Non-linear extension of interval arithmetic and exact resolution of interval equations over square regions
aa r X i v : . [ m a t h . G M ] M a r N ON - LINEAR EXTENSION OF INTERVAL AR ITHMETIC ANDEXAC T R ESOLUTION OF INTERVAL EQUATIONS OVER SQUAR ER EGIONS
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Giovanny A. Fuentes S
Universidade Federal FluminesiNiteroi,RJBrasil [email protected]
March 30, 2020 A BSTRACT
The interval numbers is the set of compact intervals of R with addition and multiplication operation,which are very useful for solving calculations where there are intervals of error or uncertainty, how-ever, it lacks an algebraic structure with an inverse element, both additive and multiplicative Thisfundamental disadvantage results in overestimation of solutions in an interval equation or also over-estimation of the image of a function over square regions. In this article we will present an originalsolution, through a morphism that preserves both the addiction and the multiplication between thespace of the interval numbers to the space of square diagonal matrices. K eywords Interval Arithmetic · Interval Equation · Interval Computation
Contents
In this article we propose a general method to solve equations with interval variables, that is to say where the unknownof the problem is a compact interval, through a ϕ application that takes intervals and takes them to 2 times 2 squarematrices. In the first part of this article we will give a brief introduction to the Artificial Interval. In the second part wewill state and demonstrate the fundamental theorem of the interval functions , which gives us the necessary conditionsso as not to have lost points, when computing the image of a square region (finite Cartesian producer of compactintervals) through the ϕ application. And finally in the third and last part of this article, through the fundamental PREPRINT - M
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30, 2020 theorem of the interval functions , we will state and demonstrate the fundamental theorem of the interval equations,that will give us necessary and sufficient conditions to solve the interval equations.
Recall that the closed interval demoted by [ a, b ] is the real numbers given by [ a, b ] = { x ∈ R , a ≤ x ≤ b } (1)We say that an interval is Degenerate if a = b . Such an interval contains a single real number a . By convention, weagree to identify a degenerate interval [ a, a ] . In this sense, we may write such equation as , (2)We will denote by K c ( R ) the set of compact intervals real.We are about to define the basic arithmetic operations between intervals. The key point in these definition isthat computing with set. For example when we add two interval, the resulting interval is set containing the sums of allpair of number, one form each of the initial sets. By definition then, the sum of two intervals X and Y is the set X + Y = { x + y ; x ∈ X, y ∈ Y } (3)The difference of two intervals X and Y is the set X − Y = { x − y ; x ∈ X.y ∈ Y } (4)The product of X and Y is given by XY = { xy ; x ∈ X, y ∈ Y } . (5)Finally, the quotient X/Y with Y is defined as X/Y = { x/y ; x ∈ X, y ∈ Y } (6) Let us that an operational way to add intervals. Since x ∈ X = [ x , x ] means that x ≤ x ≤ x and y ∈ Y = [ y , y ] means that y ≤ y ≤ y , we see by addition of inequalities that the numerical sums x + y ∈ X + Y must satisfy x + y ≤ x + y ≤ x + y . Hence, the formula X + Y = [ x + y , x + y ] Example 1
Let X = [0 , and Y = [ − , . Then X + Y = [0 − , − , Subtraction
