin(mx) x Poisson summation dt for m = 1, (nh) + 2(h) e D n e D 1(nh) + 2(h) e D n D π g(x) = e a cosh(x), ĝ(x) = 2K i2πx (a) e π2 x x 1(nh) + 2(h) e D n D log(d) π 2

30 Bad news I g and ĝ cannot have both arbitrary decay. Theorem (Uncertainty principle) Let g : with g = 0. { g(x) = O(e α 1 x β1 ) If ĝ(x) = O(e α 2 x β2 ) then 1 β β 2 1. In particular e D n D 1 β β 2. Optimal case : the gaussian g(x) = exp( σx 2 ).

31 Paley-Wiener theory The decay of ĝ corresponds to the regularity of g : ĝ has finite support g entire of order 1 ĝ has more than exponential decay g entire g holomorphic on a strip + i[ t, t] ĝ(x) = O(e 2πtx )

32 Control of ĝ Theorem If g : C has holomorphic continuation to a strip τ = + i[ τ, τ] such that g( iτ) 1 + g( + iτ) 1 = M 2 (τ) < ; g(x ± it) x ± 0 uniformly on t < τ ; then ĝ(x) M 2 (τ)e 2πτ x.

33 Control of ĝ Theorem If g : C has holomorphic continuation to a strip τ = + i[ τ, τ] such that g( iτ) 1 + g( + iτ) 1 = M 2 (τ) < ; g(x ± it) x ± 0 uniformly on t < τ ; then ĝ(x) M 2 (τ)e 2πτ x. e 2πxτ e 2iπxt g(t + iτ) e 2iπxt g(t) iτ

34 Control of ĝ Theorem If g : C has holomorphic continuation to a strip τ = + i[ τ, τ] such that g( iτ) 1 + g( + iτ) 1 = M 2 (τ) < ; g(x ± it) x ± 0 uniformly on t < τ ; then ĝ(x) M 2 (τ)e 2πτ x. Goal : find best decay for g such that it remains holomorphic on a strip.

35 Bad news II Théorème If g : C satisfies g(x) M 1 e αe β x on (DE) ; g has bounded holomorphic continuation on a strip τ with βτ > π 2 then g = 0. In particular cannot expect more than DE decay for g for DE functions : e D n D log(d) π 2

36 Good news DE decay extends near the axis. Theorem (Phrägmen-Lindelöf) If g is holomorphic on a strip τ and satisfies g = O(e αe β x ) sur ; g M sur τ ; then for all t < τ, g(x ± it) Me α te β x, with α t = α(cos(βt) sin(βt) tan(βτ) ). easy to control ĝ from DE hypothesis

37 Main theorem Let g : such that 1 g has analytic continuation to a strip τ = + i] τ, τ[ ; 2 g(x) M 1 e αe β x on with α, β > 0 ; 3 g(z) M 2 on τ ; then for any D > 0 there are explicit values n, h with n (D + log M 2) log(d + log M 1 ) 2πτβ and h 2πτ D + log M 2 such that g h n k= n g(kh) e D.

38 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ τ iτ I ϕ( τ ) τ f holomorphic on τ.

39 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 f (x) M 1 e α x β τ iτ ϕ(t) = sinh(t) = et e t ϕ (t) = cosh(t) 2

40 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ sin(τ) τ τ iτ ϕ(t) = sinh(t) = et e t ϕ (t) = cosh(t) 2

41 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ f (x) M 1 e α x β sin(τ) τ f (z) M 2 e A z γ τ iτ ϕ(t) = sinh(sinh(t)) ϕ (t) = cosh(t) cosh(sinh(t))

42 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ iτ f (x) M1 1+ x α ϕ(t) = sinh(sinh(t)) ϕ (t) = cosh(t) cosh(sinh(t))

43 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ iτ ϕ(t) = tanh(λ sinh(t)) ϕ (t) = λ cosh(t) cosh 2 (λ sinh(t))

44 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ Ym(τ) M 2 f (z) (1+ z ) 1+γ τ iτ X m(τ) f (x) M1 (1+ x ) α ϕ(t) = tanh(λ sinh(t)) ϕ (t) = λ cosh(t) cosh 2 (λ sinh(t))

45 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 a f M 1 b τ iτ ϕ(t) = e t αe βt ϕ (t) = (1 + αβe βt )ϕ(t)

46 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ iτ ϕ(t) = e t αe βt ϕ (t) = (1 + αβe βt )ϕ(t)

47 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 Y m(τ) f M 2 f M 1 a b X m(τ) τ iτ

48 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ iτ 0 f M 1 e αxβ

49 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ iτ

50 DE method : regularity hypothesis Problem : DE setting : f, 1, 2 g = f (ϕ).ϕ, 1, 2 τ τ f (z) M 2 e λ z 0 τ iτ f M 1 e αxβ Xm

51 Summary τ Ym(τ) M 2 f (z) (1+ z ) 1+γ f (x) M 1 e α x β sin(τ) τ f (z) M 2 e A z γ τ X m(τ) f (x) M1 (1+ x ) α τ f (z) M 2 e λ z f M 2 Ym(τ) τ 0 f M 1 a f M 1 e αxβ b Xm(τ) Xm Under these hypothesis, DE method rigorously evaluates f to D digits using O( D log D 2πτ ) points.

