On polynomials of least deviation from zero in several variables

0.26 MB·English

by  Yuan Xu

To appear in J. Experimental Math.
ON POLYNOMIALS OF LEAST DEVIATION FROM ZERO IN
SEVERAL VARIABLES
4
0
YUANXU
0
2
n Abstract. Apolynomialoftheformxα−p(x),wherethedegreeofpisless
a thanthetotaldegreeofxα,issaidtobeleastdeviationfromzeroifithasthe
J smallestuniformnormamongallsuchpolynomials. Westudypolynomialsof
leastdeviationfromzeroovertheunitball,theunitsphereandthestandard
9
2 simplex. For d = 3, extremal polynomial for (x1x2x3)k on the ball and the
sphereisfoundfork=2and4. Ford≥3,afamilyofpolynomialsoftheform

] (x1···xd)2−p(x)isexplicitgiven andproved tobethe leastdeviation from
A zeroford=3,4,5,anditisconjectured tobetheleastdeviationforalld.
C
.
h
t 1. Introduction
a
m
Let Πd denote the space of polynomials of degree at most n in d variables and
n
[ we write Π = Π1. For d = 1, it is well-known that the 21−n multiple of the
n n
1 Chebyshev polynomial of the first kind
v T (x)=cosn(arccosx)=2n−1xn+q(x), q Π ,
6 n ∈ n−1
1 is the monic polynomial of least deviation from zero in Π in the space C[ 1,1];
n
−
4 that is,
1 inf xn p(x) =21−n T =21−n.
0 p∈Πn−1k − kC[−1,1] k nkC[−1,1]
4
Equivalently,we saythatxn 21−nT is the bestapproximationto xn inC[ 1,1].
0 Let Ω be a regionin Rd. F−or f Cn(Ω), the best approximationof f from−Πd in
h/ ∈ n
the uniform norm is the quantity
t
a
(1.1) E (f;Ω)= inf f p ,
m n p∈Πdn−1k − kC(Ω)
v: where f C(Ω) = maxx∈Ω f(x). We call p∗ an extremal polynomial for f if
k k | |
i E (f;Ω) = f p∗ . For x = (x ,...,x ) Rd and α = (α ,...,α ) Nd,
X wen define thke m−onokmC(iaΩl) xα = xα1 1xαd. dTh∈e degree of the m1 onomidal ∈xα 0is
r 1 ··· d
a α = α1 + ... + αd. If p∗(x) is an extremal polynomial for the monomial xα,
| |
wecallxα p∗(x)thepolynomialofleastdeviationfromzero. ForΩbeingaregion
in Rd, poly−nomials of least deviation are known only in the case that Ω is a cube.
We are interested in the case of the unit ball Bd = x: x 1 , where x is the
{ k k≤ } k k
usualEuclideannormof x, the unit sphereSd−1 = x: x =1 and the standard
{ k k }
simplex Td = x:x 0,...,x 0,1 x ... x 0 .
1 d 1 d
{ ≥ ≥ − − − ≥ }
Ford=2,theleastdeviationofxnym fromΠ2 inthespaceC(Ω)hasbeen
n+m−1
studiedforB2andT2(see,forexample,[Gearhart, 73],[Reimer, 77],[Newman and Xu,93],
Date:Feburary25,2003.
1991 Mathematics Subject Classification. 41A10,41A50,41A63.
Keywordsandphrases. Leastdeviationfromzero,Chebyshevpolynomial,bestapproximation,
severalvariables.
WorkpartiallysupportedbytheNationalScienceFoundationunderGrantDMS-0201669.
1
2 YUANXU
[Bojanov,Haussmann and Nikolov, 01]). Ford>2,theonlycaseknownisx x
1 d
···
on Bd and Sd−1, which is a polynomial of least deviation by itself. This is shown
recently in [Andreev and Yudin, 01],
(1.2) p∈iΠnddf−1kx1···xd−p(x)kC(Bd) =p∈iΠnddf−1kx1···xd−p(x)kC(Sd−1)
=kx1···xdkC(Sd−1) =d−d/2.
