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An Interval Polynomial Interpolation Problem and its Lagrange Solution ¤ Chenyi Hu, Angelina Cardenas Stephanie Hoogendoorn, and Pedro Sepulveda Jr.

Abstract Numerical interpolation and approximation are powerful tools to solve real world application problems. To take data measurement errors into considerations of interpolating a d degree unknown polynomial with n+ 1 knots, we de¯ned an interval interpolation problem1 in section 1. Algorithms based on Lagrange interpolation formula are proposed to solve the problem in section 2. Average overestimation is de¯ned in section 3. Numerical experiments are presented in section 4. Section 5 concludes the paper.

1

An Interval Interpolation Problem

Let us clarify notations used in this paper ¯rst. Throughout this paper, boldfaced letters such as x; y represent interval variables while x; y represent real numbers. For an interval x, we use x and x to represent its greatest lower bound and least upper bound respectively, i.e. x = [x; x]. The general problem to be addressed in this paper is to reliably bound the value of a polynomial f (x) = a0 + a1 x + : : : + ad xd , where ad 6= 0, for ¤

This research was partially supported by NSF grants CCR-9503757, CDA-9522157 and DoD/ARO grant DAAH-0495-1-0250. Among the last three undergraduate co-authors, Hoogendoorn and Sepulveda are research assistants supported by the grants. Comments about this paper can be directed to the ¯rst author at [email protected]. 1 The authors wish to express their great appreciations to the anonymous referee who explicitly formed the problem in his review report.

1

x 2 x with unknown coe±cients ak , (0 · k · d); provided interval knots (xi ; yi ) such that for each i 2 f0; 1; 2; ¢ ¢ ¢ ; ng, n ¸ d, there is at least an xi 2 xi with f (xi ) 2 yi ; and xi \ xj = ; if i 6= j. Without loss of generality, we assume that x0 < x1 < x2 < ¢ ¢ ¢ < xn in the rest of this paper. This problem is an extension of classical numerical interpolation problem in which xi s and yi s are degenerate, i.e., xi = xi and y i = y i . The classical polynomial interpolation problem has been very elegantly solved in 18th century. It is very well known that there is an unique polynomial Pn (x) of degree at most n for given (xi ; yi ); 0 · i · n, provided xi 6= xj if i 6= j, such that Pn (xi ) = yi for i = 0; 1; 2; ¢ ¢ ¢ ; n. Therefore, the unknown polynomial f (x) = a0 + a1 x + ¢ ¢ ¢ + ad xd can be uniquely determined when n ¸ d. Newton provided an algorithm to ¯nd the unknown coe±cients a0 ; a1 ; ¢ ¢ ¢ ; ad . Lagrange proved that f (x) equivalent to Pn (x) =

n X i=0

0

yi B @

n Y j=0 j6=i

1

x ¡ xj C A xi ¡ xj

(1)

The polynomial (1) is called the Lagrange interpolation polynomial. One may use the Lagrange interpolation polynomial Pn (x) to evaluate f (x) for n ¸ d if the point knots (xi ; yi ) are precise. In applications, the point knots (xi ; yi ) are obtained from direct or indirect measurements which have system errors and rounding errors. The inaccuracy of measurements will cause Lagrange interpolation polynomial (1) far away from the unknown polynomial f (x). For example, the 24 knots in Table 1 result a 23rd degree polynomial with (1). However, the data were come from measuring a 3rd degree polynomial with moderate errors. xi yi xi yi xi yi

.01 1.9 6.6 36.1 13.2 -48.6

1.1 7.9 7.1 23.7 14.1 -40.2

1.6 24.9 8.2 13.0 15.6 -51.6

2.4 24.9 9.1 20.5 16.1 -30.5

2.5 34.9 9.4 -3.10 17.6 -34.6

4.1 42.7 11.1 -13.0 17.9 -16.4

5.2 29.7 11.4 -28.7 19.1 -13.4

6.1 49.8 12.2 -39.5 20.0 -1.10

Table 1: Twenty-four knots (xi ; yi ); 0 · i · 23. In the literature, there are some numerical methods to address the problem of inaccurate data. For example, one may use spline interpolation to 2

