The Mystery of the Shape Parameter

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by  Lin-Tian Luh

The Mystery of the Shape Parameter
Lin-Tian Luh
Department of Mathematics, Providence University
0
Shalu, Taichung
1
0 email:[email protected]
2
phone:[phone] ext. 15126

n fax:[phone]
a
J
January 28, 2010
8
2

] Abstract. It’s well known that the multiquadrics ( 1) β (c2+ x 2)β, β > 0, and the inverse
A − ⌈ ⌉ k k
multiquadrics (c2+ x 2)β, β <0, are very useful radialbasis functions for generatingapproximat-
N k k
ing functions. However the optimal choice of the shape parameter c is unknown. This question has
h. beenperplexingmanypeopleformanyyearsandisregardedasoneofthemostimportantquestions
t inthetheoryofRBF.Hithertothereareveryfewtheoreticalworksdealingwiththis topic. Instead,
a
m sporadicresults ofexperimentcanbe seenin the literature. The purposeofthis paper is to uncover
its mystery.
[

1 keywords: radial basis function, multiquadric, shape parameter
v
7
8 1 Introduction
0
5
Before entering the core of our theory, we would like to make a clarification of the definition. We
.
1 are not going to use the simple forms of the multiquadrics and inverse multiquadrics as mentioned
0
in the abstract. Instead, we define
0
1
β
v: h(x):=Γ (c2+ x2)β2, β R 2N 0, c>0 (1)
−2 | | ∈ \ ≥
i (cid:18) (cid:19)
X
,where x istheEuclideannormofxinRn,Γistheclassicalgammafunctionandc,β areconstants.
r | |
a The function h is called multiquadric or inverse multiquadric, respectively, depending on β > 0 or
β <0. The reasonofouradoptingΓisthatitwillmaketheFouriertransformofhandouranalytic
work simpler.
Ifβ >0,thenh(x)willbeconditionallypositivedefinite[8]oforderm= β where β denotes
⌈2⌉ ⌈2⌉
the smallest integer greater than or equal to β. If β < 0, h(x) is c.p.d.(conditionally positive
2
definite) of order m=0. All these can be found in [8].
For any approximated function f, our interpolating function will be of the form
N
s(x):= c h(x x )+p(x) (2)
i i
−
i=1
X
1
where p(x) P , the space of polynomials of degree less than or equal to m 1 in Rn, X =
m 1
∈ − −
x , ,x is the set of centers(interpolation points). For m = 0, P := 0 . We require that
1 N m 1
s{() i·n·t·erpol}ate f() at data points (x ,f(x )), ,(x ,f(x )). There−fore a{lin}ear system of the
1 1 N N
· · ···
form
N Q
c h(x x )+ b p (x )=f(x ), j =1, ,N (3)
i j i i i j j
− ···
i=1 i=1
X X
and
N
c p (x )=0 , j =1, ,Q (4)
i j i
···
i=1
X
, where p , ,p is a basis of P , has to be satisfied. Since h is c.p.d., this requirement will
1 Q m 1
be theor{etica·l·l·y sat}isfied [8]. Howev−er if c is very large, h will be numerically constant, making
the linear system (3),(4) numerically unsolvable. Moreover,if c is very large, the coefficient matrix
of the linear system will have a very large condition number, making the interpolating function s
unreliable when f(x ), ,f(x ) are not accurately evaluated, as pointed out by Madych in [7].
1 N
···
Our approach for choosing c is based on Theorem2.4 and Corollary2.5 of [4] which we will cite
directly but make a slight modification to make them easier to understand.
Before introducing the main theorem, we need some fundamental definitions. Let (Rn) denote
D
the space of complex-valued functions on Rn that are compactly supported and infinitely differen-
tiable. For any integer m 1, let
≥
= ϕ (Rn): xαϕ(x)dx=0 for all α <m .
m
D { ∈D | | }
Z
If m=0, := (Rn).
m
D D
Definition 1.1 Let h be as in (1) and m = max 0, β . We write f (Rn) if f C(Rn)
{ ⌈2⌉} ∈ Ch,m ∈
and there is a constant c(f) such that for all ϕ in ,
m
D
1/2
f(x)ϕ(x)dx c(f) h(x y)ϕ(x)ϕ(y)dxdy . (5)
≤ −
(cid:12)Z (cid:12) (cid:26)Z Z (cid:27)
(cid:12) (cid:12)
If f h,m(Rn), we(cid:12)(cid:12)let f h to den(cid:12)(cid:12)ote the smallest constant c(f) for which (5) is true.