Let X = [ x , x ] and Y = [ y , y ] . We add the inequalities x ≤ x ≤ x and − y ≤ − y ≤ − y (7)to get x − y ≤ x − y ≤ x − y . It follows that X − Y = [ x − y , x − y ] . Note that X − Y = X + ( − Y ) where − Y = [ − y , − y ] Example 2
Let X = [ − , and Y = [1 , . Then X − Y = [ − − , −
1] = [ − , − Multiplication
In terms of endpoint, the product XY of two intervals X and Y is given by XY = [min S, max S ] , where S = { x y , x y , x y , x y } (8) Example 3
Let X = [ − , and Y = [1 , . Then S = {− , − , } and XY = [ − , . The multiplication of intervals is given in terms of the minimum and maximum of four products of endpoint, this canbe broken into nine spacial cases. Let X = [ x , x ] and Y = [ y , y ] then XY = Z = [ z , z ] , thencase z z ≤ x , y x y x y x < < x and ≤ y x y x y x ≤ and ≤ y x y x y ≤ x and y < < y x y x y x ≤ and y < < y x y x y ≤ x and y ≤ x y x y x < < x and y ≤ x y x y x ≤ and y ≤ x y x y x < < x and y < < y min { x y , x y } max { x y , x y } PREPRINT - M
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Division
As with real number, division can accomplished via multiplication by the reciprocal of the second operand.That is, we can implement equation using
X/Y = X (cid:18) Y (cid:19) , (9)where Y = (cid:26) y ; 1 y ∈ Y (cid:27) . (10)Again, this assume Y . For more information about interval numbers, see [1] The main result of this section is the fundamental theorem of the interval functions, which allows us to calculate thevalue of a function on a square region, that is, on a finite Cartesian product of compact intervals, this can be interpretedas determining the image of a Interval function where each variable detects a unique value under the same symbol.Consider M x ( R n ) as the space of the square matrices two by two over R n . Let k . k n a norm in R n and k . k s thenorm of sum in M x ( R ) then we define the following norm in M x ( R n ) as (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ( a , . . . , a n ) ( b , . . . , b n )( c , . . . , c n ) ( d , . . . , d n ) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) = k ( a , . . . , a n ) k n + k ( b , . . . , b n ) k n + k ( c , . . . , c n ) k n + k ( d , . . . , d n ) k n (11)We have this application, it satisfies the norms of norm, then the space R n is a normed space. Let k . k be a matrix normin M x ( R n ) . In this section we will consider D ( R n ) the space of the square diagonal matrices two by two over R n with the following norm k . k ∗ = k . k . Obviously k . k ∗ is a norm in D ( R n ) . Proposition 4 M x ( R n ) with a norm defined is complete. Proof
Let { A j } j an Cauchy sequence in M x ( R n ) and let ε > , then exist a N > such that for p, m > N wehave k A p − A m k < ε , then (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) ( a p , . . . , a pn ) − ( a m , . . . , a mn ) ( b p , . . . , b pn ) − ( b m , . . . , b mn )( c p , . . . , c pn ) − ( c m , . . . , c mn ) ( d p , . . . , d pn ) − ( d m , . . . , d mn ) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) (12) = k ( a p , . . . , a pn ) − ( a m , . . . , a mn ) k n + k ( b p , . . . , b pn ) − ( b m , . . . , b mn ) k n (13) + k ( c p , . . . , c pn ) − ( c m , . . . , c mn ) k n + k ( d n , . . . , d pn ) − ( d m , . . . , d mn ) k n < ε (14)this implique that k ( x p , . . . , x pn ) − ( x m , . . . , x mn ) k n < ε for x = a, b, c, d , as R n is complete, then ( x p , . . . , x pn ) is convergent. Therefore { A j } j is convegent. Which proves that M x ( R n ) is complete. Proposition 5
Let { A j } j a sequence in M x ( R n ) , If the series ∞ X j =0 k A j k converges for any norm in, so the series ∞ X j =0 A j is converges. (cid:4) Proof