Inotherword,thebestapproximationofx x fromΠd isthezeropolynomial.
1··· d d−1
Finding polynomials of least deviation on these regions appear to be a difficult
problem. Only a handful of explicit non-trivial examples of extremal polynomials
for d 3 are known in the literature.
≥
In the present paper, we study the least deviation from zero for monomials of
lower degrees. We found extremal polynomials for (x x x )2 and (x x x )4 on
1 2 3 1 2 3
B3 and S2 and a family of extremal polynomials for x2 x2 on Bd and Sd−1,
1··· d
which are derived from the extremal polynomials for x x x and (x x x )2 on T3
1 2 3 1 2 3
and x x on Td, respectively. We give an explicit construction of this family
1 d
···
of polynomials and conjecture that they are the least deviation polynomials. The
conjecture is proved for d = 3,4,5. The result provides, we believe, the first non-
trivial example of polynomials of least deviation on these domains. For example,
we have
pi∈nΠf35kx21x22x23−p(x)kC(B3) =pi∈nΠf35kx21x22x23−p(x)kC(S3) =72−2
and the minimum is attained by the extremal polynomial R (x,y) defined by
3
R (x ,x ,x )=72x2x2x2 4(x2+x2+x2)+4(x4+x4+x4)2+1.
3 1 2 3 1 2 3− 1 2 3 1 2 3
Theleastdeviation,72−2,issurprisinglysmallinviewofthevalue33/2 forx x x .
1 2 3
Our proof is based on a general result for the Chebyshev approximation in
[Rivlin and Shapiro, 61], in which the best approximationelement is characterized
in terms of extremal signature. The most difficult part, however, is to identify a
correct extremal polynomial. There is no general method for this purpose. We
relied heavily on the computer algebrasystem Mathematica to test and verify con-
jectures. In retrospect, the explicit construction is natural and rather suggestive.
Forexample,R (x)agreeswiththeChebyshevpolynomialT (x)onthethreeedges
3 2
of the face of T3 defined by x +x +x = 1. The result allows first glimpse of
1 2 3
what an extremal polynomial in more than two variables may look like.
Thepaperisorganizedasfollows. Inthefollowingsectionwerecallthetheoretic
background needed to prove our result. The results for d = 3 are discussed in
Section 3 and those for d>3 are in Section 4.
2. Extremal signature and best approximation
We recall the characterization of the extremal polynomials in terms of the ex-
tremal signature. The study in [Rivlin and Shapiro, 61] is given in the general
setting of approximationfrom a finite dimensional subspaceof C(Ω) on a compact
Hausdorff space Ω. We shall restrict the statement to our setting.
Let Ω be aninfinite compact set in Rd. A signature σ on the setΩ is a function
with finite support, whose nonzero values are either +1 or 1. A signature σ is
−
3
called extremal with respect to Πd if there exists a subset in the support of σ
n S
and positive numbers λ , v , such that
v
∈S
λ σ(v)p(v)=0, for all p in Πd.
v n
vX∈S
Let r >0 be a fixed number. For each p Πd , we denote by (p;f) the set
∈ n−1 Sr
(p;f)= x Ω: f(x) p(x) =r .
r
S { ∈ | − | }
If r = f p , (p;f) is the set of extremal points of f p and we denote it
C(Ω) r
k − k S −
by (p;f).
S
The characterization of the best approximation of f from Πd is given by the
n
following theorem in [Rivlin and Shapiro, 61]:
Theorem2.1. Apolynomialp∗ inΠd satisfies f p∗ =E (f;Ω)ifandonly
n k − kC(Ω) n
if there exists an extremal signature σ with support in (p∗;f) such that σ(v) =
S
sign(f p∗)(v) for all v (p∗;f).
− ∈S
Thesufficientpartofthetheoremprovidesamethodtoverifyifapolynomialp∗
is extremal. One needs, however, to know the extremal polynomial in advance, as
the extremal signature is supported on the set (p∗;f) which depends on p∗. The
S
sufficientpartofthetheoremcanbeextendedtothe signaturesupporton (p;f),
r
S
inwhichr isnotnecessarily f p . Wewillusethisslightlyextendedversion,
C(Ω)
k − k
which we state in the following. A simple proof is included for completeness; see
[Rivlin and Shapiro, 61] for more details.
