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Optimal choice of linear interval extension
Optimal Choice of l&au
Interval Extension*
Renpu Get
Zntemutbnul Znstitute for Applied System Analysis A-2361 -4g, Austria
Transmitted by John Casti
ABSTRACT A basic problem in interval analysis is to find more accurate interval extension of a function on a given interval. The more accurate intewal extension is, the less computation is needed in the solution of a problem, and the more useful the analysis is in other applications. As the first step, this paper investigates the optima choice of linear interval extension. It is found that one only needs to calculate function values at two particular points in order to find the optimal interval extension generated by linear interval extensions.
1.
INTRODUCTION
R. E. Moore [l, 21 and others introduced a new tool for computational mathematics-interval analysis. Assume that f(x) isa function defined in R” X,JT is an interva?.in -%“,where and X=(X1,...,
xi = [xiZ,xiS,], xZ G %SS
i=1,2 ,..., n,
. (10
and Xi1 and xis are real numbers. Then we have the range interval of fix),
*Supportedjointly b y the North-South Dialogue Program of Austria and Mzrnational Institute for Applied System Analysis. ‘Cue& research scholar in the System aud Decision Science Program of International Institute for Applied System Analysis, henburg, Austria,and on leave from the Institute of Computational aud AppW Mathematics of Xi’an Jiaotong University, Xi’an, China. APPLIED MATZZZMATZCS AND COMPUT’ATZON30: MS- 189 (1989) Q Elsevies Science Publishing Co., Inc., 1989 655 Avenue of the Americas, New York, NY 10010
165
RENEW GE
166
which will be denoted by
f(X)and be defined to be
f(x) = (fb)lx
= xl
0 .2)
l
Any set to set mapping F(X) from R” to R such that and
f(X)sF(X)
f(x)=I;‘(x)
for XEX
. (13)
is called an inclusive interval extension of f(x) on X. Furthermore, if for two intervals X, and X, such that X, G X, we have
F(q) c F(X,),
0
0
then we say that F(X) is a monotonically inclusive interval extension of f(x) on X. In this paper we shall simply call a monotonically inclusive interval extension an interval extension. Clearly, the endpoints of F(X) are respectively a lower bound and an upper bound of the range of f(x) on X. An interval extension for a given function f(x) on a certain interval X is not unique. To obtain an interval extension of a certain elementary function f(x), some interval arithmetics [l, 21 have been introduced. Assume X = [xl, x,] and Y = [zJ~,ys]. Then we define
X+Y = -
[r,+yz,x,+ys] = (x+yJxEX
x = [-
xs, -
qz]= ( -
p-Y=[r,- y,,x,--yJ =
yEY}.
EY),
0 .5)
x},
0 .0)
(x+y[xa,Y-}s
. (17)
x(x E
xy = [En{ XZYZ, XSYS, X,Ys, = (rylxEX,
and y
“SY,)
a=(
XZYZ, XSYS, XZYS, XSYZ
>I 0 .9)
Linear Interval Eden&on
167
For the same interval X we define
X “E
[XI”&],
x+Oor
[x&x;],
xs
[OJXl”],
0 E X ad n is even;
n isodd, n iseven,
(1.10)
(1.11)
IV= m4lx,L Id
is called the absolute value of the interval X. It is easy to see that the following operation rules hold:
X+(Y+Z)=(X+Y)+Z,
(1.12)
X(YZ) =(XY)Z,
(1.13)
X+Y=Y+X,
(1.14)
XY=YX,
(1.15j
where X, Y, and 2 are intervals in R”. A real number a can be thought of as an interval [a, a]. Using this, we have
x*u=
[x,*u,x,*a]
(1.16)
which shifts X to the right (or left) by a units. We have
aX=
[q,a+J,
a 24 a
l[axs,q],
(1.17)
In particular, we have 0+x=x+0=x,
0x=x0=0,
1x=x1-x.
(1.18)
However, distributivity does not always hold. But we always have socaued
168
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subdistributivity, x(Y+z)~xY+xz.
(1.19)
in some special cases, the subdistributivity becomes distributivity: x(Y+Z)=xY+xZ
for real x,
X(Y+Z)=XY+XZ
if
(1.20)
YZ>O.
(1.21)
Note that in general we have X-X+0
and
X/X+1,
(1.22)
but the cancellation law holds, that is, X+Y=Z+Y
*
x=2.
(1-B)
By a symmetric interval X, we mean that x=-x;
(1.24
by the middle point m(X) of an interval X, we mean that m(x)
Xz+ Xs = 2;
(1.25)
and by the width w(X) of X, W(X) = xs - XI.
(1.26)
For other basic elementary functions sin t, e’, and log,(l+ t) we have the following interval operations: (1.27)
ex= [ err, log,(l+
(1.28)
e's],
X) = [log,(l+ x,),log,(l+
%)I
if
1+x+0.
(1.29)
Some simple ways to construct an interval extension of a function f(x) are: (1) Directly use intervals X, Y, 2,. . . respectively in the place of x, y, 2,. . . , and use interval operations. The resulting interval extension is called the natural interval extension, and is denoted by F(X). (2) Use as many as possible socalled nested forms, that is, use as many as possible the operations on the left in (1.19) rather than the ones on the right. Usually we can get better interval extension in this way because of the inclusion. The resulting ir$erval extension is called the nested interval extension, and is denoted by F(X). (3) If f(x) EC then we can use the interval Taylor formula, for instance, in the l-dimensional case with c E X,
F,(x)=fW+
“-‘p(c)(x-c)’
c i=
i!
1
+f(n)(X)(X-C)n n!
.
(1.30)
A simple interval extension of f(x) is
Fl(X)=f(c)+f’(X)(X-c)
(c=X),
(1.31)
and we shall call it the linear interval extension of f(x). The ndimensional interval Taylor form& will be given later. Even though for the basic elementary functions we have their interval extensions with exact lower and upper bounds of their ranges on an interval X, the interval extensions of the elementary functions consisting of the basic elementary functions with arithmetic do not necessarily have exact lower and upper bounds of the ranges of those functions on X, but are more inaccurate in most cases. Therefore we need to divide X into more and more smaller and smaller intervals in order to obtain more accurate lower and upper bounds of the functions, and this increases the amount of computation very quickly. Thus, it is important to find an optimal interval extension of a function in a class of its interval extensions, either for theoretical analysis or for practical application. This paper is ad&s& to constructing an optimal interval extension from the linear interval extensions and the natural extension as well as the nested interval extension. Some research on improving the interval extension has been done by Ratschek and Rokne [3], Krawzcyk [4], and others. They are concerned with how to choose the formula rather than how to choose c E X.
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170
In the next section we give some basic theorems as the basis of our work. We consider this problem for the l-dimensional case and the ndimensional case in Section 3. Section 4 is a short discussion. 2.
BASIC THEORY
We shall present some basic theorems which are the basis for the investigation of the optimal choice of linear interval eA”lension of a function f(x) E cr. First, we notice that finding the exact f(X) is as difficult as finding the exact j(X) in the general case. Thus, usually we use an interval extension F’(X) of f(x) in the place of f(X) in (1.31). That is, we use ~(X)EF,(X)=~(C)+F'(X)(X-c).
(2.1)
The following theorem is basic.
THEOREM 2.1. If F’(X) is improved, then F,(X) will be improved for ~_yll~ c E X and any jb2cti4m f(x) E Cl as well as any jhed .