Let ε > , We will see that there is an integer N > such that p, q ≥ N , then k S p − S q k < ε , where S p = p X j =0 A j This shows that the succession S n of partial sums is a Cauchy sequence, As M x ( R n ) it’s complete,then { S n } is converges. Indeed: k S p − S q k = k q X j = p +1 A j k ≤ q X j = p +1 k A j k = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X j =0 k A j k − q X j =0 k A j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (15)3 PREPRINT - M
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30, 2020Now, if ∞ X j =0 k A j k converges, exist N > such that p, q ≥ N , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X j =0 k A j k − p X j =0 k A j k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ε. (16)what we wanted to demonstrate. (cid:4) Definition 6
Let f : X ⊂ R n → R analytic function and a = ( a , . . . , a n ) ∈ int ( X ) and ε > such that B ( a, ε ) ⊂ int ( X ) , and ( α , . . . α n ) , ( β , . . . β n ) ∈ B ( a, ε ) , we defined f n Y j =1 (cid:18) x j y j (cid:19) (17) = ∞ X j =0 j ! X j + ... + j n = j (cid:18) jj . . . j n (cid:19) ∂ n f ( a , . . . , a n ) ∂x j . . . ∂x j n n (cid:18) a − α a − β (cid:19) j . . . (cid:18) a n − α n a n − β n (cid:19) j n (18) where (cid:0) jj ...j n (cid:1) is an mutinomial coeficent. Definition 7
Let f : X ⊂ R n → R analytic function and a = ( a , . . . , a n ) ∈ int ( X ) and ε > such that B ( a, ε ) ⊂ int ( X ) , and ( α , . . . α n ) , ( β , . . . β n ) ∈ B ( a, ε ) , we defined ϕ : Kc ( R ) → D ( R ) by ϕ ([ a, b ]) = (cid:18) a b (cid:19) and ϕ : F ( Kc ( R ) n ) → F ( D ( R ) n ) by ϕf n Y j =1 [ a j , b j ] := f n Y j =1 ϕ [ a j , b j ] = f n Y j =1 (cid:18) a j b j (cid:19) (19) Proposition 8
Let f : X ⊂ R n → R analytic function and a = ( a , . . . , a n ) ∈ R n such that B ( x , ε ) ⊂ X for any ε > and n Y j =1 [ α j , β j ] ⊂ B ( x , ε ) , then f n Y j =1 (cid:18) α j β j (cid:19) = f n Y j =1 α j f n Y j =1 β j (20) Proof
Consider any norm k . k n of R n . Let a ∈ int X , as f is analytic in X , then exist ε > such that B ( x , ε ) ⊂ X such that for all x ∈ B ( x , ε ) there is a series of Taylor that converges for f ( x ) . Let n Y j =1 [ α j , β j ] ⊂ B ( x , ε ) , considera matrix (cid:18) ( a , . . . , a n ) 00 ( b , . . . , b n ) (cid:19) that is matrix representation associated with the interval n Y j =1 [ α j , β j ] and (cid:18) ( a , . . . , a n ) 00 ( a , . . . , a n ) (cid:19) the representation of the matrix associated with the interval n Y j =1 [ a j , a j ] . Consider anorm od in space M x ( R n ) . Note that: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ( a , . . . , a n ) 00 ( a , . . . , a n ) (cid:19) − (cid:18) ( α , . . . , α n ) 00 ( β , . . . , β n ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∗ (21) = 12 ( k ( a , . . . , a n ) − ( α , . . . , α n ) k n + k ( a , . . . , a n ) − ( β , . . . , β n ) k n ) < ε. (22)4 PREPRINT - M
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30, 2020then by proposition 5 the series of matrix power ∞ X j =0 j ! X j + ... + j n = j (cid:18) jj . . . j n (cid:19) ∂ n f ( a , . . . , a n ) ∂x j . . . ∂x j n n (cid:18) a − α a − β (cid:19) j . . . (cid:18) a n − α n a n − β n (cid:19) j n (23)is convergent.Now consider the partial sum m X j =0 j ! X j + ... + j n = j (cid:18) jj . . . j n (cid:19) ∂ n f ( a , . . . , a n ) ∂x j . . . ∂x j n n (cid:18) a − α a − β (cid:19) j . . . (cid:18) a n − α n a n − β n (cid:19) j n (24) = (cid:18) Γ m ( a − α ) j . . . ( a n − α n ) j n