Theorem 2.2. Suppose there exists a polynomial p∗ Πd and an extremal signa-
∈ n
ture σ supported on (p∗;f). Then E (f;Ω) r.
r n
S ≥
Proof. We can normalize the measure λ for the extremal signature so that it is
µ
a probability measure; that is, λ = 1. Let S(r) = S (p∗,f) in this
v∈Sr(p∗;f) v r
proof. Since λ p(v)=0 for anPy polynomial p Πd, we have
v ∈ n
P
f(x) p(x) λ f(v) p(v)
C(Ω) v
k − k ≥ | − |
X
v∈S(r)
λ σ f(v) λ σ p(v)
v v v v
≥(cid:12) X − X (cid:12)
(cid:12)v∈S(r) v∈S(r) (cid:12)
(cid:12) (cid:12)
= λ σ f(v) = λ σ (f(v) p∗(v))
v v v v
(cid:12) X (cid:12) (cid:12) X − (cid:12)
(cid:12)v∈S(r) (cid:12) (cid:12)v∈S(r) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
= λ f(v) p∗(v) =r λ =r.
v v
| − |
X X
v∈S(r) v∈S(r)
where we have used the fact that f(v) p∗(v)=σ r for v S(r). (cid:3)
v
− ∈
The extension allows us to apply the result to the situation where a good can-
didate for p∗ is identified but the norm of f p∗ is hard to determine. This is
−
precisely our case in Section 4.
Ourconstructionismotivatedbytherecentstudyin[Andreev and Yudin, 01],in
whichitisshownthatiff isinvariantunderafinitegroupG(thatis,f(xg)=f(x)
for all g G), then the best approximation E (f;Sd−1) is attained at G invari-
n
∈
ant polynomials. (This result appeared early in [Ganzburg and Pichugov,81] as
4 YUANXU
pointedoutbyareferee.) Moreprecise,westatetheresultin[Andreev and Yudin, 01]
as follows:
Proposition 2.3. Let G be a subgroup of the rotation group O(d) and let GΠd
n
denote the polynomials in Πd that are invariant under G. If f is invariant under
n
G, then
p∈iΠndnf−1kf(x)−p(x)kC(Sd−1) =p∈GinΠfdn−1kf(x)−p(x)kC(Sd−1).
Using this fact, the best approximation of several invariant functions are given
in [Andreev and Yudin, 01], including the case x x in (1.2) (invariant under
1 d
···
thesymmetricgroup). Theproofin[Andreev and Yudin, 01]canbeappliedtoany
region Ω and f that are invariant under a finite group G. In particular, if f is
invariant under a subgroup G of the symmetric group S (Td) of the simplex Td,
d
then an extremal polynomial of f can be taken as a G-invariant polynomial.
If f is even in each of its variables, then f is invariant under the sign changes
of each variable (invariant under the group Zd); the extremal polynomial can be
2
taken as a polynomial even in each of its variables. Furthermore, instead of Bd or
Sd−1, we can work with Td and Td−1 in this case. In fact, the following general
proposition holds:
Proposition 2.4. Let α Nd and write 2α = (2α ,...,2α ) and α = n.
∈ 0 1 d | |
If p∗(x) is an extremal polynomial for E (xα;Td) then p∗(x2,...,x2) is an ex-
n 1 d
tremal polynomial for E (x2α;Bd); conversely, if q∗ is an extremal polynomial for
n
E (x2α;Bd) in the form q∗(x)= p∗(x2,...,x2), then p∗(x) is an extremal polyno-
n 1 d
mial for E (xα;Td). Furthermore, let f (x ,...,x ) = xα1 xαd−1(1 x
n α 1 d−1 1 ··· d−1 − 1 −
... x )αd; then theabove conclusion holds for E (f ;Td−1)and E (x2α;Sd−1).
d−1 n α n
−
We note that f (x2,...,x2 ) = x2α on Sd−1. The proposition follows easily
α 1 d−1
from the fact that x (x2,...,x2) is one-to-one from Td to Bd = x Bd :x
7→ 1 d + { ∈ i ≥
0 ,andthemapalsoinducesaone-to-onemappingfromΠd toGΠd withG=Zd,
} n 2n 2
thatis,thesubspaceofpolynomialsthatareevenineachofitsvariables. Ford=2
the proposition has been used in [Bojanov, Haussmann and Nikolov, 01]. The cor-
respondence between polynomials on these domains also works for other problems
involving polynomials, such as orthogonal polynomials and cubature formulae, see
for example [Xu, 98].