52 Use cases

53 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ τ

54 Well placed poles

55 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > τ

56 Small tau value

57 Small tau value? \ p1000 realprecision = 1001 significant digits (1000 digits displayed )? f(z) =1/(1+ z ^2) ;? D =1000* log (10) ;? show ( asinh ( asinh (10+ I))) % 6 = * I? tau =0.03; h= pas_h_shsh (tau,1,1,d); show (h) % 7 = E -5? n= ceil ( asinh ( asinh (2*10^ ) )/h); show (n) %8 = ? Pi - integration_shsh (z->f(z -10),h,n,1) % 9 = E -1001

58 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 τ

59 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 e z2 dz sh(t) pole 1+e τ, τ iz 1 τ

60 Line of poles

61 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 e z2 dz sh(t) pole 1+e τ, τ iz 1 Γ(1 + iz) dz sh(t) Γ for τ < 0 τ

62 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 e z2 dz sh(t) pole 1+e τ, τ iz 1 Γ(1 + iz) dz sh(t) Γ for τ < 0 e ch(z)+i ch(2z) dz t huge L 1 ( ± iτ) sin(z) dz z sh(sh(t)) sin(ie t ) explodes τ

63 Extensions

64 Meromorphic functions If has a pole ρ τ, with residue r ρ : ĝ(x ) = ±iτ For f ĝ( k h ) 2iπ k Z ρ Z τ where ε z is the sign of Im z. e 2iπτXt g(t) dt + 2iπe 2iπX ρ r ρ ϕ 1 (ρ) ε z r ρ e ε z 2iπz/h 1 2M 1 e 2πτ/h 1

65 Taking residues into account? \ p1000 realprecision = 1001 significant digits (1000 digits displayed )? f(z) =1/(1+ z ^2) ;? tau =Pi /2.2;? h =2* Pi*tau /(1000* log (10) + log (2/ cos ( tau ))); show (h) % 6 = ? n= ceil ( asinh ( asinh (2*10^1000) )/h) %7 = 2168? Pi - integration_shsh (f,h,n,1) %8 = 0.E -1001? diff =Pi - integration_shsh (z->f(z -15),h,n,1) ; show ( diff ) % 9 = E -14? ho =[15+I,15 -I]; es =[ -I/2,I /2];? diff + poles_shsh (ho,es,h) % 11 = E E -1014* I

66 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 e z2 dz sh(t) pole 1+e τ, τ iz 1 Γ(1 + iz) dz sh(t) Γ for τ < 0 e ch(z)+i ch(2z) dz t huge L 1 ( ± iτ) sin(z) dz z sh(sh(t)) sin(ie t ) explodes τ

67 Shifting path 0 0 θ δ gamma function on vertical lines

68 Shifting path θ gamma function on vertical lines δ integrate on the middle of a nice strip

69 A list of integrals z dz sh(sh(t)) poles / 1+z 2 τ dz sh(sh(t)) pole 1+(z 10) τ, τ > dz 100 th( π 1+z 2 2 sh(t)) pole τ, τ > 0.06 e z2 dz sh(t) pole 1+e τ, τ iz 1 Γ(1 + iz) dz sh(t) Γ for τ < 0 e ch(z)+i ch(2z) dz t huge L 1 ( ± iτ) sin(z) dz z sh(sh(t)) sin(ie t ) explodes τ

70 Examples

71 erfc(x) = i e z2 π e x2 z ix dz Error function

72 ( erfc(x) = hx 1 π e x2 erfc(x) = i π e x2 x k 1 comparison with MPF e k2 h 2 k 2 h 2 + x 2 Error function e z2 z ix dz ) e 2πx/h + O(e π2 /h 2 ) gp/mpfr 4 MPF 50 times faster precision integration 400 times faster

73 Incomplete gamma function Γ inc (s, x) = x ts e t dt t for s C and x /. X 0

74 Incomplete gamma function Γ inc (s, x) = x ts e t dt t for s C and x /. X τ 0

75 Incomplete gamma function Γ inc (s, x) = x ts e t dt t for s C and x /. X τ 0

76 Incomplete gamma function Γ inc (s, x) = x ts e t dt t for s C and x /. X 0

77 Incomplete gamma function Γ inc (s, x) = x ts e t dt t for s C and x /. t s 1 e t M(X, τ, s) explicite τ X θ 0 Γ inc (s, X ) to prec D using n 2D log D π 2 evaluations.

78 Conclusion DE integration is not always the best method (better convergence for Gauss, provided big precomputations and well known integrand) still very easy to implement and nice convergence rigorous integration possible with basic bounds on the integrand otherwise efficient heuristic procedures (guess the error term from first iterations)

79 eferences Hidetosi Takahasi and Masatake Mori. Double exponential formulas for numerical integration. Kyoto University. esearch Institute for Mathematical Sciences. Publications, 9 : , Masatake Mori. Discovery of the double exponential transformation and its developments. Publ. es. Inst. Math. Sci., 41(4) : , David H. Bailey, Karthik Jeyabalan, and Xiaoye S. Li. A comparison of three high-precision quadrature schemes. Experiment. Math., 14(3) : , Pascal Molin. Numerical integration and L functions computations. Theses, Université Sciences et Technologies - Bordeaux I, October 2010.

Experimental mathematics and integration

Experimental mathematics and integration David H. Bailey http://www.davidhbailey.com Lawrence Berkeley National Laboratory (retired) Computer Science Department, University of California, Davis October