.
GOOF.
AssumeF’(X) = [g,, gs]. Then
F,(x) = [f(c)+~(&b,
- C)Yg,(XS - 4 &I
- 4ds(% - 4)Y
(2.2) Denote the lower and upper bounds of F,(X) respectively as b(gl, g,, c) and Mg,, gs, c). Then
b(g,,g,Yc)=f(c)+~(g,(~,-c)Ygl(xs-c)YgS(~I-C),gS(XS-C)),
(2.3) u(gIygs,c j = f(c)+m=( gh - 4, gAx,- 4, i&(x,- 4 id% - C>)-
(2.4)
Linear Intenaal Extension then 02gz(xz-c)2g&-C)
lf O
-
171
and OGdXs-Ch
c). Thus,
min(&(Xz- c), g,(xs - c), g,(xl- 49 &s - 4) = dxz - 4, (2.5) =gs(xs-C)s C2-6)
max(gz(r,-C),g~(XS-C)~~S(X~-C),gS(XS-C))
and so
(2.7) (29 0
Therefore
ab(gz’gs’c) agz
ab(&&,c)
a4gzsid
aa,, g,,c)
-_O
=o
and
md
agz
ags
=xz
(2.9)
-c
-cao. =xs
(2.10)
These mean that if 0 G g, \< gs, then b(gz, gs,c) and u(gz, gs,c) do nof depend on g,, and b(gz, g,,c) is a decreasing function of gs while u(g,, g,, c) is an increasing function of g,. Thus, when gs is decreased, that is, F’(X) is improved, then F,(X) will be improved.
If gz
then O
and Oags(xs-c)a
min(gz(xz-c),g~(x~-c),g~(x~-~),g~(x~-~)}
=gAxs-C),
(2.11)
max(gz(xz-c),g~(x~-c),g~(;,-~)~g~(x~-~))
“gkxz-C),
(2.12)
and so
b(g,, g,,c) = f(c) + gzbs
-
ch
(2.13) (2.14)
REIWU GE
wit,,&4
-c
agz
auk,, gs,4
and
ags
=%
(2.16)
o
=
l
TINS mean that if g, < gs G 0, then b(gz, gs,c) ad u(gz, gs, ~1 do not depend on gs, and &(gz,gs,e; is a~ ~hm&izg fwxtim of g; whik u(g,, g,, C) is a decreasing function of g,. Thus, when gz is increased, that is, F’(X) is improved, then F,(X)will be improved. If g, 6 0 ~g,, then g,(x, -c) < 0, gs(xz - c) G 0 and g& - C) 2 0, g,(x, - C) 2 0.When gz6 0 and g, is increasing, then gz(xs - c) is increashg and gz(rz -c) is decreasing, while when g, 2 0 and gs is decreasing, &en g,(x, - C) is increasing and g,jx, - c) is decreasing. That is, when F’(X) is improved, then I;,(X) will be improved. Combining the above conclusions in the three cases, we can conclude that if P’(X) in improved then F’(X) wiqp‘be improved for any fixed c E X md any function f(x) E Cl as well as any interval X c pi. m Theorem 2.1 is completely proved now.
COROUAIW. Forajbedc~X~Rwe
haue (2.17)
For a function Ax) from R” to R and an interval X=(X1,..., X#, where Xi (i=l,2,..., n) are given in (l.l), the linear interval extension of flIx) on X is
F,(X)=f(c)+VF(x)=(X-c), where c =(x1,...,
c,)~EX,
CiEXi, i=l,2,...,
r~,Ed [ X11 -
W(X) =
..
k”z:p I nS
(2.18)
Cl, XlS --
1X2z -C2,X2s .
4
-c2l
. .
1 X nZ -cn~xnS-cnl
.
(2.19)
Linear IntervalExth
F,(x)=
173
[b(g,,g,,c),u(g,,g,,c)l,
(3.20) where
b(gz,gs,c)=f(C)+
i
~(giz(Xiz-C,)~giz(Xis-C,)~
i=l
giStxiZ - ciJ¶giStxiX of giz giZtxiS
< gis <
-
o for a
C~) -< ,esS!XiS
=
giZ(xiS
-
-
-
ci)}
l
t2*=)
C~M.II i, then giz(xiz - ci) >, gis(xis - ci) >, 0 0. Thus,
and
Ci) ~
Cij,
@23)
and so
ab(gz,gs*4 agiz
=
MS?,, .-_ &s, cj
S&z
xiS
= xis
-ci>,O
and
wgz~ iw)
z
0,
(2.25)
0.
(2.26)
a&S
- ci G 0
and
~f4gz~ g,,c)
~g,s =
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144
Therefore, if fora certain C, g:; < gis < 0, then b(g,, gs, C) and u(g,, gs, C) do not depend on gis, wwe if gi, is increasing, that is, v I;(X ) is improved, then b(g,, g,, c) is increasing and u(g,, g,, c) is decreasing, that is, F,(X) is improved. Similarly, we can prove the same conclusions in the cases g:,z\< 0 < gis and gis >, giz >, 0 for a certain i. Thus, we have proved the fouowing theorem.
THEOREM 2.2. Assume VI;(X) and X - c have the fm in (2.19) and F,(X) is d@& in (2.18). If vF(X) is irnpmed, that is, SOWUS OT all giz am i-d while some or cl.! _gfs are dmmmd, tkn Fe(X) will be inam&. Theorem 2.3 is very useful in the optimal choice of linear interval extensions:
THEOREM 2.3.
Fm a fixed v F(X) we have
f(X)c
n F,(x)-
(2.27)
CEX
PROoF. This is a straightfoxward consequence of (2.1).
REMARk An extended conclusion of Theorem 2.3 is that if there are intervals Fa(X) 2 flX) for (YE A and A may be any set, then (2.28)
holds for any X E a”.
3.
THE BEST INTERVAL EXTENSION
NOWby using (2.1), we can find many linear interval extensions Fc(X ) of jlr) on X for different c G X. Clearly, the best interval extension constructed
Linear &ma!
Extension
by Iinear intend
extensions in the l-dimensional case is
Fb(X) = n F,(X) CEX cm~x(f(c)+min(g,(r,-c),g,(r,-c), g,k4gs<“s--4})~ min (f(c) CEX
+max(g,(x,
- 4
g,(xs - 4,
(3 1) l
About FJX) we have the following theorem.
THEOREM~.~. Zfg+g,>,O, tIwn
[f(x,),f(x,)l
(3.2)
fbW= [fbs)9fh>l
(3.3)
&SW
=
l
If g,
If g, G 0 G g,, then &.DW = 4bW
(3.4)
n c%(X) n &b(X),
where fib(x)
= [f(c,) [f(c,)
+ g,bs + gsh
- 4s
fh)
+ gsbs -
(3.5)
cdl9
-cl),f~c,)+g,(xs-c,)l(I&lQd~
F2b(x)= i [f(c~)+g,(~,-c,),ffc,)+g,(~,-~2)1
(&
(3.6)
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Cl =
min(ci, ci),
c, = max[ci, c$
MO.
and
gzxs gs-gz ’
&ix, Ci =
c; =
gsxs- gzxz gs-gz ’
(3.9)
PROOF. If g, >,g, >,0, then f(x) >,0 for every x E X = [x,, xs], and so f(x) is monotonically increasing on X. Thus, (3.2) holds, similarly, if gs < gz < 0, then fix) is monotonically decreasing on X, and so (3.3) holds. We should point out that a similar derivation PXTKSSto the following derivation gives the same results. Ifgz
gz(xz- 4
20
g,(xs
gs(xs - 4 3 0
-
c) G 0
gs(xz - 4 G 0
(3.10)
and so
w( g,(x, - 49 g,(x, - 49&z - 4 iA% - 4) =
(3.11)
d{ gzbs- c),gg(x,-c)}.