00 Γ m ( a − β ) j . . . ( a n − β n ) j n (cid:19) , (25)where Γ m = m X j =0 j ! X j + ... + j n = j (cid:18) jj . . . j n (cid:19) ∂ n f ( a , . . . , a n ) ∂x j . . . ∂x j n n . As ( α , . . . , α n ) , ( β , . . . , β n ) ∈ B ( a, ε ) , then Γ m ( a − α ) j . . . ( a n − α n ) j n → f n Y j =1 α j and Γ m ( a − β ) j . . . ( a n − β n ) j n → f n Y j =1 β j as m → ∞ , then (cid:18) Γ m ( a − α ) j . . . ( a n − α n ) j n
00 Γ m ( a − β ) j . . . ( a n − β n ) j n (cid:19) → f n Y j =1 α j f n Y j =1 β j (26)as m → ∞ , then f n Y j =1 (cid:18) α j β j (cid:19) = f n Y j =1 α j f n Y j =1 β j (27) (cid:4) Definition 9
Let f : X ⊂ R n → R differential function and n Y j =1 [ a j , b j ] ⊂ X , we say that n Y j =1 [ a j , b j ] is free ofsingularity if the components of the gradient vector are different from zero in all n Y j =1 [ a j , b j ] , that is to say ∂f ( x ) ∂x j = 0 for all x ∈ n Y j =1 [ a j , b j ] and j = 1 , . . . , n. (28) Definition 10
Let f : X ⊂ R n → R analytic function, n Y j =1 [ a j , b j ] ⊂ X free of singularity and [ a j , b j ] an interval ofthe j -variable. Defined the switch \ [ a j , b j ] of the interval as [ a j , b j ] if ∂f∂x j > and [ b j , a j ] if ∂f∂x j < , and denote by φf n Y j =1 [ a j , b j ] = ϕf n Y j =1 \ [ a j , b j ] . Note that φ is well defined, since the swapper depends only on the sign of the directional derivative on the square region,and this does not depend on how the function is formulated and also how the square region is free of singularities, then5 PREPRINT - M
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30, 2020this sign is constant in the square region. And on the other hand, since f is an analytical function, we have that thematrix series is convergent. Theorem 11 (Fundamental Theorem of Interval Functions)
Let f : X ⊂ R n → R analytic function and x ∈ R n such that B ( x , ε ) ⊂ X for any ε > and n Y j =1 [ a j , b j ] ⊂ B ( x , ε ) free of singularity, then f n Y j =1 [ a j , b j ] = ϕ − φf n Y j =1 [ a j , b j ] (29) . Proof
Let f : X ⊂ R n → R analytic function and x ∈ R n such that B ( x , ε ) ⊂ X for any ε > and n Y j =1 [ a j , b j ] ⊂ B ( x , ε ) free of singularity, then φf n Y j =1 \ [ a j , b j ] = ϕf n Y j =1 [ x j , y j ] (30)where [ x j , y j ] is the result of applying the switch to [ a j , b j ] , then by Proposition 8 since f is analytic function, we have ϕf n Y j =1 [ x j , y j ] = f n Y j =1 x j f n Y j =1 y j . (31)Applying ϕ − , we have the following interval, f n Y j =1 x j , f n Y j =1 y j . (32)Now we will prove that the interval above corresponds to the image of f on R = n Y j =1 [ a j , b j ] . First we observethat both f n Y j =1 x j and f n Y j =1 x j are elements of f ( R ) , since R is connected and closed, we have that f n Y j =1 x j , f n Y j =1 y j is a subset of f ( R ) .On the other hand. As we have that the gradient is different from zero within R , we have that the functionreaches its extreme points at the border of R . First we will prove that the maximum and minimum point are not in theedges of R , but are in one of the vertex. Indeed, since the gradient has non-zero components in all R in particularlyin the edge of R , we have that the extreme points of f can not be in the edges, since f restricted to the edges, it is acontinuous function over a compact interval where its derivative is not null, then its extreme points are the extremesof the interval. Since the argument is valid for all edges, we conclude that the extreme points of f are in the vertex.Let ( z , . . . , z j , . . . , z n ) be a vertex of R . Since f is monotonous on the edges, we have the following inequality in z j leaving the other variables fixed f ( z , . . . , x j , . . . , z n ) ≤ f ( z , . . . , z j , . . . , z n ) ≤ f ( z , . . . , y j , . . . , z n ) . (33)Taking this inequality inductively on each variable, we have f ( z , . . . , z j , . . . z n ) ≥ f ( x , . . . , z j , . . . z n ) ≥ f ( x , . . . , x j , . . . z n ) ≥ f ( x , . . . , x j , . . . x n ) (34)6 PREPRINT - M