piecewisely approximate f (x) with d degree polynomials. Or, one may use the least square method to obtain a d degree polynomial Pd (x) to ¯t the knots \best" in terms of minimizing

n X i=0

(Pd (xi ) ¡ f (xi ))2 . Applying the

least square method to the data in Table 1, the 3rd degree polynomial ¡6:38689 + 24:103x ¡ 3:61389x2 + 0:122286x3 was obtained by using MATHEMATICA. But, it is still an unreliable approximation. Another approach assumes that the xi s can be precisely measured, i.e. the intervals xi are degenerate: xi = [xi ; xi ], then ¯nding the bounds of the interval of possible values of f(x) can be reduced to solve the following two linear programming problems: max a0 + a1 x + : : : + ad xd and min a0 + a1 x + : : : + ad xd under n conditions: yi¡ · a0 + a1 xi + ¢ ¢ ¢ + ad xdi · yi+ ; 1 · i · n;

where yi§ are the bounds of yi = [yi¡ ; yi+ ]. This approximation works only if the xi s are degenerate, and the result may not mathematically reliable either. Di®erently with the above approaches, we take measurement errors into considerations by using interval knots (xi ; yi ) instead of point knots (xi ; yi ) to form an interval interpolation problem. The interval interpolation problem starts when: ² we know that the unknown function f is a polynomial of the given degree d; and ² we know the results of measuring the unknown function f in n points (n ¸ d); but we do not know the exact values of xi , and we do not know the exact values of yi = f(xi ). Instead, due to the inevitable inaccuracy of the measurements, we know the intervals xi and yi that contain the unknown values xi and yi .

2

Lagrange Solution

In the general case, when we only know the intervals xi s and yi s, to reliably ¯nd the interval of possible values of f (x), we really need to ¯nd the interval 3

contains all possible values of an interpolation polynomial (1) with degree d for any (xi ; yi ) 2 (xi ; yi ). This approach is NP-hard and impractical for traditional point arithmetic. Interval computations provides a powerful tool to ¯nd an enclosure interval of f (x). If we apply naive interval computations, i.e., replace each operation with numbers in (1) by an operation with intervals, we obtain the formula Pn (x) =

n X i=0

0

n Y yi B @

j=0 j6=i

1

x ¡ xj C A xi ¡ xj

(2)

We call (2) an interval Lagrange interpolation formula. It is obvious that Pn (x) ¾ f (x). However, the naive interval Lagrange interpolation formula (2) may severely overestimate f (x) if n >> d. We need to reduce the ovrestimation. Proposition 1: If f(x) is a polynomial of degree d, and for each of n + 1 interval knots there is a point knot (xi ; f(xi )) 2 (xi ; yi ); then for any set S = fs0 ; s1 ; ¢ ¢ ¢ ; sd g ½ f0; 1; ¢ ¢ ¢ ng and x 2 x, we have f(x) 2 PSd (x) =

X i2S

0

Y yi B @ j2S j6=i

1

x ¡ xj C A xi ¡ xj

(3)

Proof: From classical interpolation theory, 0 1 we have f (x) = PdS (x) =

X i2S

2

X i2S

0

f(xi ) B @

Y yi B @ j2S j6=i

Y

j2S j6=i

1

x ¡ xj C A xi ¡ xj

x ¡ xj C S A = Pd (x). 2 xi ¡ xj

Equation (3) contains much less amount of interval computations than (2) when n >> d, therefore, it should reduce the overestimation due to interval arithmetic as well. Based on Proposition 1, we propose algorithms below to reliably ¯nd an enclosure interval of f (x). Algorithm 2: (General Intersection Method) (2.0) Input x 4

(2.1) (2.2) (2.3)

µ

¶

n+1 cardinality d+1 subset S of f0; 1; ¢ ¢ ¢ ; ng, d+1 ¯nd PSd (x)\with (3) Find y = PSd (x).