Since the straight line expres& by y = gt(xs - c) in the c-y coordinate system increases from 0 at xs Mik the one expressed by y = gs(xz - c) decreases from 0 at xz, they intemt at a point ci which is the solution of the equations with respect to ct
gz(x,- 4 = dx, - 4,
(3.12)
i.e,
Ci =
wz - gzxs gs-g,
.
(3.13)
Note that since -g,zO, g,*,
xz =
-
gs>/oI
iad
XI<%
gz*z~ gs*z- gz*s
gs-gz
gs-gz
gs*s- gz*s
=ci<
a--gz
=x
S?
(3.14)
i.e. .
Ci E [XI, Xs]
E
X.
(3.15)
If follows from (3.10) that
ma{ gz(*z- 4, gzbs- 4, g&z - 4
d*s - c,)
= m={gzbz- 4, gsh - c))*
(3.16)
A similar analysisshows that the two strtight lines corresponding to the expressions in ( } in (3.16) intersect at cz’=
g,*s - gz*z 4%-
(3.17)
gz
and c,l E
[XI, xs] =x.
(3318)
Furthermore, if g, >/ lg,l, then -
Xrb
(3.19)
gz*z,
(3-W
g,(*, - XI) G gs(*s-
g,*, -
g,*s G gs*s-
and so gsxz - gzxs Ci =
gs-gz
G
ws - gz*z= ce’, gs-gz
(3.21)
while if g, < /gz(, then -
gzbs- *z)> f&s - *z)r
WW
and so
g,x, > &Xs -
gsxz -
Ci =
wz = C&
gs-gz
Gr-gz
(3-N
SuppOSegs >, lgzl, i.e. Ci < C;l.Then if C E [XI, Ci], then Ci =
4wz -
is,%> I
&s-gz
gs(x,- 4
2
c,
gzbs- 4,
(3W
and so
min{ g,(xs- 4,
&IX, - 4) = gzbs- c).
(3.26)
sixuihuly, c,l =
gsxs- gzxz >,C; Z C, gs-gz
(3.27)
i.e.
g,(x,- 4
2
k&q- 4,
(3.28)
andso m=fgz(zz - c-), gs(xs - c j) = g,(x, - c).
(3.29)
Thus, assuming &,(X) = [b,, ZQ],we have
since f’(c)-g+o
(3.31)
Linear IntervalExtension
179
always holds, while
since
c) - g,
fq
That
is, if
cE
[x,,ci], gz G 0 G gs,
G 0.
and lgzl
G gs, then
&(X) = [Aci) + gz(x,- cih f(ci)~+gs(xsIfg,a lgzl ad c E
ci)l*
@*W
[ci,c,11,then Ci =
g,(x,
wz - gzxs< \ c, gs-gz
-
c j G gz(xs-- c),
(3-w
(3.36j
and so
mh( gz(q - 4, g&z - cl} =:g&z - 4;
(3.37)
while
cz’ =
ws - gzxz
gs-gz
>
c,
(3.38)
i.e.
g,(xs- c) 3 f&; - 4s
(3.39)
and so m~(g,(lz-c),g~(~~-c))
=d%-4.
(3-m)
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180
Thus, assnming
F,,(X) = [be, u,],we
have
b, = c~~~~~lIf(c)+gs(~,-c)) ?
9
=f(G)+&r-G)>
@a
f'(c) - g,G 0; while (3.42)
also implies that
f&J=
That
(3.41)
cE~~l(f(c)+&(~s-c)) ,
is,if c E [ci,c;], g, F,(x)=
6
=f(c~)+gs(rs-c,‘)*
@*a
0 G g,, ad lgzl d gs, then
(3.44)
[f(ci~+gs(~z-c;),f(c~)+gs(~s-c~~l~
If g, 2 lgzl and C E [c& xs], then (3.36) implies that mit+z(xs
- CL gs(xz - c)) = &z
- c)*
(3.45)
Now
ci =
ws - gzxz( \ c, gs-gz
i.e. (3.47j and so
(3.48) Thus, assuming
u,],we have
Fab(X) = [bar
Now we turn cE
[xz,c,l],
to investigate
case in which
the
lgzl > g,,
5.
ci )
CL, If
then
ci =
wz -
lzz% > c
gs-gz
’
(3.52)
i.e.
g,(x,- 4
’
gz(xs- 4,
(3.53)
and so
min(g,(x, - c), g&z - 4) = gz(xs- 4;
(3.54)
while c,l =
g,x, - gzxz 4%-
gz
2
c,
(3.55)
i.e.
and so
m={gzbz- 49 i!ik(xs - 4) = gsh - c). Assuming F,,(X)
(3s7)
= [b,, u,], we have
since
f(c)- g,z 0;
(3.59)
FiENPU GE
18% while
FdO = In 4 + gzbs- 4,
+ gsbs-
f(G)
41
l
(3.61)
If c E [c& ci], then (3.54) still holds, but
c,l =
Ws - gzxz
(3.62)
G c,
gs-gz
4,
g&s - c) Q gz(xz-
(3-W
andso
mu{ gz(x,- 4 Assuming
g&s - c)>= gh - 4
(3-W
F2b(X) = [b,, u2], we have b2 =
~~~~~~l(Ac)+gz(xs-c)) =f(c;)+gz(%-cf), I
I
(3.65)
,
us=
That
is, if c E
c~~c;lm+gz(~z-c)~ L5
=f(c~)+gz(xr-c;).
[c;,ci], g, \<0 G g,, andigzl r
F,Jx)=
gs,
(3.66)
then
[ffci)+gz(~s-cf),f(c,')+gz(~z-c~)l~
(3.67)
Zf c E [ci, xs], then (3.35) to (3.37) hold and (3,46) to (3.48) hold, that is, we have
min{ gz(#s- 4, g&z - 4) = 4&z - 4
(3.68)
rnax(gz(~~-c),g~(x,-c))=g,(x,-c)~
(3.69)
Linetw interval Etiension
Assuming F’&,(X) = [b,,
183
u,],then
That is, if c E [Cf, Xs], gz < 6 G gs, and ]gll> gs9 then
&b(x)= [f(ci)+gs(x,-ci)~Aci)+g,(r,-ci)l*
(3.72)
Combining (3.34) W;th (3.61), (3.51) with (3.72), and (3.44) with (3.67), we see that (3.5) and (3.7) are true. Theorem 3.1 is completely proved now. q Theorem 3.1 shows that we do not need to caMate t-~elinear interval extension at every point c in [xz, Q], but only need to calculate f(x) at four points in order to obtain the best linear interval extension on an interval X = [x,, xs]. This analysis greatly reduces the amount 2-fcomputation.
EXAMPLE3.1. f(x)is
Suppose f(x) = 1 - x + x’, X = [6,2]. The exact range of
f(X) = [6.75,3].