3. Least deviation from zero for d=3
We considerbestapproximationto the monomialsx x x and(x x x )2 inthis
1 2 3 1 2 3
section. The main task is to identify an extremal polynomial. The results in the
previous section provide some guidance, but there is no general method for this
purpose. Our first example, R (x) given below, was found after many attempts.
3
See the comments after the proof.
Theorem 3.1. Define the polynomial R (x) by
3
R (x)=72x x x 4(x +x +x )+4(x +x +x )2 8(x x +x x +x x )+1.
3 1 2 3 1 2 3 1 2 3 1 2 2 3 1 3
− −
Then 72−1R (x) is a polynomial of least deviation from zero and
3
E2(x1x2x3;T3)=E2(x1x2(1 x1 x2);T2)=72−1 R3 C(T3) =72−1
− − k k
5
Furthermore, 72−2R (x2,x2,x2) is a polynomial of least deviation from zero and
3 1 2 3
E (x2x2x2;B3)=E (x2x2x2;S2)=72−2 R (x2,x2,x2) =72−2.
5 1 2 3 5 1 2 3 3 1 2 3 C(B3)
(cid:13) (cid:13)
Proof. ByProposition2.4weonlyneedtowork(cid:13)withthesimple(cid:13)x. Itiseasytoverify
that R (0,0,0) = 1 and R (1/2,1/2,0) = 1. Solving the equations ∂ R (x) =
3 3 i 3
−
0, i = 1,2,3, shows that R has 4 critical points inside T but none of them is
3 3
maximum or minimum, since the values of R (x) at these points are less than 1.
3
| |
Thus, R (x) attainsits maximumonthe boundaryofT3. It is easyto verifythat
3
| |
the polynomial R (x) satisfies
3
R (x,y,0)=R (x,0,y)=R (0,x,y)=(1 2x)2+(1 2y)2 1
3 3 3
− − −
whichis boundedby1 inabsolutevalue. Hence,weonly needtoshowthat R (x)
3
| |
isboundedbyoneonthefaceofT3 definedbyx +x +x =1;thatis,weneedto
1 2 3
showthatU (x ,x )=R (x ,x ,1 x x )is boundedby 1inabsolutevalue on
3 1 2 3 1 2 1 2
− −
T2. Taking derivatives of U (x ,x ) and solving for the critical points shows that
3 1 2
it has 4 critical points inside T2, of which only the point (1/3,1/3) is a maximal,
U(1/3,1/3)=1. Furthermore, it is easy to verify that
U (x,0)=U (0,x)=U (x,1 x)=T (x);
3 3 3 2
−
that is, it agrees with Chebyshev polynomial of degree 2 on the boundary of the
triangle. Hence, U (x) 1. Furthermore, the above analysis also shows that
3
| |≤
:= x: R (x)=1 = (0,0,0),(1,0,0),(0,1,0),(0,0,1),(1/3,1/3,1/3)
+ 3
S { | } { }
:= x: R (x)= 1 = (1/2,1/2,0),(1/2,0,1/2),(0,1/2,1/2) .
− 3
S { | − } { }
Let σ(v) = 1 on (0,0,0) and σ(v) = 1 on . We show that σ is an
+ −
S \{ } − S
extremal signature. Define L f and L f by
1 2
3 1 1 1 1
L f = f , , + (f(1,0,0)+f(0,1,0)+f(0,0,1))
1
4 (cid:18)3 3 3(cid:19) 12
and
1 1 1 1 1 1 1
L f = f , ,0 +f ,0, +f 0, , .