For each of the

S

Output y 2

From Proposition 1, we know that Algorithm 2 provides a reliable enclosure of f (x). Because of the intersection step, it is expected that the overestimation of the result provided by Algorithm 2 will be less than (or at most the same as) the least overestimation with any single application of (3). However, when n is signi¯cantly larger than d, steps (2.1) and (2.2) would be very expensive. To reduce the amount of computations, we propose Algorithm 3 below. Algorithm 3: (Selective Intersection Method) (3.0) Input x (3.1) Let S = fs0 ; s1 ; ¢ ¢ ¢ ; sd g be a subset of f0; 1; ¢ ¢ ¢ ; ng. We assume that xs0 < xs1 < xs2 < ¢ ¢ ¢ < xsd . Find PSd (x) with (3) for those S ¤ s such that xs¤0 < x < xs¤d . \ (3.2) Find y = PSd (x). (3.3)

S¤

Output y 2

The selection procedure in (3.1) is based on a heuristic, observed through numerical experiments, that for the set S if x 6½ [xs0 ; xsd ] then it is most likely that PSd (x) very severely overestimate f (x). To most signi¯cantly reduce computational complexity, one may piecewisely interpolate f (x) with (3). However, without an intersection step, it may result large overestimation. What we really want is to reduce computations as much as possible as long as we are satis¯ed with the result. To measure the satisfaction of an estimation, we de¯ne an index of satisfaction below. De¯nition 4: Let the interval y¤ be the estimated bounds of f(x) at x = x¤ provided f(x¤ ) 2 y¤ . Then, the satisfaction index of the estimation is de¯ned

5

as

8 > > >
> > :

1¡ 0; 1;

y ¤ ¡ y¤ ; 2 maxfjy ¤ j; jy ¤ jg

if y¤ 6= 0 and y¤ is bounded if y¤ is unbounded otherwise

Example: The satisfaction index of the bound estimation of f(1) 2 [99; 101] is 99%. The satisfaction index of the bound estimation of f(2) 2 [10; 10] is 100%. The satisfaction index of the bound estimation of f (3) 2 [3; 5] is 80%. Algorithm 5: (Satisfaction Intersection Method) (5.0) Input x, and e¤ user determined satisfaction lower bound (5.1) Find a subset S = fs0 ; s1 ; ¢ ¢ ¢ ; sd g of f0; 1; ¢ ¢ ¢ ; ng; and then ¯nd y = PSd (x) and the satisfaction index e. (5.2) Repeat: If e < e¤ and there is a new subset Snew with cardinality d + 1; then y à y \ PdSnew (x), and ¯nd e for the new y. (5.3) Output y 2 In applying Algorithm 5, the user provided lower bound of satisfaction index, e¤ , may never be reached if it is unrealisticly high. In that case, the result produced would be the same as Algorithm 2.

3

Average Overestimation

We have mentioned the term \overestimation" in previous sections. Here, we de¯ne an index to quantify the average overestimation for an algorithm due to interval arithmetic. De¯nition 6: The width of an interval x is de¯ned as w(x) = x ¡ x: De¯nition 7: Let y¤i be the estimated bound of an algorithm with xi as input. Then, the relative over estimation of yi by that algorithm is de¯ned as

6

8 < w(y¤i ))¡w(yi )

£ 100%; if w(yi ) 6= 0 w(yi ) ri = : , unde¯ned; if w(yi ) = 0 De¯nition 8: The average overestimation for an algorithm is de¯ned as the average of ri 's, which are de¯ned, for i = 0; 1; 2; ¢ ¢ ¢ ; n. We use R to denote the average overestimation for that algorithm. To study the average overestimations for di®erent algorithms is a interesting problem. Through numerical experiments, we have some conjectures which are presented in section 4 with numerical examples.

4

Numerical Experiments

We have programmed the algorithms in section 2 of this paper in FORTRAN 77. The programs have been run on a VAX 4000-300 computer with VAX/VMS version V5.5-2 operating system for some test data set. Rigorous interval computations are performed by calling routines of the portable interval software library INTLIB [5].

4.1

Test Problem 1

4.1.1

The Test Problem

Provided f(x) is a fourth degree polynomial, and the ¯ve knots in table 2 below have maximum §0:1% error on both xi s and yi s. We want to reliably bound the ranges of f (x) at x = 1:8; 2:4; 3:0; 3:2; 5:0, and 7:5. i xi yi

0 1.0 14.2

1 2.7 17.8

2 3.2 22.0

3 4.8 38.3

4 5.6 51.7

Table 2: Five point knots (xi ; yi ); 0 · i · 4 Since the data in Table 2 have §0:1% percent measurement error for both xi s and yi s, the interval knots can be represented as follow.