F(X) = [ - 1,5]. The nested interval extension is P(x) = [ - 1,3]. We have the exact range interval of f(x),
f(‘x 1= Thus,
-1+2x,
xz=o, x,=2, g,= -
1, g, = 3,
Ci = 0.5,
F’(X) = [ - 1,3]. and so we obtain c; = 1.5.
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184
Using the formulae (3.4), we find the best linear interval extension Fb(X) = [ - 0.75,3.25]. If we combiue the nested interval extension and Fb(X), then we obtain a much better interval extension,
F,(X) = e(X)n
F&X) = [ - 0.75,3].
According to Theorem 2.1, if F’(X) is improved, then every F,( X ) will be improved and so Fb(X) will be improved. Thus, we should use the above procedure first to find a better F’(X) and then to fhd &(X ). IGrthermore, we cau repeat this procedure iu turn to find better F(“)(X), F(“-l)(X),..., F’(X) and finally find Fb(X). Now let us extend the above conclusion to the hensional case. Assume now
X=
1x219. x2sl .
I
and
vF(X)=
(3.73)
.
I
X”S1 LX “I: Then the liuear interval extension for a fixed C E X, C = (Cl, . . 2 Cn)T, is l
&(X)
= [b(gls
g,, 6, uk,,
b(g,sgssc)=f(c)+
gs, dL
i i=l
u(gIsgS9 ‘I= f(‘) + C i-l
ad
~~~gi~(xi~-ci)~gi~(xis-C~)~ .
maw;{ gil(xj, - ci),gi,(xiS- Ci),
Linear Interval Extension
18s
where
g,=
&z
as
g2z
g2, . .
I
and g,=
:
(3.76)
is
B’nz
For convenience, we denote
mj(c)=~(gjZ(XjZ-cj)~gjZ(rj~-cj)~gj~(”jZ-cj)~gj~(rj~-cj)}~ (3.77) Mi(c) = max( giZ(x,Z - ci)* giZ(%iS- ci)s
giS(%iZ
+ ci)9
giS(xiS
-
ci)}’
(3.78) As before, we know that
mi(
C)
=
i
min(giZ(xu-c,),g,s(xil-c~))y
g4SagiZa0s
~~(gis(ri~-c4),gis(xiZ-ci)),
giZG”Gg4S>
min(giZ(xis-ci),gis(~is-ci)},
giZ~giS~“~
max(giZ(ris-ci),gis(xi~-Ci)},
gi!$agiZaO*
M,(c)= max{Izi!dx4Z max(
&z(X4z
-
cj),giS(xiS-ci))~
giSaO&giZs
-
ci),giS(“iZ-cj)}~
giZGgiSGO*
13*7g)
(3e8o)
Noting that
ab(g,, g,d
af(c)
or
=--g,,20
aCi
aft4
-
- gis 6 0,
ac
(3.81)
i
iiC4
when gis>, O>,gjz wehave
C,l, =
gisX4z- g4zx4s !34s -
g4z
and
A
c,
42
=-
g,s*rs
-
g4zx4z (=9
&S -
g4z
’
186
RENPUGE
A longer derivation similar to that in the I-dimensional case and a combination of all possible cases give the following theorem.
THEOREM 3.2.
5iuppme
(3.83)
(3.84) where
Fdx)
= ftri,sl
ci j 1’ ‘i,Z)
+ CgijZfrijS
- ci iI),
.I t-
- cijl)
. 1. 1
[
f( X-k I I*‘ijl’ ‘t,S)
F26(x)= I ff xikSs cijIp
+ Cgy(rip
BijS(X4Z - ‘ijl)
ff yrsacz’ cij2s xi,S)
=
ff xi$$s cij2s “i,l) [
(3-s)
9
(gijS ’ lgijZl)
rimZ) + C
ij
L
F36(x)
1
+ C . tj
( gijZ(‘ijZ
(
- cijl)
(gijS G
igijd) i ’
gijS(‘ijS
- ‘ij2)
(gijS 2 I&,Zl)
gijZtxijS
- Ci,e)
( gijS G IgijZl)
+ C gijS(Xy
- cij2)}
)I ’
(3*86)
9
9
f(x. IkZ' cd529xi,!3 ) +
CgijZt’ijZ ij
- ‘ij2)
1 9
(3.87)
mad Ci,l=
m%;,, $21,
ci,2 = m4c
$2 ) ’
(3.88)
Linear hsvtd
Extension
187
AS proved in Section 2, Fb(X) will be improved if at kt one of [gil, gis] is improved, and we can use the same procedure as above to improve [gil, g,s] and even the interval extensbs of higher-order derivatives.
EXAMPLE3.2. Suppose f(+ xz) = x,(1 - x1 + x2) - xt and X = [O,l] x [0, 11.The exact range of f(z) on X is
f(X)=
[-kg*
The natural and nested interval extensions se the same: F(X)=
[ -1,2]
and
R(X)=
L-1,2].
We have
af
-=l-22w,+xs, ax,
af
-=xX,-2x,. a%
Their natural extensions are respectively
S(x)= 1
[ -1,23,
Z(x)=
[ -2,1].
2
Therefore we have
and
Using the formuia (3.84), we find the best linear interval extension
188
RENPU
GE
If we combine the nested interval extension and the best interval extension, we obtain a much better interval extension,
F,(X)=E(X)nF,(X)= [ -1,g. which has the exact lower bound of the range of $(x) on X.
4.
DISCUSSION
We have given the best linear interval extension of a function in the class C’. Note that here “the best” only means that among possible linear interval extensions constructed with the formula (2.1) it is the best one. There exist other ways to construct an interval extension of a function in CP, where p > 0 is an integer. Therefore, FE(X) is not the best interval extension of the function on X. The following questions naturally arise: (1) Can we find dim&~ the M stance, in the ldimensional case
quadratk interval extension-for
Fqc(X)=flc)+(f(c)+BF”(X)(X-c)}(X-c),
in-
(4..1)
where P”(X) is an interval extension of f’(x)? (2) We have seen that sometimes the natural interval extension and the nested interval extension may ‘be better than the best linear interval extension, or one of the lower and upper bounds may be better, as shown in Example 3.1. Thus, we should calculate the natural and nested interval extensions and use
F,(X)=F,(X)n&X)nF(X)
(4.2)
instead of Fb( X) as we did in Example 3.1, b a more accurate interval extension is a very important problem in both theoretical analysis and practical applications. The more accurate the interval extension is, the less computation is needed for a given tolerance. Thus, we should pay more attention on this problem. c&)S&iG
Linear Znterval Extensh
180
REF’ERENCES 1 R. E. Moore, Znfemt Andysis, Prentice-Ha& Englewood Cl& NJ., 1968. 2 R. E. Moore, M&hods and Applications of Interval Analysis, SIAM, Philade@ia, 1070. 3 H. Ratschek and J. Rokne, Computer Methods for the Range of hzctions, Ellis Horwood, 1984. 4 R. Krawczyk, Intervalhteigungen ti rationale Fm&iogim and zugeordnete zen5
trische Formen, Freiburger IntervaU-Rerichte,83/2, Fre!burg. R. Krawczyk and IL Nickel, Die zentrische Form in der Intemllarithmetik, ihre quadratische Konvergenz und ihre hkhsionsisotonie, Computing 28~117-132
(1082).