∈C k k
Thefunctionspace (Rn),abbreviatedas ,iscallednativespacewhosecharacterization
h,m h,m
C C
can be found in [2],[3],[5],[6] and [8].
Definition 1.2 For n = 1,2,3, , the sequence of integers γ is defined by γ = 2 and γ =
n 1 n
···
2n(1+γ ) if n>1.
n 1
−
Definition 1.3 Let n and β be as in (1). The numbers ρ and ∆ are defined as follows.
0
(a) Suppose β <n−3. Let s=⌈n−β2−3⌉. Then
(i) if β <0, ρ= 3+3s and ∆0 = (2+s)(ρ12+s)···3;
(ii) if β >0, ρ=1+ 2 βs+3 and ∆0 = (2m+2+s)(2ρm2m++1+2s)···(2m+3)
⌈2⌉
where m= β .
⌈2⌉
(b) Suppose n 3 β <n 1. Then ρ=1 and ∆ =1.
0
− ≤ −
2
(c) Suppose β n 1. Let s= n−β−3 . Then
≥ − −⌈ 2 ⌉
1 β
ρ=1 and ∆ = where m= .
0 (2m+2)(2m+1) (2m s+3) ⌈2⌉
··· −
Theorem 1.4 Let h be defined as in (1) and m = max 0, β . Then given any positive number
{ ⌈2⌉}
b , there are positive constants δ and λ, 0<λ<1, which depend completely on b and h for which
0 0 0
the following is true: For any cube E in Rn of side length b , if f and s is the map defined
0 h,m
∈C
as in (2) which interpolates f on a finite subset X of E, then
f(x) s(x) 2n+4β+1πn+41√nαncβ2 ∆0(λ)1δ f h (6)
| − |≤ k k
holds for all 0 < δ δ and all x in E provided that δ = d(pE,X) := sup inf y x. Here,
α denotes the volu≤me0of the unit ball in Rn, and c, ∆ were defined yi∈nE(1)xa∈nXd|D−efin|ition1.3
n 0
respectively. Moreover δ0 = 6Cγn(1m+1), and λ=(23)6C1γn where
2 ρ
C =max 2ρ′√ne2nγn, , ρ′ = .
3b c
(cid:26) 0(cid:27)
The integer γ was defined in Definition1.2, and f is the h-norm of f in , as defined in
n h h,m
k k C
Definition1.1. The constant ρ was defined in Definition1.3.
Remark: Theorem1.4 is cited directly from [4] with only a slight modification. The proof is very
technical. The main contribution of the theorem is that it uncovers the mystery of λ and δ whose
0
values were unknown. Both numbers appear in the currently used exponential-type error bound
for multiquadric interpolation which was only an existence theorem. Obviously the domain E in
Theorem1.4 can be extended to a more general set Ω Rn which can be expressed as the union of
⊆
rotations and translations of a fixed cube of side b .
0
In (6) it’s clearly seen that the error bound is greatly influenced by the shape parameter c.
However, in order to present useful criteria for the choice of c, we still need a few theorems.
Definition 1.5 For any σ >0, the class of band-limited functions f in L2(Rn) is defined by
B = f L2(Rn): fˆ(ξ)=0 if ξ >σ
σ
{ ∈ | | }
, where fˆdenotes the Fourier transform of f.
Theorem 1.6 Let h be as in (1) with β >0. Any function f in B belongs to and
σ h,m
C
f h m!S(m,n)2−n−1+4βπ−n−14σ1+β4+nec2σc1−β4−n f L2(Rn) (7)
k k ≤ k k
where c, β are as in (1) apnd S(m,n) is a constant determined by m and n.
Proof. We are going to show B by Corollary3.3and Theorem5.2 of [6].