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i 0 1 2 3 4

xi = xi ¨ 0:1% [0.9990, 1.0010] [2.6973, 2.7027] [3.1968, 3.2032] [4.7952, 4.8048] [5.5944, 5.6056]

yi = yi ¨ 0:1% [14.1858, 14.0142] [17.7822, 17.8178] [21.9780, 22.0220] [38.2617, 38.3383] [51.6483, 51.7517]

Table 3: Five interval nodes (xi ; yi ) 4.1.2

Interval Lagrange Interpolation

In the test problem, n = d; therefore, Algorithms 2, 3 and 5 provide the same answer. With interval Lagrange interpolation polynomial (2), we approximate f(x) by P4 (x) at x = [1:8; 1:8]; [2:4; 2:4]; [3:0; 3:0]; [3:2; 3:2]; [5:0; 5:0], and [7:5; 7:5]. Numerical results are presented in Table 4 below. P4 ([1:8; 1:8]) = [9:296083176166190; 15:77494025419015] P4 ([2:4; 2:4]) = [13:85341377069415; 17:32090226755574] P4 ([3:0; 3:0]) = [19:37142780369701; 21:19415011318991] P4 ([3:2; 3:2]) = [21:11269538401268; 22:91659595228631] P4 ([5:0; 5:0]) = [38:88218216754060; 43:28603602440520] P4 ([7:5; 7:5]) = [55:61688114587412; 222:3519238000697]

Table 4: Reliable Interval Estimation With classical Lagrange interpolation polynomial (1), we approximate f (x) by P4 (x) at x = 1:8; 2:4; 3:0; 3:2; 5:0 and 7:5 and obtain the following result. P4 (1:8) = 12:53272123964755 P4 (3:0) = 20:26722169264469 P4 (5:0) = 41:05450997838501

P4 (2:4) = 15:58180662834920 P4 (3:2) = 22:00000000000000 P4 (7:5) = 138:9658143497962

Table 5: Point Approximation Since rigorous interval computations are performed with INTLIB, for any x 2 x we should have Pn (x) 2 Pn (x). Table 4 provides reliable estimations. The fact of the large width of P4 ([7:5; 7:5]) has caused our attention. By checking input data, we found that 7:5 > x4 . We have found that largewidth estimation is produced if x < x0 or x > xn for other input data.

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4.1.3

About Overestimation

To invest how measurement errors in (xi ; yi )s e®ect the average overestimation, we used di®erent relative error bounds for input data in Table 2. The numerical results show that for ¯xed yi error bound the average overestimation R increases when the error bound of xi increases; and for ¯xed xi error bound, the average overestimation decreases when the error bound of yi increases. We only list a few data to demonstrate the result. xi ; (0 · i · 4) xi ¨ 0:1% xi ¨ 1% xi ¨ 5%

yi ; (0 · i · 4) yi ¨ 0:1% yi ¨ 0:1% yi ¨ 0:1%

Avg. Over Est. 69.67812033626889 % 806.5148796025809 % 15016.85358718145 %

xi ; (0 · i · 4) xi ¨ 0:1% xi ¨ 0:1% xi ¨ 0:1%

yi ; (0 · i · 4) yi ¨ 0:1% yi ¨ 1% yi ¨ 5%

Avg. Over Est. 69.67812033626889 % 7.896479496892357 % 2.404778088948134 %

Table 6: R increases when relative error of xi increasing

Table 7: R decreases when relative error of yi increasing Although the average over estimation R does not represent exact overestimation of P(x) for any interval x, it does provide a measurement for users to approximate overestimation. To reduce the average overestimation, it is important to measure the independent variable x as precise as possible. However, in obtaining the knots, one should measure y precisely also. It is because of that the larger the widths of yi 's are, the less the satisfaction indices are. 4.1.4

Approximate f(x) When x \ xi 6= ; for Some i

In this subsection, we study how to estimate f(x) if x \ xi 6= ; for some i 2 f0; 1; 2; ¢ ¢ ¢ ; ng. Let k 2 f0; 1; 2; ¢ ¢ ¢ ; ng and x \ xk 6= ;, then we propose the follow algorithm to approximate f (x). Algorithm 9: 1. Input an interval x 2. Find the set K = fkjx \ xk 6= ;; 0 · k · ng.