Interval Extension*
Renpu Get
Zntemutbnul Znstitute for Applied System Analysis A-2361 -4g, Austria
Transmitted by John Casti
ABSTRACT A basic problem in interval analysis is to find more accurate interval extension of a function on a given interval. The more accurate intewal extension is, the less computation is needed in the solution of a problem, and the more useful the analysis is in other applications. As the first step, this paper investigates the optima choice of linear interval extension. It is found that one only needs to calculate function values at two particular points in order to find the optimal interval extension generated by linear interval extensions.
1.
INTRODUCTION
R. E. Moore [l, 21 and others introduced a new tool for computational mathematics-interval analysis. Assume that f(x) isa function defined in R” X,JT is an interva?.in -%“,where and X=(X1,...,
xi = [xiZ,xiS,], xZ G %SS
i=1,2 ,..., n,
. (10
and Xi1 and xis are real numbers. Then we have the range interval of fix),
*Supportedjointly b y the North-South Dialogue Program of Austria and Mzrnational Institute for Applied System Analysis. ‘Cue& research scholar in the System aud Decision Science Program of International Institute for Applied System Analysis, henburg, Austria,and on leave from the Institute of Computational aud AppW Mathematics of Xi’an Jiaotong University, Xi’an, China. APPLIED MATZZZMATZCS AND COMPUT’ATZON30: MS- 189 (1989) Q Elsevies Science Publishing Co., Inc., 1989 655 Avenue of the Americas, New York, NY 10010
165
RENEW GE
166
which will be denoted by
f(X)and be defined to be
f(x) = (fb)lx
= xl
0 .2)
l
Any set to set mapping F(X) from R” to R such that and
f(X)sF(X)
f(x)=I;‘(x)
for XEX
. (13)
is called an inclusive interval extension of f(x) on X. Furthermore, if for two intervals X, and X, such that X, G X, we have
F(q) c F(X,),
0
0
then we say that F(X) is a monotonically inclusive interval extension of f(x) on X. In this paper we shall simply call a monotonically inclusive interval extension an interval extension. Clearly, the endpoints of F(X) are respectively a lower bound and an upper bound of the range of f(x) on X. An interval extension for a given function f(x) on a certain interval X is not unique. To obtain an interval extension of a certain elementary function f(x), some interval arithmetics [l, 21 have been introduced. Assume X = [xl, x,] and Y = [zJ~,ys]. Then we define
X+Y = -
[r,+yz,x,+ys] = (x+yJxEX
x = [-
xs, -
qz]= ( -
p-Y=[r,- y,,x,--yJ =
yEY}.
EY),
0 .5)
x},
0 .0)
(x+y[xa,Y-}s
. (17)
x(x E
xy = [En{ XZYZ, XSYS, X,Ys, = (rylxEX,
and y
“SY,)
a=(
XZYZ, XSYS, XZYS, XSYZ
>I 0 .9)
Linear Interval Eden&on
167
For the same interval X we define
X “E
[XI”&],
x+Oor
[x&x;],
xs
[OJXl”],
0 E X ad n is even;
n isodd, n iseven,
(1.10)
(1.11)
IV= m4lx,L Id
is called the absolute value of the interval X. It is easy to see that the following operation rules hold:
X+(Y+Z)=(X+Y)+Z,
(1.12)
X(YZ) =(XY)Z,
(1.13)
X+Y=Y+X,
(1.14)
XY=YX,
(1.15j
where X, Y, and 2 are intervals in R”. A real number a can be thought of as an interval [a, a]. Using this, we have
x*u=
[x,*u,x,*a]
(1.16)
which shifts X to the right (or left) by a units. We have
aX=
[q,a+J,
a 24 a
l[axs,q],
(1.17)
In particular, we have 0+x=x+0=x,
0x=x0=0,
1x=x1-x.
(1.18)
However, distributivity does not always hold. But we always have socaued
168
RENPUGE
subdistributivity, x(Y+z)~xY+xz.
(1.19)
in some special cases, the subdistributivity becomes distributivity: x(Y+Z)=xY+xZ
for real x,
X(Y+Z)=XY+XZ
if
(1.20)
YZ>O.
(1.21)
Note that in general we have X-X+0
and
X/X+1,
(1.22)
but the cancellation law holds, that is, X+Y=Z+Y
*
x=2.
(1-B)
By a symmetric interval X, we mean that x=-x;
(1.24
by the middle point m(X) of an interval X, we mean that m(x)
Xz+ Xs = 2;
(1.25)
and by the width w(X) of X, W(X) = xs - XI.
(1.26)
For other basic elementary functions sin t, e’, and log,(l+ t) we have the following interval operations: (1.27)
ex= [ err, log,(l+
(1.28)
e's],
X) = [log,(l+ x,),log,(l+
%)I
if
1+x+0.
(1.29)
Some simple ways to construct an interval extension of a function f(x) are: (1) Directly use intervals X, Y, 2,. . . respectively in the place of x, y, 2,. . . , and use interval operations. The resulting interval extension is called the natural interval extension, and is denoted by F(X). (2) Use as many as possible socalled nested forms, that is, use as many as possible the operations on the left in (1.19) rather than the ones on the right. Usually we can get better interval extension in this way because of the inclusion. The resulting ir$erval extension is called the nested interval extension, and is denoted by F(X). (3) If f(x) EC then we can use the interval Taylor formula, for instance, in the l-dimensional case with c E X,
F,(x)=fW+
“-‘p(c)(x-c)’
c i=
i!
1
+f(n)(X)(X-C)n n!
.
(1.30)
A simple interval extension of f(x) is
Fl(X)=f(c)+f’(X)(X-c)
(c=X),
(1.31)
and we shall call it the linear interval extension of f(x). The ndimensional interval Taylor form& will be given later. Even though for the basic elementary functions we have their interval extensions with exact lower and upper bounds of their ranges on an interval X, the interval extensions of the elementary functions consisting of the basic elementary functions with arithmetic do not necessarily have exact lower and upper bounds of the ranges of those functions on X, but are more inaccurate in most cases. Therefore we need to divide X into more and more smaller and smaller intervals in order to obtain more accurate lower and upper bounds of the functions, and this increases the amount of computation very quickly. Thus, it is important to find an optimal interval extension of a function in a class of its interval extensions, either for theoretical analysis or for practical application. This paper is ad&s& to constructing an optimal interval extension from the linear interval extensions and the natural extension as well as the nested interval extension. Some research on improving the interval extension has been done by Ratschek and Rokne [3], Krawzcyk [4], and others. They are concerned with how to choose the formula rather than how to choose c E X.
RENPUGE
170
In the next section we give some basic theorems as the basis of our work. We consider this problem for the l-dimensional case and the ndimensional case in Section 3. Section 4 is a short discussion. 2.
BASIC THEORY
We shall present some basic theorems which are the basis for the investigation of the optimal choice of linear interval eA”lension of a function f(x) E cr. First, we notice that finding the exact f(X) is as difficult as finding the exact j(X) in the general case. Thus, usually we use an interval extension F’(X) of f(x) in the place of f(X) in (1.31). That is, we use ~(X)EF,(X)=~(C)+F'(X)(X-c).
(2.1)
The following theorem is basic.
THEOREM 2.1. If F’(X) is improved, then F,(X) will be improved for ~_yll~ c E X and any jb2cti4m f(x) E Cl as well as any jhed .