σ h,m
⊆C
Letm= β . Thentherequirementa =0forall γ =2minCorollary3.3of[6]isanimmediate
⌈2⌉ γ | |
resultof Theorem5.2of [6]. Since B L2(Rn), any f B is a member of where denotes the
σ σ ′
⊆ ∈ S S
Schwarzspace. Now, let ρ(ξ) be the Borelmeasurementioned in Corollary3.3of[6]. By (3.9)of[6],
it suffices to show that
1/2
m!
f := (Dαf)ˆ 2 <
k kh  α! k kL2(ρ) ∞
|αX|=m(cid:18) (cid:19) 
 
3
for all f B . We proceed as follows.
σ
∈
1/2
m!
(Dαf)ˆ(ξ)2dρ(ξ)
 α! | | 
|αX|=m ZRn 
1/2
 m! 
= imξαfˆ(ξ)2dρ(ξ)
 α! | | 
|αX|=m ZRn 
1/2
 m! 
= ξ2α fˆ(ξ)2dρ(ξ)
 α! | | 
|αX|=m ZRn 
1/2
 1  1
= (m!)1/2 ξ2α fˆ(ξ)2 dξ (by (3.8) and Theorem5.2 of [6])
|αX|=mα! ZRn | | (2π)2n|ξ|2mhˆ(ξ) 
1/2
√m!  1 ξ2α fˆ(ξ)2 1 
= | | dξ (by Theorem8.15 of [8])
(2π)n |αX|=mα!ZRn |ξ|2m · 21+β2(|ξc|)−β2−n2Kn+2β(c|ξ|) 
≤ (√2πm)n! ·2−(12+β4)(S(m,n)ZRn|fˆ(ξ)|2· |ξc|nn+2+2ββ · Kn+2β1(c|ξ|)dξ)1/2 where S(m,n) denotes the
number of terms in the “ ”
X 1/2
m!S(m,n) c−(n4+β) fˆ(ξ)2 ξ n+2β c|ξ| dξ (by Corollary5.12 of [8])
≤ p(2π)n221+β4 · (ZRn| | | | · pπ2e−c|ξ| )
1
= p(2mπ)!Sn2(m21+,nβ4) ·c1−(n4+β) rπ2!2 (cid:26)ZRn|fˆ(ξ)p|2|ξ|1+n2+βec|ξ|dξ(cid:27)1/2
m!S(m,n) 2−n−14−β4π−14−nσ1+n4+βec2σc1−(n4+β) f L2(Rn)
≤ · k k
< .
p
∞
ThusB and(7)follows. ♯
σ h,m
⊆C
Ifin(1)β <0,thenorm isdefinedinaslightlydifferentway. Hencewehandleitseparately
h
k·k
and present the following theorem.
Theorem 1.7 Let h be as in (1) with β < 0 such that n+β 1 or n+β = 1. Any function f
≥ −
in B belongs to and satisfies
σ h,m
C
f h 2−n−1+4βπ−n−14σ1+β4+nec2σc1−(n4+β) f L2(Rn). (8)
k k ≤ k k
Proof: If β <0, then h is conditionally positive definite of order m =0 [8]. By Theorem5.2 of [6],
we know that in Corollary3.3of [6] a =0 for γ =2m. Obviously any f B belongs to since
γ σ ′
| | ∈ S
f L2(Rn). In order to apply Corollary3.3 of [6] to show that B , it remains to show that
σ h,m
∈ ⊆C
f B implies fˆ L2(ρ) where dρ(ξ):=r(ξ)dξ = 1 dξ as stated in [6].
∈ σ ∈ (2π)2nhˆ(ξ)
4
Now, let f B . By (3.9) of [6], it suffices to show that f := fˆ < . We proceed as
σ h L2(ρ)
∈ k k k k ∞
follows.