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3. Find the closure of the relative complement of the set [k2K xk to x.

We call the result set Z. It is easy to see that Z is a union of at least jKj ¡ 1 and at most jKj + 1 disjointed intervals (call them zj 's, could be empty if x ½ xk ) if K 6= ;.

4. f (x) can be approximated by ([j P(zj )) [ ([k2K yk ).

Numerical examples show that Algorithm 9 obtains sharper interval than that provided by (2). The following two tables provide a comparison2 . x [2:96; 3:28] [3:024; 3:344]

P(x) [1.882575585789230, 46.50709081525388] [1.593975892369722, 47.26975070954503]

Table 8: Approximation with (2) x [2:96; 3:28] [3:024; 3:344]

y2 [ P(z) [6.986095925934057, 38.75010019712957] [10.63413137086082, 33.99451863461799]

Table 9: Approximation with Algorithm 9

4.2

Test Problem 2

4.2.1

The Test Problem

Provided f(x) is a third degree polynomial, and the knots in Table 10 below have maximum §0:5% error on both xi s and yi s. We want to reliably bound the ranges of f (x) at x = ¡0:5; 0; 2:5; 5:0; 9:5, and 16. i xi yi

0 -2.002 13

1 0.1 3.1092

2 1.003 4.005

3 4.01 -41

4 9.03 -638.5

5 12 -1563

Table 10: Twelve knots (xi ; yi ); 0 · i · 11. 2

In Tables 8 and 9, we used the interval node (x2 ; y2 ) where x2 = [3:1968; 3:2032] and y2 = [21:9780; 22:0220].

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4.2.2

Interval Lagrange Interpolation

From data in Table 10, we can easily form the six interval knots with §0:5 percent relative error for both xi and yi . There are 15 di®erent combinations of four interval knots out of the six. With Algorithm 2, general intersection method, we estimated the function values for each of the 15 di®erent combinations by using (3). The table below lists 15 di®erent estimations for f (¡0:5) and the satisfaction index e of each. 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2,

S 1, 2, 1, 2, 1, 2, 1, 3, 1, 3, 1, 4, 2, 3, 2, 3, 2, 4, 3, 4, 2, 3, 2, 3, 2, 4, 3, 4, 3, 4,

3 4 5 4 5 5 4 5 5 5 4 5 5 5 5

f (¡0:5) [2.612002926124677, 3.144788996778373] [2.623674100684147, 3.132972389826237] [2.621686644955066, 3.129230831413292] [2.408493541735073, 3.344470933455444] [2.452755710827451, 3.277314110417823] [-0.5275140007137611, 6.137604997543502] [1.612031602444530, 4.135186112936243] [1.787119450737507, 3.897784143568561] [-6.737121646719446, 12.10422766488884] [-44.46832175142903, 48.96193423928143] [1.977559671746458, 3.781051276836537] [2.124357724869077, 3.663872363998053] [-1.497851857575839, 7.428047427199745] [-18.30986206064202, 24.62161566033456] [-59.11950428186875, 66.30039002342567]

e 91.5% 91.9% 91.9% 86.0% 87.4% 45.7% 69.5% 72.9% 22.2% 4.59% 76.1% 79.0% 39.9% 12.8% 5.42%

Table 11: Estimations of f (¡0:5). By intersecting the above 15 di®erent estimations, we obtain the best possible estimation of f(¡0:5), which should be in the interval [2.623674100684147, 3.129230831413292], with the 6 interval knots. This estimation has the satisfaction index of e = 92%. The table below presents the result intervals and the satisfaction index, e, of the best estimation at x = 0; 2:5; 5:0; 9:5, and 16 with the provided 6 interval knots. x 0 2:5 5:0 9:5 16

Pn (x) [2.862125998097987, 3.141873945468383] [-4.538754635417469, -3.023376028735940] [-94.53963750604516, -88.01108954014208] [-768.3403741001655, -731.4218307608894] [-3894.304770984470, -3718.191133811501]

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e 95.5% 83.3% 96.5% 97.6% 97.7%