.
GOOF.
AssumeF’(X) = [g,, gs]. Then
F,(x) = [f(c)+~(&b,
- C)Yg,(XS - 4 &I
- 4ds(% - 4)Y
(2.2) Denote the lower and upper bounds of F,(X) respectively as b(gl, g,, c) and Mg,, gs, c). Then
b(g,,g,Yc)=f(c)+~(g,(~,-c)Ygl(xs-c)YgS(~I-C),gS(XS-C)),
(2.3) u(gIygs,c j = f(c)+m=( gh - 4, gAx,- 4, i&(x,- 4 id% - C>)-
(2.4)
Linear Intenaal Extension then 02gz(xz-c)2g&-C)
lf O
-
171
and OGdXs-Ch
c). Thus,
min(&(Xz- c), g,(xs - c), g,(xl- 49 &s - 4) = dxz - 4, (2.5) =gs(xs-C)s C2-6)
max(gz(r,-C),g~(XS-C)~~S(X~-C),gS(XS-C))
and so
(2.7) (29 0
Therefore
ab(gz’gs’c) agz
ab(&&,c)
a4gzsid
aa,, g,,c)
-_O
=o
and
md
agz
ags
=xz
(2.9)
-c
-cao. =xs
(2.10)
These mean that if 0 G g, \< gs, then b(gz, gs,c) and u(gz, gs,c) do nof depend on g,, and b(gz, g,,c) is a decreasing function of gs while u(g,, g,, c) is an increasing function of g,. Thus, when gs is decreased, that is, F’(X) is improved, then F,(X) will be improved.
If gz
then O
and Oags(xs-c)a
min(gz(xz-c),g~(x~-c),g~(x~-~),g~(x~-~)}
=gAxs-C),
(2.11)
max(gz(xz-c),g~(x~-c),g~(;,-~)~g~(x~-~))
“gkxz-C),
(2.12)
and so
b(g,, g,,c) = f(c) + gzbs
-
ch
(2.13) (2.14)
REIWU GE
wit,,&4
-c
agz
auk,, gs,4
and
ags
=%
(2.16)
o
=
l
TINS mean that if g, < gs G 0, then b(gz, gs,c) ad u(gz, gs, ~1 do not depend on gs, and &(gz,gs,e; is a~ ~hm&izg fwxtim of g; whik u(g,, g,, C) is a decreasing function of g,. Thus, when gz is increased, that is, F’(X) is improved, then F,(X)will be improved. If g, 6 0 ~g,, then g,(x, -c) < 0, gs(xz - c) G 0 and g& - C) 2 0, g,(x, - C) 2 0.When gz6 0 and g, is increasing, then gz(xs - c) is increashg and gz(rz -c) is decreasing, while when g, 2 0 and gs is decreasing, &en g,(x, - C) is increasing and g,jx, - c) is decreasing. That is, when F’(X) is improved, then I;,(X) will be improved. Combining the above conclusions in the three cases, we can conclude that if P’(X) in improved then F’(X) wiqp‘be improved for any fixed c E X md any function f(x) E Cl as well as any interval X c pi. m Theorem 2.1 is completely proved now.
COROUAIW. Forajbedc~X~Rwe
haue (2.17)
For a function Ax) from R” to R and an interval X=(X1,..., X#, where Xi (i=l,2,..., n) are given in (l.l), the linear interval extension of flIx) on X is
F,(X)=f(c)+VF(x)=(X-c), where c =(x1,...,
c,)~EX,
CiEXi, i=l,2,...,
r~,Ed [ X11 -
W(X) =
..
k”z:p I nS
(2.18)
Cl, XlS --
1X2z -C2,X2s .
4
-c2l
. .
1 X nZ -cn~xnS-cnl
.
(2.19)
Linear IntervalExth
F,(x)=
173
[b(g,,g,,c),u(g,,g,,c)l,
(3.20) where
b(gz,gs,c)=f(C)+
i
~(giz(Xiz-C,)~giz(Xis-C,)~
i=l
giStxiZ - ciJ¶giStxiX of giz giZtxiS
< gis <
-
o for a
C~) -< ,esS!XiS
=
giZ(xiS
-
-
-
ci)}
l
t2*=)
C~M.II i, then giz(xiz - ci) >, gis(xis - ci) >, 0 0. Thus,
and
Ci) ~
Cij,
@23)
and so
ab(gz,gs*4 agiz
=
MS?,, .-_ &s, cj
S&z
xiS
= xis
-ci>,O
and
wgz~ iw)
z
0,
(2.25)
0.
(2.26)
a&S
- ci G 0
and
~f4gz~ g,,c)
~g,s =
RENPUGE
144
Therefore, if fora certain C, g:; < gis < 0, then b(g,, gs, C) and u(g,, gs, C) do not depend on gis, wwe if gi, is increasing, that is, v I;(X ) is improved, then b(g,, g,, c) is increasing and u(g,, g,, c) is decreasing, that is, F,(X) is improved. Similarly, we can prove the same conclusions in the cases g:,z\< 0 < gis and gis >, giz >, 0 for a certain i. Thus, we have proved the fouowing theorem.
THEOREM 2.2. Assume VI;(X) and X - c have the fm in (2.19) and F,(X) is d@& in (2.18). If vF(X) is irnpmed, that is, SOWUS OT all giz am i-d while some or cl.! _gfs are dmmmd, tkn Fe(X) will be inam&. Theorem 2.3 is very useful in the optimal choice of linear interval extensions:
THEOREM 2.3.
Fm a fixed v F(X) we have
f(X)c
n F,(x)-
(2.27)
CEX
PROoF. This is a straightfoxward consequence of (2.1).
REMARk An extended conclusion of Theorem 2.3 is that if there are intervals Fa(X) 2 flX) for (YE A and A may be any set, then (2.28)
holds for any X E a”.
3.
THE BEST INTERVAL EXTENSION
NOWby using (2.1), we can find many linear interval extensions Fc(X ) of jlr) on X for different c G X. Clearly, the best interval extension constructed
Linear &ma!
Extension
by Iinear intend
extensions in the l-dimensional case is
Fb(X) = n F,(X) CEX cm~x(f(c)+min(g,(r,-c),g,(r,-c), g,k4gs<“s--4})~ min (f(c) CEX
+max(g,(x,
- 4
g,(xs - 4,
(3 1) l
About FJX) we have the following theorem.
THEOREM~.~. Zfg+g,>,O, tIwn
[f(x,),f(x,)l
(3.2)
fbW= [fbs)9fh>l
(3.3)
&SW
=
l
If g,
If g, G 0 G g,, then &.DW = 4bW
(3.4)
n c%(X) n &b(X),
where fib(x)
= [f(c,) [f(c,)
+ g,bs + gsh
- 4s
fh)
+ gsbs -
(3.5)
cdl9
-cl),f~c,)+g,(xs-c,)l(I&lQd~
F2b(x)= i [f(c~)+g,(~,-c,),ffc,)+g,(~,-~2)1
(&
(3.6)
RENPUGE
Cl =
min(ci, ci),
c, = max[ci, c$
MO.
and
gzxs gs-gz ’
&ix, Ci =
c; =
gsxs- gzxz gs-gz ’
(3.9)
PROOF. If g, >,g, >,0, then f(x) >,0 for every x E X = [x,, xs], and so f(x) is monotonically increasing on X. Thus, (3.2) holds, similarly, if gs < gz < 0, then fix) is monotonically decreasing on X, and so (3.3) holds. We should point out that a similar derivation PXTKSSto the following derivation gives the same results. Ifgz
gz(xz- 4
20
g,(xs
gs(xs - 4 3 0
-
c) G 0
gs(xz - 4 G 0
(3.10)
and so
w( g,(x, - 49 g,(x, - 49&z - 4 iA% - 4) =
(3.11)
d{ gzbs- c),gg(x,-c)}.