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30, 2020and f ( z , . . . , z j , . . . z n ) ≤ f ( y , . . . , z j , . . . z n ) ≤ f ( y , . . . , y j , . . . z n ) ≤ f ( y , . . . , y j , . . . y n ) . (35)Hence f ( x , . . . , x j , . . . , x n ) ≤ f ( z , . . . , z j , . . . , z n ) ≤ f ( y , . . . , y j , . . . , y n ) . (36)finally, we have to for all u = ( u , . . . , u n ) in R , f n Y j =1 x j ≤ f n Y j =1 u j ≤ f n Y j =1 y j (37)or equivalent f ( R ) ⊂ f n Y j =1 x j , f n Y j =1 y j . (38)Therefore f ( R ) = f n Y j =1 x j , f n Y j =1 y j (39)which means that f ( R ) = ϕ − φf ( R ) . (cid:4) An observation to the above theorem, is that it remains true if the gradient is annulled at the most in two vertexes sothat gradient does not cancel out in other points of R . For example, there is no problem if it is only annulled in asingle vertex , in the case of two vertex, one of the points must correspond to a local maximum and the other a localminimum, or also that one of the two points is a saddle point. Now it can not happen that there are two maximumpoints or two minimum points, because, this would imply that there must exist a point of R between those points suchthat the gradient is zero. To this condition we will call free of sigularity except for the most in two vertexes. Corollary 11.1
Under the same hypothesis of the above theorem. Let R = m [ j =1 R j where R j are free of sigularityexcept for the most in two vertexes, then f ( R ) = m [ j =1 ϕ − φf ( R j ) . (40) Proof
Indeed f ( R ) = f m [ j =1 R j = m [ j =1 f ( R j ) = m [ j =1 ϕ − φf ( R j ) (cid:4) PREPRINT - M
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30, 2020A consequence of the theorem is the following corollary.
Corollary 11.2
Let the following sets;1. K c ( R ) +0 = { X ∈ K c ( R ) , x ≥ } ,2. K c ( R ) − = { X ∈ K c ( R ) , x ≤ } ,3. K c ( R ) + = { X ∈ K c ( R ) , x > } ,4. K c ( R ) − = { X ∈ K c ( R ) , x < } .Then 1. ϕ ( X + Y ) = ϕ ( X ) + ϕ ( Y ) for all X, Y ∈ Kc ( R ) ,2. ϕ ( kX ) = kϕ ( X ) for all X ∈ Kc ( R ) and k ≥ ,3. ϕ ( kX ) = kϕ ( b X ) for all X ∈ Kc ( R ) and k ≤ ,4. ϕ ( XY ) = ϕ ( X ) ϕ ( Y ) for all X, Y ∈ Kc ( R ) +0 ,5. ϕ ( XY ) = ϕ ( X ) ϕ ( b Y ) for all X ∈ Kc ( R ) − and Y ∈ Kc ( R ) +0 ,6. ϕ ( XY ) = ϕ ( b Y ) ϕ ( b Y ) for all X, Y ∈ Kc ( R ) − . Proof
By simple inspection. (cid:4)
We can observe that the ϕ application preserves the addition and multiplication for intervals intervals with a singlesign. besides that we can observe the first point is that due to 2) and 3) we have that ϕ is nonlinear. And on theother hand, in general these properties are not combinable, for example by combining the properties of additivityand multiplication, we obtain the following inconsistency ϕ ([1 , − [0 , (cid:18) (cid:19) and on the other hand ϕ ([1 , − [1 , , (cid:18) − (cid:19) , from where we come to a contradiction. Suppose we have an interval equation, for example a linear equation AX + B = C , where all the components areintervals, what should be the procedure to solve this equation ?, assuming that there is some solution. we could forexample consider the equation ax + b = c , where the values of this equation are defined over their correspondingintervals, that is to say that a ∈ A , and clear x of the equation and then determine the image of the square region, usingthe fundamental theorem, however, what we will obtain is a region that contains the solution of the equation. Example 12