2
3(cid:18) (cid:18)2 2 (cid:19) (cid:18)2 2(cid:19) (cid:18) 2 2(cid:19)(cid:19)
Then Lf :=L f L f satisfies Lf =0, f Π3. Thus, σ is an extremal signature
1 − 2 ∈ 2
forx x x inC(T3). Furthermore,thesupportsets and ofσ areontheface
1 2 3 + −
S S
of T3, which is identified with T2. This shows that σ is also an extremal signature
for x x (1 x x ) in C(T2). (cid:3)
1 2 1 2
− −
Letusmentionaconnectionbetweencubatureformulaeandthe extremalsigna-
ture for R (x). A cubature formula is a linear combination of function evaluations
3
that gives an approximation to an integral ([Stroud, 71]). Let dµ be a positive
measure on Ω Rd. If
⊂
N
f(x)dµ= λ f(x ), f Πd,
Z k k ∈ n
Ω kX=1
and there is at least one f Πd such that the equality fails, then the cubature
∈ n+1
formulaissaidtobeofdegreen. Itiscalledpositive,ifallλ arepositivenumbers.
k
FortheextremalsignatureforR (x), itiseasytoverifythatbothL f andL f are
3 1 2
6 YUANXU
cubature formulae of degree 2 for dx on the set Σ2 = x T3 :x +x +x =1 ;
1 2 3
{ ∈ }
that is,
f(x)dx=L f =L f, for all f in Π2.
ZΣ2 1 2 2
Since we identify Σ2 with T2, one can write L and L as linear combinations of
1 2
function evaluations for functions oftwo variables. Thus, the extremalsignature is
given by the difference of two positive cubature formulae.
During our searchfor R (x), we found U (x) first. In retrospect, the formula of
3 3
U (x), which can be written as
3
U (x ,x )=72x x (1 x x ) 3+4(x2+x2+(1 x x )2),
3 1 2 1 2 − 1− 2 − 1 2 − 1− 2
is quite natural since it agrees with the Chebyshev polynomials of degree 2 on the
boundary of T2. This also suggests the possibility that other monomials may also
have extremal polynomials that agree with Chebyshev polynomials on the bound-
ary of T3. For example, for (x x x )n, one may look for a polynomial that agrees
1 2 3
with Chebyshev polynomials of n-th degree on the boundary of the simplex. One
example of such a polynomial is T (R (x)), where T (t) denotes the Chebyshev
n 3 n
polynomial of degree n. Although this function is not a polynomial of least devia-
tion, it helps us to find a solution for the monomial (x x x )2.
1 2 3
Theorem 3.2. Define polynomial R (x) by
5
R (x)=272b(x x x )2 1+2(x +x +x ) 2(x +x +x )2
5 1 2 3 1 2 3 1 2 3
− −
+2 1 4(x +x +x )+4(x2+x2+x2) 2
− 1 2 3 1 2 3
27(cid:2)x x x (32/9 2a+b)(x +x +x )(cid:3)2+6a(x x +x x +x x ) .
1 2 3 1 2 3 1 2 1 3 2 3
− −
(cid:2) (cid:3)
Then 27−2b−1R (x) is a polynomial of least deviation from zero,
5
E5(x21x22x23;T3)=E5(x21x22(1−x1−x2)2;T2)=27−2b−1kR5kC(T3) =27−2b−1,
where the constant a,b and the reciprocal of the least deviation is given by
a=28.5926243, b=21.8935834, 272b=15960.4223.
Furthermore, 72−4b−2R (x2,x2,x2) is a polynomial of least deviation from zero,
3 1 2 3
E (x2x2x2;B3)=E (x2x2x2;S2)=72−4b−2 R (x2,x2,x2) =27−4b−2.
5 1 2 3 5 1 2 3 5 1 2 3 C(B3)
(cid:13) (cid:13)
Just like the case of R (x), the proof amoun(cid:13)ts to showing(cid:13)that R (x) 1 on
3 5
| | ≤
T3 andthereexistsanextremalsignature. Itisnotdifficultoncethe formulaofR
5
is identified. We first give an account on how R is discovered.