Since the straight line expres& by y = gt(xs - c) in the c-y coordinate system increases from 0 at xs Mik the one expressed by y = gs(xz - c) decreases from 0 at xz, they intemt at a point ci which is the solution of the equations with respect to ct
gz(x,- 4 = dx, - 4,
(3.12)
i.e,
Ci =
wz - gzxs gs-g,
.
(3.13)
Note that since -g,zO, g,*,
xz =
-
gs>/oI
iad
XI<%
gz*z~ gs*z- gz*s
gs-gz
gs-gz
gs*s- gz*s
=ci<
a--gz
=x
S?
(3.14)
i.e. .
Ci E [XI, Xs]
E
X.
(3.15)
If follows from (3.10) that
ma{ gz(*z- 4, gzbs- 4, g&z - 4
d*s - c,)
= m={gzbz- 4, gsh - c))*
(3.16)
A similar analysisshows that the two strtight lines corresponding to the expressions in ( } in (3.16) intersect at cz’=
g,*s - gz*z 4%-
(3.17)
gz
and c,l E
[XI, xs] =x.
(3318)
Furthermore, if g, >/ lg,l, then -
Xrb
(3.19)
gz*z,
(3-W
g,(*, - XI) G gs(*s-
g,*, -
g,*s G gs*s-
and so gsxz - gzxs Ci =
gs-gz
G
ws - gz*z= ce’, gs-gz
(3.21)
while if g, < /gz(, then -
gzbs- *z)> f&s - *z)r
WW
and so
g,x, > &Xs -
gsxz -
Ci =
wz = C&
gs-gz
Gr-gz
(3-N
SuppOSegs >, lgzl, i.e. Ci < C;l.Then if C E [XI, Ci], then Ci =
4wz -
is,%> I
&s-gz
gs(x,- 4
2
c,
gzbs- 4,
(3W
and so
min{ g,(xs- 4,
&IX, - 4) = gzbs- c).
(3.26)
sixuihuly, c,l =
gsxs- gzxz >,C; Z C, gs-gz
(3.27)
i.e.
g,(x,- 4
2
k&q- 4,
(3.28)
andso m=fgz(zz - c-), gs(xs - c j) = g,(x, - c).
(3.29)
Thus, assuming &,(X) = [b,, ZQ],we have
since f’(c)-g+o
(3.31)
Linear IntervalExtension
179
always holds, while
since
c) - g,
fq
That
is, if
cE
[x,,ci], gz G 0 G gs,
G 0.
and lgzl
G gs, then
&(X) = [Aci) + gz(x,- cih f(ci)~+gs(xsIfg,a lgzl ad c E
ci)l*
@*W
[ci,c,11,then Ci =
g,(x,
wz - gzxs< \ c, gs-gz
-
c j G gz(xs-- c),
(3-w
(3.36j
and so
mh( gz(q - 4, g&z - cl} =:g&z - 4;
(3.37)
while
cz’ =
ws - gzxz
gs-gz
>
c,
(3.38)
i.e.
g,(xs- c) 3 f&; - 4s
(3.39)
and so m~(g,(lz-c),g~(~~-c))
=d%-4.
(3-m)
RENPUGE
180
Thus, assnming
F,,(X) = [be, u,],we
have
b, = c~~~~~lIf(c)+gs(~,-c)) ?
9
=f(G)+&r-G)>
@a
f'(c) - g,G 0; while (3.42)
also implies that
f&J=
That
(3.41)
cE~~l(f(c)+&(~s-c)) ,
is,if c E [ci,c;], g, F,(x)=
6
=f(c~)+gs(rs-c,‘)*
@*a
0 G g,, ad lgzl d gs, then
(3.44)
[f(ci~+gs(~z-c;),f(c~)+gs(~s-c~~l~
If g, 2 lgzl and C E [c& xs], then (3.36) implies that mit+z(xs
- CL gs(xz - c)) = &z
- c)*
(3.45)
Now
ci =
ws - gzxz( \ c, gs-gz
i.e. (3.47j and so
(3.48) Thus, assuming
u,],we have
Fab(X) = [bar
Now we turn cE
[xz,c,l],
to investigate
case in which
the
lgzl > g,,
5.
ci )
CL, If
then
ci =
wz -
lzz% > c
gs-gz
’
(3.52)
i.e.
g,(x,- 4
’
gz(xs- 4,
(3.53)
and so
min(g,(x, - c), g&z - 4) = gz(xs- 4;
(3.54)
while c,l =
g,x, - gzxz 4%-
gz
2
c,
(3.55)
i.e.
and so
m={gzbz- 49 i!ik(xs - 4) = gsh - c). Assuming F,,(X)
(3s7)
= [b,, u,], we have
since
f(c)- g,z 0;
(3.59)
FiENPU GE
18% while
FdO = In 4 + gzbs- 4,
+ gsbs-
f(G)
41
l
(3.61)
If c E [c& ci], then (3.54) still holds, but
c,l =
Ws - gzxz
(3.62)
G c,
gs-gz
4,
g&s - c) Q gz(xz-
(3-W
andso
mu{ gz(x,- 4 Assuming
g&s - c)>= gh - 4
(3-W
F2b(X) = [b,, u2], we have b2 =
~~~~~~l(Ac)+gz(xs-c)) =f(c;)+gz(%-cf), I
I
(3.65)
,
us=
That
is, if c E
c~~c;lm+gz(~z-c)~ L5
=f(c~)+gz(xr-c;).
[c;,ci], g, \<0 G g,, andigzl r
F,Jx)=
gs,
(3.66)
then
[ffci)+gz(~s-cf),f(c,')+gz(~z-c~)l~
(3.67)
Zf c E [ci, xs], then (3.35) to (3.37) hold and (3,46) to (3.48) hold, that is, we have
min{ gz(#s- 4, g&z - 4) = 4&z - 4
(3.68)
rnax(gz(~~-c),g~(x,-c))=g,(x,-c)~
(3.69)
Linetw interval Etiension
Assuming F’&,(X) = [b,,
183
u,],then
That is, if c E [Cf, Xs], gz < 6 G gs, and ]gll> gs9 then
&b(x)= [f(ci)+gs(x,-ci)~Aci)+g,(r,-ci)l*
(3.72)
Combining (3.34) W;th (3.61), (3.51) with (3.72), and (3.44) with (3.67), we see that (3.5) and (3.7) are true. Theorem 3.1 is completely proved now. q Theorem 3.1 shows that we do not need to caMate t-~elinear interval extension at every point c in [xz, Q], but only need to calculate f(x) at four points in order to obtain the best linear interval extension on an interval X = [x,, xs]. This analysis greatly reduces the amount 2-fcomputation.
EXAMPLE3.1. f(x)is
Suppose f(x) = 1 - x + x’, X = [6,2]. The exact range of
f(X) = [6.75,3].
