certi ed solutions of polynomials with uncertainties · certi ed solutions of polynomials with...
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Certied solutions of polynomials with uncertainties
David Daney, Jean Pierre Merlet, Odile PourtallierCoprin team
INRIA Sophia Antipolis Méditerranée
December 5th, 2007
MACIS 07, Rocquencourt
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1 I. Introduction
Problem
P (z) =dX
i=0
[ai]zi,
[ai] = [ai, ai] real intervals.
Find
ΩR = z ∈ R, ∃ai ∈ [ai], i = 0 · · · d,dX
i=0
aizi= 0,
ΩC = z ∈ C, ∃ai ∈ [ai], i = 0 · · · d,dX
i=0
aizi= 0.
Certied solutions of polynomials with uncertainties O.Pourtallier
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2 I. Introduction
Outline
Motivation
Interval analysis
Edge theorem
Exclusion tests and ltering procedures
Perspectives
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3 I. Introduction
Motivation
θ, X, Y , determined by solving some real polynomial of degree 6,
6Xi=0
ai(l1, l2, l3, A1, A2, A3)γi= 0,
θ = fθ(γ), X = fX(l1, l2, l3, γ), Y = fY (l1, l2, l3, γ)
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4 I. Introduction
Direct problem : li ∈ [li, li], determine the workspace Wl1,l2,l3of the robot ?
Design problem : Determine intervals [li, li] such that Wl1,l2,l3contains a given workspace
W .
Solving the problem :
Ranges for li possibly large to be taken into account.
All possible uncertainties (hopefully small) to be taken into account
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5 I. Introduction
Certication :
For some applications (e.g. medical) there
is a strong need of certication
→ Interval Analysis techniques
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6 II. Interval Analysis
Interval Analysis
Interval Arithmetic
[a, b] + [c, d] = [a + c, b + d],
[a, b]− [c, d] = [a− d, b − d],
[a, b] [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)], · · ·
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7 II. Interval Analysis
Evaluation
f(x) = x3 + x2 + x, f(x), x ∈ [X] = [−1, 1] = [−1, 3]
[X]3+ [X]2 + [X] = [−1, 1] + [0, 1] + [−1, 1] = [−2, 3]
Overestimation
[X]2([X] + 1) + [X]) = [−1, 3]
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8 II. Interval Analysis
Bisection Algorithm
exclusion tests,
Filtering procedures,
Uniqueness tests.
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9 II. Interval Analysis
Filtering (2B) : Example
Solve 1 + x + x2 + x3 = 0 in [−1, 1]
x = −1− x2 − x
3
[−1, 1] −1− [−1, 1]2 − [−1, 1]
3= [−3, 0],
Necessarily x ∈ [−1, 0]
x = −x2(x + 1)− 1
[−1, 1] −[−1, 1]2([−1, 1] + 1)− 1 = [−1, 0],
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10 II. Interval Analysis
Necessarily x = −1 !
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11 II. Interval Analysis
Exclusion test
Evaluation if 0 6∈ P (X) then exclude X
Example
P (x) = (x− 1)(x− 2)(x− 3)(x− 4)(x− 5)
= x5 − 15x
4+ 85x
3 − 225x2+ 274x− 120
= −120 + x(274 + x(−225 + x(85 + x(−15 + x)))).
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12 II. Interval Analysis
X = [4.9, 4.99]
(X5 − 15X
4+ 85X
3 − 225X2+ 274X − 120) = [−855.229710, 853.060670]
(−120 + x(274 + x(−225 + x(85 + x(−15 + x))))) = [−133.5775355, 132.1841260]
We cannot conclude
X = [4.99990, 4.99999]
(X5 − 15X
4+ 85X
3 − 225X2+ 274X − 120) = [−0.879868..0.877236]
(−120 + x(274 + x(−225 + x(85 + x(−15 + x))))) = [−0.1373968..0.1347577)]
We cannot conclude
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13 II. Interval Analysis
−→ need to use adapted exclusion tests
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14 III. Polynomial solving
use of polynomials with 1 interval coecient
If the number of interval coecient is more than 1, the complex solution set has
an non empty interior set.
Exclusion test always fail in the interior of solution test ( → cut).
−→ Interior tests or Determine the boundary of solution set
−→ Edge theorem
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15 III. Polynomial solving
Edge theorem(Bartlett, Hollot, Huang, 88)
It suces to solve edge polynomials :
Pi(x) =Xk 6=i
bkxk+ [ai]x
i,
bk ∈ ai, ai
Solution set of 3rd degree interval polynomial
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16 III. Polynomial solving
1-D uncertainty. The solution is a 0-D space. no need to have interior
test.
d.2d polynomials with one interval coef to consider.
The solutions of a part of the previous polynomials are in the interior of
the solution set.
How to select the polynomials whith solutions on the boundary of the
solution set ?
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17 III. Polynomial solving
Edge theorem revised
Polar representation.
N = lcm(2, 3, · · · , d). Ik = [k2πN , (k + 1)2π
N ]
We can determine 2(d + 1) edge polynomials necessary to determine
the solution set boundary in R+ × Ik.
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18 III. Polynomial solving
2(d + 1) edge polynomials to consider in each angular sector,
lcm(2, 3, · · · , d) angular sectors to consider....
2(d + 1) is not the minimal number of polynomials to consider. We
may improve further edge theorem.
Allows to solve only low degree uncertain polynomials.
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19 III. Polynomial solving
Edge theorem for real solutions
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20 III. Polynomial solving
x ≥ 0, P (x) = [P+− (x), P
++ (x)]
x ≤ 0, P (x) = [P−− (x), P
−+ (x)]
P+− (x) = a0 + a1x + a2x
2 · · ·+ adxd,
P++ (x) = a0 + a1x + a2x
2 · · ·+ adxd,
P−− (x) = a0 + a1x + a2x
2+ a3x
3 · · · ,
P−+ (x) = a0 + a1x + a2x
2+ a3x
3 · · · ,
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21 III. Polynomial solving
Problem to solve :
Certied real solutions of non interval polynomials :
positive zeros of P+− and P+
+
negative zeros of P−− and P−
+ .
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22 III. Polynomial solving
Complex solutions for one interval coecient polynomials
Exclusion tests :
Weyl exclusion test :
P (z) =dX
i=0
aizi=
dXi=0
bi(z − z0)i,
If |b0| >Pd
i=1 |bi|δi then no solution in B(z0, δ).
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23 III. Polynomial solving
Yakoubsohn test :
Mz0(δ) = |P (z0)| −
Xk
|pk(x)|k
δk.
If Mz0(δ) > 0, then no solution in B(z0, δ).
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24 III. Polynomial solving
Real solutions for one interval coecient polynomials
Exclusion tests:
Weyl
Yakoubsohn
Filtering procedures
2-B,
Interval Newton algorithm
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25 III. Polynomial solving
Uniqueness tests:
based on Rouche theorem,
Krawczyk operator,
Kantorovitch theorem.
Deation
Once a solution is known, division by (z − z0).
First numerical results
Find the 19 solutions of Wilkinson 19.
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26 III. Polynomial solving
Perspectives
There is probably a room for improving Edge theorem for complex solutions
Implementation for complex solutions,
Consider correlated Interval coecients.
Certied solutions of polynomials with uncertainties O.Pourtallier