5
Following the construction of R (x), we look for a polynomial in the form of
3
U (x,y)=27xy(1 x y) 27bxy(1 x y)+3a(x2+y2+(1 x y)2) c
5
− − − − − − −
+2 3+4(x2+(cid:2) y2+(1 x y)2) 2 1 (cid:3)
− − − −
(cid:2) (cid:3)
thatwill be a polynomialof leastdeviationonT2 with leading monomialx2y2(1
−
x y)2. Note that the Chebyshev polynomial of degree 2 is T (t) = 2t2 1 and
2
− −
T (2t 1)= 3+4(t2+(1 t)2)). TheformofU ischosensothatontheboundary
2 5
− − −
of T2 it satisfies
U (x,0)=U (0,x)=U (x,1 x)=T (T (2x 1))=T (2x 1).
5 5 5 2 2 4
− − −
7
Wethenchoosec=2/9+a+bsothatU (1/3,1/3)=1. Itfollowsthat1 U (x,x)
5 5
−
can be factored as
1 U (x,x)=x(1 2x)(1 3x)2(64 54ax+27bx+162bx2).
5
− − − −
We need to choose a and b so that the last factor is positive for 0 x 1/2. One
≤ ≤
choice is to make this factor 2b(9x+d)2. This leads to a = 16(3 4d)/(3d2) and
−
b=32/d2. Atthispoint,itbecomesapparentthatthereneedtobemorepointson
whichU (x,y)= 1insideT2. WethereforesolvetheequationsU (x,x)= 1and
5 5
− −
U′(x,x)=0. Thisleadstod= 1.208972894,whichgivesthe valuesforaandbin
5 −
the theorem. It turns out that this choice does work out and U (x,y) 1 on T2.
5
| |≤
The final step is to identify the formula of R (x ,x ,x ) from that of U (x ,x )
5 1 2 3 5 1 2
with the requirement that R (x ,x ,1 x x ) = U (x ,x ) and R (x,y) 1
5 1 2 1 2 5 1 2 5
− − | | ≤
on T3. This step is not trivial since an additional multiple of (x +x +x )k to
1 2 3
any term in R (x) does not change the value of the polynomial on the face of T3
5
defined by x =1 x x . The polynomial U (x ,x ) is an extremal polynomial
3 1 2 5 1 2
− −
on the triangle T2 that agrees with the Chebyshev polynomials of degree 4 on the
three boundary lines of T2, its graph is depicted below.
1
0.5 1
0
0.8
-0.5
-1 0.6
00
00..22 0.4
00..44
00..66 0.2
00..88
0
1
The polynomial U
3
Let us point out that there does not seem to be a closed form for the values of
a and b. In fact, the value of d in the above paragraph is one of the real roots of
the following polynomial:
612220032 1365527808t 835528041t2 101556504t3
− − − −
[phone]t4+26037504t5+7670016t6+929280t7+41984t8.
This polynomial has 4 real roots and 4 complex roots, and it cannot be factored
over the integers.
We now give a formal proof of Theorem 3.2.
8 YUANXU
Proof of Theorem 3.2. First of all, we need to show that R (x) 1 for x T3.
5
| | ≤ ∈
Solving ∂ R (x) = 0, i = 1,2,3, numerically for critical points shows that R (x)
i 5 5
| |
attains its maximum on the boundary of T3. Furthermore,
R (x,y,0)=R (x,0,y)=R (0,x,y)
5 5 5
= 1+2(x+y) 2(x+y)2+2(1 4(x+y)+4(x2+y2))2,
− − −
andthe polynomialhas no criticalpoint insideT2. Consequently,the maximum of
R (x) is attained on the face of T3 defined by x +x +x = 1. In other words,
5 1 2 3
| |
we only need to show that U (x ,x ) 1 on T2. Again this can be proved by
5 1 2
| | ≤
solving ∂ U (x ,x ) = 0, i = 1,2, and the maximum is attained on the boundary.