F(X) = [ - 1,5]. The nested interval extension is P(x) = [ - 1,3]. We have the exact range interval of f(x),
f(‘x 1= Thus,
-1+2x,
xz=o, x,=2, g,= -
1, g, = 3,
Ci = 0.5,
F’(X) = [ - 1,3]. and so we obtain c; = 1.5.
RENPU GE
184
Using the formulae (3.4), we find the best linear interval extension Fb(X) = [ - 0.75,3.25]. If we combiue the nested interval extension and Fb(X), then we obtain a much better interval extension,
F,(X) = e(X)n
F&X) = [ - 0.75,3].
According to Theorem 2.1, if F’(X) is improved, then every F,( X ) will be improved and so Fb(X) will be improved. Thus, we should use the above procedure first to find a better F’(X) and then to fhd &(X ). IGrthermore, we cau repeat this procedure iu turn to find better F(“)(X), F(“-l)(X),..., F’(X) and finally find Fb(X). Now let us extend the above conclusion to the hensional case. Assume now
X=
1x219. x2sl .
I
and
vF(X)=
(3.73)
.
I
X”S1 LX “I: Then the liuear interval extension for a fixed C E X, C = (Cl, . . 2 Cn)T, is l
&(X)
= [b(gls
g,, 6, uk,,
b(g,sgssc)=f(c)+
gs, dL
i i=l
u(gIsgS9 ‘I= f(‘) + C i-l
ad
~~~gi~(xi~-ci)~gi~(xis-C~)~ .
maw;{ gil(xj, - ci),gi,(xiS- Ci),
Linear Interval Extension
18s
where
g,=
&z
as
g2z
g2, . .
I
and g,=
:
(3.76)
is
B’nz
For convenience, we denote
mj(c)=~(gjZ(XjZ-cj)~gjZ(rj~-cj)~gj~(”jZ-cj)~gj~(rj~-cj)}~ (3.77) Mi(c) = max( giZ(x,Z - ci)* giZ(%iS- ci)s
giS(%iZ
+ ci)9
giS(xiS
-
ci)}’
(3.78) As before, we know that
mi(
C)
=
i
min(giZ(xu-c,),g,s(xil-c~))y
g4SagiZa0s
~~(gis(ri~-c4),gis(xiZ-ci)),
giZG”Gg4S>
min(giZ(xis-ci),gis(~is-ci)},
giZ~giS~“~
max(giZ(ris-ci),gis(xi~-Ci)},
gi!$agiZaO*
M,(c)= max{Izi!dx4Z max(
&z(X4z
-
cj),giS(xiS-ci))~
giSaO&giZs
-
ci),giS(“iZ-cj)}~
giZGgiSGO*
13*7g)
(3e8o)
Noting that
ab(g,, g,d
af(c)
or
=--g,,20
aCi
aft4
-
- gis 6 0,
ac
(3.81)
i
iiC4
when gis>, O>,gjz wehave
C,l, =
gisX4z- g4zx4s !34s -
g4z
and
A
c,
42
=-
g,s*rs
-
g4zx4z (=9
&S -
g4z
’
186
RENPUGE
A longer derivation similar to that in the I-dimensional case and a combination of all possible cases give the following theorem.
THEOREM 3.2.
5iuppme
(3.83)
(3.84) where
Fdx)
= ftri,sl
ci j 1’ ‘i,Z)
+ CgijZfrijS
- ci iI),
.I t-
- cijl)
. 1. 1
[
f( X-k I I*‘ijl’ ‘t,S)
F26(x)= I ff xikSs cijIp
+ Cgy(rip
BijS(X4Z - ‘ijl)
ff yrsacz’ cij2s xi,S)
=
ff xi$$s cij2s “i,l) [
(3-s)
9
(gijS ’ lgijZl)
rimZ) + C
ij
L
F36(x)
1
+ C . tj
( gijZ(‘ijZ
(
- cijl)
(gijS G
igijd) i ’
gijS(‘ijS
- ‘ij2)
(gijS 2 I&,Zl)
gijZtxijS
- Ci,e)
( gijS G IgijZl)
+ C gijS(Xy
- cij2)}
)I ’
(3*86)
9
9
f(x. IkZ' cd529xi,!3 ) +
CgijZt’ijZ ij
- ‘ij2)
1 9
(3.87)
mad Ci,l=
m%;,, $21,
ci,2 = m4c
$2 ) ’
(3.88)
Linear hsvtd
Extension
187
AS proved in Section 2, Fb(X) will be improved if at kt one of [gil, gis] is improved, and we can use the same procedure as above to improve [gil, g,s] and even the interval extensbs of higher-order derivatives.
EXAMPLE3.2. Suppose f(+ xz) = x,(1 - x1 + x2) - xt and X = [O,l] x [0, 11.The exact range of f(z) on X is
f(X)=
[-kg*
The natural and nested interval extensions se the same: F(X)=
[ -1,2]
and
R(X)=
L-1,2].
We have
af
-=l-22w,+xs, ax,
af
-=xX,-2x,. a%
Their natural extensions are respectively
S(x)= 1
[ -1,23,
Z(x)=
[ -2,1].
2
Therefore we have
and
Using the formuia (3.84), we find the best linear interval extension
188
RENPU
GE
If we combine the nested interval extension and the best interval extension, we obtain a much better interval extension,
F,(X)=E(X)nF,(X)= [ -1,g. which has the exact lower bound of the range of $(x) on X.
4.
DISCUSSION
We have given the best linear interval extension of a function in the class C’. Note that here “the best” only means that among possible linear interval extensions constructed with the formula (2.1) it is the best one. There exist other ways to construct an interval extension of a function in CP, where p > 0 is an integer. Therefore, FE(X) is not the best interval extension of the function on X. The following questions naturally arise: (1) Can we find dim&~ the M stance, in the ldimensional case
quadratk interval extension-for
Fqc(X)=flc)+(f(c)+BF”(X)(X-c)}(X-c),
in-
(4..1)
where P”(X) is an interval extension of f’(x)? (2) We have seen that sometimes the natural interval extension and the nested interval extension may ‘be better than the best linear interval extension, or one of the lower and upper bounds may be better, as shown in Example 3.1. Thus, we should calculate the natural and nested interval extensions and use
F,(X)=F,(X)n&X)nF(X)
(4.2)
instead of Fb( X) as we did in Example 3.1, b a more accurate interval extension is a very important problem in both theoretical analysis and practical applications. The more accurate the interval extension is, the less computation is needed for a given tolerance. Thus, we should pay more attention on this problem. c&)S&iG
Linear Znterval Extensh
180
REF’ERENCES 1 R. E. Moore, Znfemt Andysis, Prentice-Ha& Englewood Cl& NJ., 1968. 2 R. E. Moore, M&hods and Applications of Interval Analysis, SIAM, Philade@ia, 1070. 3 H. Ratschek and J. Rokne, Computer Methods for the Range of hzctions, Ellis Horwood, 1984. 4 R. Krawczyk, Intervalhteigungen ti rationale Fm&iogim and zugeordnete zen5
trische Formen, Freiburger IntervaU-Rerichte,83/2, Fre!burg. R. Krawczyk and IL Nickel, Die zentrische Form in der Intemllarithmetik, ihre quadratische Konvergenz und ihre hkhsionsisotonie, Computing 28~117-132
(1082).