1/2
fˆ = fˆ(ξ)2dρ(ξ)
L2(ρ)
k k | |
(cid:26)ZRn (cid:27)
1/2
= fˆ(ξ)2r(ξ)dξ
| |
(cid:26)ZRn (cid:27)
1 1/2
= fˆ(ξ)2(2π)−2n dξ where w()=hˆ() by Theorem5.2 of [6]
| | · w( ξ) · ·
(cid:26)ZRn − (cid:27)
1/2
= fˆ(ξ)2(2π)−2n2−1−β2(|ξ|)β+2n 1 dξ by (8.7), p.109 of [8]
(ZRn| | c · Kβ+2n(c|ξ|) )
1/2
2−1−β2−2nπ−2nc−(n+2β) |fˆ(ξ)|2|ξ|n+2β dξ if n+β 1 by Corollary5.12
≤  ZRn π2 · e√−cc|ξξ|  | |≥
| |
of [8] p
 
1/2
= 2−21−β2−2nπ−2n−12c21−n+2β fˆ(ξ)2 ξ n+2β+1ec|ξ|dξ
| | | |
(cid:26) ZRn (cid:27)
1/2
2−12−β−2nπ−2n−12c21−n+2βσn+2β+1 fˆ(ξ)2ec|ξ|dξ if n+β+1 0
≤ | | ≥
(cid:26) ZRn (cid:27)
1/2
2−12−β−2nπ−2n−12σn+2β+1c21−n+2βecσ fˆ(ξ)2dξ
≤ ( Z|ξ|≤σ| | )
2−14−β−nπ−n−41σn+4β+1c1−(n4+β)ec2σ f L2(Rn)
≤ k k
< ♯
∞
In the preceding deduction we put the restriction n+β 1 on it. This is a drawback because
| |≥
the frequently seen case n=1 and β = 1 is not covered. This gap will be discussed shortly.
−
Corollary 1.8 For any positive number σ and any function f B , under the conditions of Theo-
σ
∈
rem1.4,
f(x) s(x) m!S(m,n)(2π)−34n√nαnσ1+β4+n ∆0c1+β4−nec2σ(λ)1δ f L2(Rn) (9)
| − |≤ k k
p p
if n+β 1 or n+β = 1. The number S(m,n) is determined by m and n, and is one whenever
≥ −
β <0. The other constants are as in Theorem1.4.
2 How to choose c?—a more practical approach
In (9) the only things influenced by c are c1+β4−n, ec2σ and (λ)1δ. The number δ is the famous fill
distance of the data points, as explained in Corollary2.5 of [4]. The constants β, σ are defined in
(1) and Definition1.5 respectively. Theoretically (λ)1δ is very influential. However it contributes
little to the error estimate in practice due to two reasons. First, λ is quite large and near one
as shown in Theorem1.4. Second, δ can not be very small(near zero). As shown by Madych in
5
[7], whenever δ is too small, the condition number of the linear system (3),(4) will be very large,
making its solution meaningless. Note that γ very fast as n . The first examples are
n
→ ∞ → ∞
γ =2, γ =12, γ =78 and γ =632. Thus λ reacts to the change of c very slowly. The number
1 2 3 4
λ will become small enough to make (λ)1δ influential only when c is extremely large. For example,
if the dimension n=2 and the domain cube E has side length b =1, then
0
1
2 6Cγ2
λ=
3
(cid:18) (cid:19)
where
2 ρ
C =max 2ρ′√2e4γ2, 3 , ρ′ = c, γ2 =12.
(cid:26) (cid:27)
The numberρ isequaltooneorabit largerthanone,depending onβ. ThusC is usuallyverylarge
and λ is near one, unless c is very large.
Another problem is that the function
β
h(x):=Γ (c2+ x2)β2
−2 | |
(cid:18) (cid:19)
willinpracticemakethelinearsystem(3),(4)numericallyunsolvablewhenevercistoolarge. Inour
precedingexampleallcenters(interpolationpoints)lieinthecubeofside1. Therefore x x √2.
i j
| − |≤
If our computer allows seven or fourteen significant digits, then h(xi−xj) will become Γ(−β2)(c2)β2
whenever c 104 or 107. In our example λ>(2)418 and is near one even when c=1012. Therefore
≥ 3
(λ)1δ is only theoretically influential and can be ignored in practice. What’s in practice influential
is c1+β4−n and ec2σ.
In this section we assume that β, b (the side length of the cube), and δ are fixed and m =
0
max 0, β as in Theorem1.4. The optimal choice or suggested value of c is presented according
⌈2⌉
to thne valuesoof β, n, σ, δ, and b0.