i 5 1 2
This provesthat R (x) 1onT3 anditalsogivesthe set = x: R (x) =1
5 + 5
| |≤ S { | | }
and the set = x : R (x) = 1 . Let S be the symmetric group of three
− 5 3
elements. FoSr a ={(a ,|a ,a )| R−3 }we define aτ := (a ,a ,a ), τ S and
1 2 3 ∈ τ1 τ2 τ3 ∈ 3
(a) := aτ :τ S . Then
G 3
{ ∈ }
= (1/3,1/3,1/3),(0,0,0),(1,0,0) ,(1/2,1/2,0) ,(t ,t ,1 2t )
+ G G 1 1 1 G
S { − }
= ((2 √2)/4,(2+√2)/4,0) ,(t ,t ,1 2t ) ,
− G 2 2 2 G
S { − − }
where t = 0.4588164122 and t = 0.1343303216. We consider the signature σ
1 2
defined by σ(v)=1, v (0,0,0) and σ(v)= 1, v . To show that σ is
+ −
∈S \{ } − ∈S
an extremal signature, we define Lf by
Lf =c f(1/3,1/3,1/3)+c f((1,0,0)τ)
0 1
Xτ
+c f((1/2,1/2,0)τ)+c f((t ,t ,1 2t )τ)
2 3 1 1 1
−
Xτ Xτ
c f(((2 √2)/4,(2+√2)/4,0)τ) c f((t ,t ,1 2t )τ),
4 5 2 2 2
− − − −
Xτ Xτ
where the sum is taken over all distinct permutations of the base point and the
coefficients are given by
c =0.0997251873, c =0.0097228135, c =0.0621246411,
0 1 2
c =0.0243979796, c =0.0615774830, c =0.1178707075.
3 4 5
Then Lf = 0 for all f Π3, which shows that σ is an extremal signature. This
completes the proof of T∈heo4rem 3.1. (cid:3)
ThelinearfunctionalLf givenaboveisevidentlyasumoftwolinearfunctionals
withpositivecoefficients. UnlikethecaseofR ,however,thetwolinearfunctionals
3
are not cubature formulas of degree 5 with respect to the Lebesgue measure.
Thetwocasessolvedinthissectionappeartoindicateasurprisinglycomplicated
pictureforthebestapproximationofmonomialsinthreevariables,andthepicture
is remarkably different from that of one and two variables. We make two remarks
in this regard.
Remark 3.1. One surprising fact of Theorem 3.2 is that the least deviation is not
given by a reciprocal of an integer. This indicates a major difference between the
caseofthreevariablesandthatofoneandtwovariables. Inthecaseofonevariable,
the polynomial of least deviation from zero is the classical Chebyshev polynomial,
9
for which the least deviation of xn to Π in C[ 1,1] is 21−n. In the case of two
n−1
−
variables, we know for example
En(xkyn−k;B2)= inf xkyn−k p(x) C(B2) =21−n.
p∈Π2n−1k − k
For three variables, however,we do not know if the least deviation of xα to Πd
|α|−1
could be represented by a simple formula that depends only on the total degree of
themonomial. Theresultinthissectionseemstoindicatethatsuchaformuladoes
not exist.
Remark 3.2. The values of the leastdeviationin Theorems 3.1 and3.2 are surpris-
ingly small. Let us examine the case of the unit ball. We know
(3.1) E (xkxn−k;B3)=E (xkxn−k;B3)=E (xkxn−k;B3)=21−n.
n 1 2 n 1 3 n 2 3
This follows from the fact that an extremal polynomial p∗ for xmxn must be even
1 2
inx sincexmxn is invariantunder the groupZ appliedonthe thirdvariable. Let
3 1 2 2
p∗ be so chosen; then
kx1n−mxm2 −p∗(x1,x2,x3)kC(B3) ≥kx1n−mxm2 −p∗(x1,x2,q1−x21−x22)kC(B2)
≥pi∈nΠf2nkx1n−mxm2 −p(x1,x2)kC(B2) =21−n.