Let the interval linear equation [1 , X + [0 ,
1] = [1 , we can verify that the solution is [1 , however,we can consider the function f ( a, b, x ) = ax + b with a ∈ [1 , and b ∈ [0 , and solve the following equation f ( a, b, x ) = c with c ∈ [1 , in terms of x , when clearing x we get the following function g ( a, b, c ) = c − ba , nowcalculating the image of g over the square region [1 , × [0 , × [1 , is [0 , which does not correspond to thesolution of the equation. We will then give a theorem that gives us the procedure to determine the solution of an interval equation, however,this solution does not always exist, since, the matrix we obtain as a solution to the matrix equation associated with theequation does not always satisfy the condition of have the first entry less than or equal to the last entry.8
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30, 2020Below we present the main theorem of this article
Theorem 13 (Fundamental Theorem of Interval Equations)
Let f : X ⊂ R n → R analytic function, n Y i =1 X j ⊂ X free of singularity with X j = [ a j , b j ] , and X ⊂ f ( X ) an compact interval. Suppose it exists a function g : n Y i =2 X j → R be such a function that for all x ∈ X exists ( x , . . . , x n ) ∈ n Y i =2 X j such that f g n Y i =2 x j ! , n Y i =2 x j ! = x ,consider c X the switch X with respect to f . Then the equation f n Y i =1 X j ! = X has solution in X if and only if ϕ c X = g n Y i =2 ϕ b X j ! defines an matrix with the first entry less than or equal to the last entry, i.e that a ≤ b . Proof
Consider the following interval equation in X : f X , n Y i =2 X j ! = X . (41)As n Y i =1 X j ⊂ X is free of singularity, then f X , n Y i =2 X j ! = ϕ − f ϕ c X , n Y i =2 ϕ b X j ! (42)Then ϕ − f ϕ c X , n Y i =2 ϕ b X j ! = X or the equivalent f ϕ c X , n Y i =2 ϕ b X j ! = ϕX (43)On the other hand we have by hypothesis, that there exists a function g : n Y i =2 X j → R be such a function that for all x ∈ X exists ( x , . . . , x n ) ∈ n Y i =2 X j such that f g n Y i =2 x j ! , n Y i =2 x j ! = x , that is, we can clear X of the matrixequation, then ϕ b X = g n Y i =2 ϕ b X j ! (44)Suppose there is a solution to the equation (41), then (44) define a matrix solution for the equation (43) where the firstentry less than or equal to the last entry.Let’s suppose that ϕ b X = g n Y i =2 ϕ b X j ! defined a matrix such that the first entry less than or equal to the last entry,by (43), we have f g n Y i =2 ϕ b X j ! , n Y i =2 ϕ b X j ! = ϕX (45)then ϕ − f g n Y i =2 ϕ b X j ! , n Y i =2 ϕ b X j ! = X (46)9 PREPRINT - M
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30, 2020by theorem 11, we have X = ϕ − f g n Y i =2 ϕ b X j ! , n Y i =2 ϕ b X j ! = f ϕ − b g n Y i =2 ϕ b X j ! , n Y i =2 X j ! (47)as ϕ b X = g n Y i =2 ϕ b X j ! defines a matrix with the first entry less than or equal to the last entry, we have to ϕ − b g n Y i =2 ϕ b X j ! ∈ K c ( R ) , then, we have that the latter corresponds to a solution of the equation (41). (cid:4) we can generalize the previous theorem for square regions with an arbitrary amount of singularities (that is, pointswith some null component of the gradient). Corollary 13.1
