From Intervals to Taylor Models Validated Symbolic-Numeric...

From Intervals to Taylor Models

Validated Symbolic-Numeric Computation

Markus NeherInstitut fur Angewandte Mathematik

Universitat Karlsruhe

I Basic Concepts

II Dependency

III Wrapping

IV ODEs

V Challenges

Markus Neher, Universitat Karlsruhe Taylor Models / ACA 03, Raleigh 1

Basic Concepts


Taylor Model

• X ⊂ Rν, f : X → R, f ∈ Cn+1, x0 ∈ X.

• Taylor’s Theorem: f(x) = Pn,f(x− x0) + Rn,f(x− x0) for x ∈ X.

(Pn,f Taylor polynomial, Rn,f remainder term)

Definition:

• An interval In,f is called a remainder bound of order n of f on X

⇔ ∀x ∈ X : Rn,f(x− x0) ∈ In,f .

• Tn,f := (Pn,f , In,f) is called a Taylor model of order n of f

⇔ ∀x ∈ X: f(x) ∈ Pn,f(x− x0) + In,f .

(Moore 1962, Krueckeberg 1968, Kaucher & Miranker 1984, Berz et al. 1990s-today)


Addition and Multiplication

• Tn,f±g := Tn,f ± Tn,g := (Pn,f ± Pn,g, In,f ± In,g),

• Tn,α·f := α · Tn,f := (α · Pn,f , α · In,f) (α ∈ R),

• Tn,f ·g := Tn,f · Tn,g := (Pn,f ·g, In,f ·g),

where

– Pn,f(x− x0) · Pn,g(x− x0) = Pn,f ·g(x− x0) + Pe(x− x0),

– Pe(x− x0) ∈ IPe, Pn,f(x− x0) ∈ IPn,f, Pn,g(x− x0) ∈ IPn,g,

– f(x) · g(x) ∈ Pn,f ·g(x− x0) + IPe + IPn,fIn,g + In,f

(IPn,g + In,g

)︸︷︷︸=:In,f ·g

.

(x ∈ X)


Numerical Example

ν = 1, n = 3, x0 = 0, x ∈ [0, 1]:

f(x) = x : Tn,f = (Pn,f , In,f) = (x, [0, 0]),

g(x) = ex: Tn,g = (Pn,g, In,g) = (1 + x +x2

2+

x3

6, [0,

e

24]).

x · ex ∈ (x + [0, 0]) · (1 + x +x2

2+

x3

6+ [0,

e

24])

= x + x2 +x3

2︸︷︷︸=:Pn,f ·g

+x4

6+ x · [0,

e

24]︸︷︷︸

⊆In,f ·g

,

x4

6∈ [0,

16], x ∈ [0, 1],

⇒ In,f ·g = [0,16] + [0, 1] · [0,

e

24] = [0,

4 + e

24].


Polynomials

Corollary: If Tn,f is a Taylor model for f , then

Tn,P

aif i

is a Taylor model for ∑aif

i.

Tn,P

aif i can be computed using only the above TM operations.


Standard Functions

• Taylor model: Tn,f = (Pn,f , In,f) of f : X → R.

• Standard function: ϕ ∈ {exp, ln, sin, cos, . . .}.

• Taylor model for ϕ(f):

– Special treatment of the constant part in Pn,f .

– Evaluate the Taylor polynomial of ϕ for the non-constant part of Tn,f .


Taylor Model for Exponential Function

x ∈ X, ϕ(f(x)) = exp(f(x)), c := f(x0), h(x) := f(x)− c:

Pn,f(x− x0) = Pn,h(x− x0) + c, In,h = In,f ,

exp(f(x)) = exp(c + h(x)) = exp(c) · exp(h(x))

= exp(c) ·{

1 + h(x) +12(h(x))2 + . . . +

1n!

(h(x))n

}

+exp(c) · 1(n + 1)!

(h(x))n+1 exp(θ · h(x)), 0 < θ < 1

=: ϕ(x) + R.


Taylor Model for Exponential Function

Evaluation of ϕ(x): ϕ(x) ∈ Pn,ϕ(x− x0) + In,ϕ (x ∈ X).

Evaluation of R: First compute IPn,hsuch that

Pn,h(x− x0) ∈ IPn,h(x ∈ X).

Then we have

(h(x))n+1 exp(θ · h(x)) ∈ (IPn,h+ In,h)n+1 exp

([0, 1] · (IPn,h

+ In,h))

︸︷︷︸=:J

,

so thatexp(f(x)) ∈ Pn,ϕ(x− x0) + In,ϕ,

where

Pn,ϕ(x− x0) = Pn,ϕ(x− x0), In,ϕ := In,ϕ + exp(c)1

(n + 1)!· J, x ∈ X.


Taylor Models for Other Standard Functions

x ∈ X, c := f(x0), h(x) := f(x)− c:

• ln(f(x)) = ln(c + h(x)) = ln(c) + ln(1 +1ch(x))

= ln c +1ch(x) + · · ·+ (−1)n+11

n(1ch(x))n

+ (−1)n+2 1n + 1

(1ch(x))n+1 1

(1 + θh(x)/c)n+1, 0 < θ < 1

• 1f(x)

=1c

11 + h(x)/c

=1c

{1− 1

ch(x) + · · ·+ (−1)n(

1ch(x))n

}+ R

• cos(f(x)) = cos c cos(h(x))− sin c sin(h(x))

• . . .