Furthermore, the equality holds since an extremal polynomial for xn−mxm on B2
1 2
can also serve as an extremal polynomial on B3. Below is a list of other cases that
we know on the unit ball:
E (x x x ;B3)=3−3/2, E (x2x2x2;B3)=2−6 3−2,
n 1 2 3 n 1 2 3 ·
E (x4x4x4;B3)=0.5340799374 2−12 3−12,
n 1 2 3 · ·
where we rewrite the value of the third one, which is given in Theorem 3.2, for
easier comparision. The value of E (x2x2x2;B3) appears to be strikingly small.
n 1 2 3
For other degree 6 monomials given in (3.1), the value of the best approximation
is only 2−5. Also note the fast decrease shown in these three values.
4. Least deviation from zero for d>3
We consider the best approximationto (x x )2 on Bd or Sd−1, and the best
1 d
···
approximation to x x on Td or Td−1 in this section. The extremal polyno-
1 d
···
mial can be taken as symmetric polynomials by Proposition 2.3. It is well known
thatevery symmetric polynomialcan be written in terms ofelementary symmetric
polynomials ([Macdonald, 95]).
The elementary symmetric polynomials of degree k in variables x ,x ,...,x
1 2 N
are defined by
e (x)= x x x , 1 k N.
k i1 i2··· ik ≤ ≤
1≤i1<X...<ik≤N
In particular, e (x ,...,x ) = x + +x and e (x ,...,x ) = x x . As
1 1 N 1 N d 1 N 1 N
··· ···
it is often the case with the symmetric functions, we assume that N is sufficiently
large and do not write the dependence of e on the number of variables. We will
k
use the notation 1k =(1,1,...,1) Rk.
∈
10 YUANXU
Definition 4.1. Using elementary symmetric functions, define T (x) by
3
T (x)=72e (x) 4e (x)+4e2(x) 8e (x)+1
3 3 − 1 1 − 2
and T (x) for k>3 by the recursive formula
k
T (x)=r e (x) T (x),
k k k k−1
−
where the constant r is determined by r =kk[T (k−11k)+1].
k k k−1
Note that k−11k = (k−1,...,k−1) Rk; we use the evaluation of T , as a
k−1
function of Rk, at this point in the defi∈nition of r . Clearly r is uniquely deter-
k k
mined. For x Rd, the function T (x) will serve as extremal polynomials. In
d
particular, the ∈polynomial T (x) for x R3 is the same as R (x) in the previous
3 3
section. For x R4, the explicit formula∈of T is given by
4
∈
T (x)=896x x x x 72(x x x +x x x +x x x +x x x )
4 1 2 3 4 1 2 3 1 2 4 1 3 4 2 3 4
−
+4(x +x +x +x ) 4(x +x +x +x )2
1 2 3 4 1 2 3 4
−
+8(x x +x x +x x +x x +x x +x x ) 1.
1 2 1 3 1 4 2 3 2 4 3 4
−
Thevalueofr is ofparticularimportance. Itcanbe computedusingthe following
d
formula.
Lemma 4.2. For d 3,
≥
d
d
r =d kd−3 ( 1)k(9k2 32k+24)+k2 .
d
(cid:18)k(cid:19) − −
Xk=4 (cid:2) (cid:3)
In particular, r =72, r =896, r =14400, and r =283392.
3 4 5 6
Wedefertheprooftotheendofthesectionandcontinuetostateourmainresult
of this section.
Theorem 4.3. For d 3, on the d-dimensional simplex
≥
E (x x ;Td)=E (x x (1 x x );Td−1) r−1,
d−1 1··· d d−1 1··· d−1 − 1−···− d−1 ≥ d
andtheequalityholds ford=3,4,5withr−1T (x)asapolynomialofleastdeviation
d d
from zero. Furthermore, on Bd and Sd−1,
E (x2 x2;Bd)=E (x2 x2;Sd−1) r−2
2d−1 1··· d 2d−1 1··· d ≥ d
andtheequalityholds withr−2T (x2,...,x2)asapolynomial ofleastdeviation from
d d 1 d
zero.
We believe that the equality still holds for d 6. In fact, all that is missing is
≥
to prove that T (x) 1 for x Td. We state it as a conjecture.
d
| |≤ ∈
Conjecture 4.4. For d≥6, the inequality kTd(x)kC(Td) ≤1 holds. In particular,
the equality in the above two theorems holds for d 6.
≥