Table 12: Estimation by Algorithm 2. The estimation satisfaction index in each of the last ¯ve rows of Table 11 are relatively small. One thing in common is that for the last ¯ve rows the point x = ¡0:5 is not in the domain used to perform the estimation. This fact inspired Algorithm 3, the selective intersection method, with possible less computations. However, our numerical experiments shows that the estimation interval obtained by Algorithm 3 may or may not as good as that provided by Algorithm 2. For example, using Algorithm 3, we have f(9:5) is in the interval [-772.5405815166976, -731.1102497143360] which is not as good as the one in Table 12. From Table 11, one can easily see that to estimate f (¡0:5), Algorithm 3 provides the same solution as Algorithm 2 does with 1=3 less computations. With the satisfaction index e¤ = 95%, Algorithm 5 obtained answers in very few iterations in this test problem except estimating f (2:5) for which the highest possible e is 83:3 in Table 12.

5

Summary

In this paper, we studied the application problem of reliably estimating the bound of a d degree unknown polynomial f (x) at a point x or an interval x with n + 1 interval knots where n ¸ d. This problem is important in applications, because of that: a) point knots in applications are not precise; and b) other available methods do not provide reliable solutions, as far as we know. The algorithms based on Lagrange formula in this paper are reliable both mathematically and computationally with interval knots and ¯nite digit computing. The trade o® for the reliability are the overestimation and the amount of computations. It is obvious that the sharper the interval knots are, the less overestimation is. In fact, if the measurement of the knots is 100% accurate, i.e. all intervals are degenerated, then algorithms of this paper provide the same solutions as that provided by classical polynomial interpolations with satisfaction index of 100%. However, it is impractical to measure knots absolutely accurate. Our numerical experiments suggest the following conjectures:

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² Less measurement errors for independent variable intervals xi s will reduce the average overestimation de¯ned in this paper signi¯cantly. ² When x overlaps with xi for some 0 · i · n, to bound f (x) one should apply Algorithm 9 to reduce the overestimation. ² With a reasonable lower bound of satisfaction index, Algorithm 5 may well balance the amount of computation and the overestimation. ² Increasing n, the number of interval knots, may obtain sharper estimation. However, the number of computations will increase signi¯cantly. If it is necessary to make n much larger than d, parallel computation is strongly recommended. There are still some open questions for this research. For example, what would be a better selection techniqe than the one used in Algorithm 4? what would be a reasonable n if d is provided? what would be a reasonable value of e¤ for Algorithm 5 in practice? The choices of a reasonable e¤ should relate to the precision of the knots measurement. What would be the relashionships? We hope interested readers to join us study these open questions.

Acknowledgement The authors wish to express their great appreciations to the anonymous referees for their careful reading and very helpful comments. One of them has explicitly formed the problem in very clear terms. The authors also wish to express their appreciations to Professor Vladik Kreinovich for his encouragement of submitting this paper.

References [1] G. Alefeld, J. Herzberger, \Introduction to interval computations", Academic Press, New York, 1983. [2] M. A. Crane, \A Bounding Technique for Polynomial Functions", SIAM J. Appl. Math. 29 (4), pp. 751-754, 1975.

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[3] G. F. Gerald, P. O. Wheatly \Applied Numerical Analysis", 5th edition, Addison-Wesley, 1994. [4] J. Herzberger, \Note on a Bounding Technique for Polynomial Functions", SIAM J. Appl. Math. 34 (4), pp. 685-686, 1978. [5] R. B. Kearfott, M. Dawande, K. Du, C.Hu, \Algorithm 737: INTLIB: A portable Fortran-77 interval standard-function library", ACM Transactions on Mathematical Software, Vol. 20, No. 4, pp. 447-459, 1994. [6] D. Kincaid and W. Cheney, \Numerical Analysis", Brooks/Cole, 1990. [7] V. Kreinovich and A. Bernat, \Parallel Algorithms for Interval Computations: An Introduction", J. Interval Computations, No.(3), pp. 6-63, 1994. [8] S. M. Markov, \Some Interpolation Problems Involving Interval Data", J. Interval Computations, No.(3), pp. 164-182, 1993. [9] R. E. Moore, \Methods and Applications of Interval Analysis ", SIAM, 1979. Department of Computer and Mathematical Sciences University of Houston-Downtown Houston, Texas 77002 USA

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