Dependency


IA vs. TMA

• f(x) = x2 + cos x + sinx− ex, x ∈ X = [0, 1].

• Direct IA:

f(x) ∈ F (X) = X2 + cos X + sinX − eX

= [0, 1] + [cos 1, 1] + [0, sin 1]− [1, e] ≈ [−2.178, 1.842].

• Mean Value Form:

f(x) ∈ f(12) + F ′(X) ∗ (X − 1

2)

= f(12) + (2 ∗X − sinX + cos X − eX) ∗ [−1

2,12]

⊆ [−0.042,−0.041] + [−3.020, 0] ∗ [−0.5, 0.5] = [−1.552, 1.469].


IA vs. TMA

• TMA:

f(x) = x2 + 1− x2

2+

ξ41

24+ x− x3

6+

ξ42

24− 1− x− x2

2− x3

6− ξ4

3

24

= −x3

3+

ξ41

24+

ξ42

24− ξ4

3

24

∈ [−0.334, 0] + [0, 0.042] + [0, 0.042]− [0, 0.042] = [−0.376, 0.082].

• Range: [1 + cos 1 + sin 1− e, 0] ∈ [−0.337, 0].


Wrapping


Example

• f(x, y) =

(x + sin(π

2y)

cos(π2x)− y

), x, y ∈ [0, 1]; A =

1√2

(1 −1

1 1

).

A f , A (A f) =?

• IA:

(X

Y

)=

([0, 2]

[−1, 1]

),

A

(X

Y

)= 1√

2

([−1, 3]

[−1, 3]

), A (A

(X

Y

)) = 1

2

([−4, 4]

[−2, 6]

)=

([−2, 2]

[−1, 3]

).


Example

• TMA: T4,f =

x + π2y − π3y3

48 + [0,π5]3840

1− y − π2x2

8 + π4x4

384 −[0,π5]3840

, x, y ∈ [0, 1],

A T4,f = 1√2

−1 + x + (1 + π2)y + π2x2

8 − π3y3

48 − π4x4

384 + [0,π5]1920

1 + x + (π2 − 1)y − π2x2

8 − π3y3

48 + π4x4

384 + [−π5,π5]3840

,

A (A T4,f) =

−1 + y + π2x2

8 − π4x4

384 + [−π5,3π5]7680

x + π2y − π3y3

48 + [−π5,3π5]7680

, x, y ∈ [0, 1].


IA vs. TMA


ODEs


Introduction

Smooth IVP: u′ = f(t, u), u(t0) = u0.

( u = (u1, . . . , un), f = (f1, . . . , fn) ).

Moore’s enclosure method:

– Automatic computation of Taylor coefficients.

– Interval iteration: For j := 0, 1, . . . :

A priori enclosure: [uj+1] ⊇ u(t) for all t ∈ [tj, tj+1] (”Algorithm I”).

Truncation error: [zj+1] :=h(m+1)

(m + 1)!f

(m)([tj, tj+1], [buj+1]).

u(tj+1) ∈ [uj+1] := [uj] +mX

k=1

hk

k!f

(k−1)(tj, [uj]) + [zj+1]

(”Algorithm II”).


Piecewise constant a priori enclosure

0

1

2

3

4

1 2 3 4 5 6


A priori Enclosures

– Moore:

Fixed point iteration with constant enclosures:

For some h > 0 determine interval [u1] such that

u0 + [0, h] · f([t0, t1], [u1]) ⊆ [u1]

⇒ Step size restrictions: Explicit Euler steps.

– Improvements suggested by Lohner, Corliss & Rihm,

Nedialkov & Jackson, . . . .

– Berz & Makino: High order TM enclosure by FP iteration.


Modifications of Algorithm II

– Moore, Eijgenraam, Lohner, Rihm, Kuehn, Nedialkov & Jackson, . . . :

Wrapping effect

– Nedialkov & Jackson: Hermite–Obreshkov–Method

– Rihm: Implicit methods

– Petras & Hartmann: Runge–Kutta–Methods

– Berz & Makino: Taylor models

Taylor expansion with respect to initial values and parameters


IA vs. TMA

Example: u′ = sin v,

v′ = cos u;

u(0) ∈ [−0.1, 0.1], v(0) ∈ [−0.1, 0.1].

• IA: Interval enclosure at each grid point in time.

u(0) = [−0.1, 0.1], u(1) = [0.261, 0.654], u(4) = [1.027, 5.158].

• TMA: TM in initial values at each grid point in time.

u(0) = 0.1x, v(0) = 0.1y, x, y ∈ [−1, 1].

u(1) = 1.2e-4xy2 + 4.6e-1 + 9.7e-2x− 1.9e-3x2 + 5.0e-6x3 − 1.1e-4y3

+8.3e-2y − 8.6e-4xy + 1.4e-4x2y − 2.4e-3y2 + [−9.3e-7, 6.6e-7]⊆ [0.271, 0.634] (x, y ∈ [−1, 1]).

u(4) = P4(x, y) + [−6.3e-4, 3.9e-4] ⊆ [2.936, 3.220].


Challenges


Challenges

• Multivariate Taylor models

No. of Taylor coefficients: N(n, ν) =(

n + νν

)=

(n + ν)!n! ν!

Example: N(10, 4) = 1001, N(20, 6) = 230230.

• Memory management for sparse Taylor models

• Software


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