Let f : X ⊂ R n → R analytic function, n Y i =1 X j ⊂ X with X j = [ a j , b j ] , and X ⊂ f ( X ) ancompact interval. Suppose it exists a function g : n Y i =2 X j → R be such a function that for all x ∈ X exists ( x , . . . , x n ) ∈ n Y i =2 X j such that f g n Y i =2 x j ! , n Y i =2 x j ! = x . Let Γ the interval in the larger variable x such that Γ × n Y i =2 X j ⊂ X and let { R α } ⊂ Γ and { R β } ⊂ n Y i =2 X j with such that:1. [ α,β R α × R β = Γ × n Y i =2 X j ,2. each (cid:26) ∂f∂x j (cid:27) has a constant sign not null in int ( R α × R β ) ,3. R α × R β is free of sigularity except for the most in two vertexes.let’s denote by X α,β = f ( R α × R β ) ∩ X and X α = X ∩ R α , let’s take the following equation in X α f ( X α × R β ) = X α,β . (48) If X ⊂ f Γ × n Y i =2 X j ! , the following equation has a solution not empty f X × n Y i =2 X j ! = X (49) if there is any solution of not empty for (48) some α and β . Additionally, we have the solution of (49) if it exists, it isequal to X = [ β ϕ − φ b g α ( R β ) where f ( g α ( R β ) , R β ) = X α,β . Proof
Let’s prove that [ α X α is a solution to the equation (49), for simplicity, we will say that in the case that theequation (49) some α and β has no solution then we will say that X α = ∅ , under the hypothesis of the corollary, wehave f [ α X α × n Y i =2 X j ! = f [ α X α × [ β R β = f [ α,β X α × R β (50) = [ α,β f ( X α × R α ) = [ α,β X α,β = [ α,β f ( R α × R β ) ∩ X = f [ α,β R α × R β ∩ X = X (51)10 PREPRINT - M
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30, 2020then [ α X α is solution of f X , n Y i =2 X j ! = X . On the other hand as R α × R β is sigularity except for the most intwo vertexes, then by theorem 13, we have X α = ϕ − φ b g α ( R β ) . Therefore X = [ β ϕ − φ b g α ( R β ) . (cid:4) Example 14
Consider the equation [2 , X + [7 ,
10] = [ − , . To solve this equation, let’s consider the function f : R × [2 , × [7 , → R given by f ( x, a, b ) = ax + b , we have the following partial derivatives ∂f∂x = a > , ∂f∂a = x and ∂f∂b = 1 (52) then we have two regions free of singularity, in one the values of x are positivee and in the other they are negativ. inthe positive case we have (cid:18) (cid:19) ϕX + (cid:18) (cid:19) = (cid:18) − (cid:19) ⇒ X ∈ K c ( R ) − (53) contradiction. Therefore there is no positive solution. For the negative case we have (cid:18) (cid:19) ϕX + (cid:18) (cid:19) = (cid:18) − (cid:19) ⇒ X = [ − , − (54) then for the corollary 13.1 we have that a solution for the general equation is X = [ − , − . Example 15
Consider the equation X (1 − X ) = (cid:20) , (cid:21) . To solve this equation, let’s consider the function f : R → R given by f ( x ) = x (1 − x ) , the singularity-free regions of this function are A = (cid:18) −∞ , (cid:19) and B = (cid:20) , ∞ (cid:19) wherewe have that in B the variable X undergoes a switch. The solution of the restricted equation in A is (cid:20) , (cid:19) andthe solution of the restricted equation in B is (cid:20) , (cid:19) , by the Corollary 13.1 guarantees that a solution of the generalequation is [0 , . that we can observe that in particular the solution of the restricted equation to B is also a solution ofthe general equation and the solution of the restriction A adding the point is also a solution of the general equation. There is another approach the resolution of interval equations apart from using square matrices, and it is to use theconscious set of polynomials of a variable over ( x − x ) , it is easy to observe that this set is a ring with the operationsof addition and multiplication of polynomials. The elements of this set are of the form a + bh , where h = h , whichwe will call pseudo complexes because of their similarity to complex numbers. In this case the application betweenthe sets of the intervlar numbers and the complex pseudo numbers is given by [ a, b ] a + ( b − a ) h and analogouslyas we did for the case of the square matrices, we can redo them for the set of the complex pseudo number. References [1] Ramon E. Moore. Method and applications of interval analysis.