Case1 β+1 n 0 Letf B andE be a cube inRn with sidelengthb asinTheorem1.4. Let
σ 0
− ≥ ∈
hbe asin(1)withn+β 1orn+β = 1. Let0<δ < b0 befixedinTheorem1.4. Suppose
≥ − 4γn(m+1)
β+1 n 0. Then
− ≥
(a) the optimal choice of c is to let c = 12ρ√ne2nγnγn(m+1)δ if n = 1 or β = 1, where ρ
6 6
and γ were defined in Definition1.3 and 1.2 respectively;
n
(b) theoptimalchoiceofc,whenn=1andβ =1,isasfollows. Letc1 =12ρ√ne2nγnγn(m+
1)δ, c =3b e4, and η = ln23 . Then
0 0 24e4δ
(i) choose c=12ρ√ne2nγnγn(m+1)δ if σ2 +η ≥0;
(ii) if σ +η <0 and c < 1 <c , choose c=c if H(c ) H(c ) and c=c
if H2 (c0)<H(c1) w1her4e(Hσ−2+(ηc)):=c041ec2σ(23)24ec4δ;1 1 ≤ 0 0
(iii) choose c=12ρ√ne2nγnγn(m+1)δ if σ2 +η <0 and c0 ≤ 4(σ−2+1η);
(iv) choose c=c if σ +η <0 and 1 c .
0 2 4(σ−2+η) ≤ 1
6
Reason: This is a consequence of (9). However in order to satisfy the condition δ δ in Theo-
0
≤
rem1.4, we put the restriction δ < b0 from the beginning. The number b0 is just δ
4γn(m+1) 4γn(m+1) 0
obtained by letting C = 2 when c = 3b e4. Increasing c(i.e. c > 3b e4) does not change δ , but
3b0 0 0 0
decreasingc(i.e. c<3b e4) makesδ smaller. After 0<δ < b0 is fixed, c cannotbe less than
0 0 4γn(m+1)
12ρ√ne2nγnγn(m+1)δ because of the restriction δ δ0.
≤
(a) follows from (9) obviously because (λ)1δ can be ignored and the only thing influenced by c is
c1+β4−n ec2σ. The smaller c is, the smaller the error bound (9) is.
·
(b)ismorecomplicated. Althoughwepreviouslymentionedthat(λ)1δ canbeignoredpractically,
there is an exception. When n =1 and β = 1, λ is not very large and δ is not required to be very
small. Therefore the influence of (λ)1δ has to be taken into consideration.
For n=1, β =1, let
H(c) := c1+β4−nec2σ(λ)1δ
1
= c41ec2σ 2 12Cδ
3
(cid:18) (cid:19)
= ( cc4411eecc22σσ((2233))b2804δec4δ iiff cc≤≥33bb00ee44.
H(c) is increasing on [3b e4, ). For c (0, 3b e4],
0 0
∞ ∈
1 c
H(c) = c41ec2σ 2 24e4δ
3
"(cid:18) (cid:19) #
ln2
= c41ec2σeηc where η := 3 <0
24e4δ
= c14e(σ2+η)c.
Then
1 σ
H′(c)=e−43e(σ2+η)c +c +η .
4 2
(cid:20) (cid:16) (cid:17)(cid:21)
If σ +η 0, then H (c) > 0 and H is increasing on (0, 3b e4]. In this situation the optimal c is
2 ≥ ′ 0
naturally the minimal possible choice c=12ρ√ne2nγnγn(m+1)δ.
Assume σ +η < 0. Then H (c) = 0 iff c = 1 =: c . H (c) < 0 if c > c and H (c) > 0 if
2 ′ 4(σ−+η) ∗ ′ ∗ ′
2
c<c . Our criteria (ii), (iii) and (iv) thus follow immediately.
∗
Remark:(a)In Case1(b)(i) the error bound goes to infinity as c . (b)The numbers ρ and
→ ∞
γ are defined in Definition1.3 and 1.2 respectively.
n
Case2 β+1 n<0 Let f B and E be a cube in Rn with side length b as in Theorem1.4.
σ 0
− ∈
Let h be as in (1) with n+β 1 or n+β = 1. If 0 < δ < b0 in Theorem1.4 is fixed and
≥ − 4γn(m+1)
β +1−n < 0, the optimal choice of c is to let c = max n−2βσ−1, 12ρ√ne2nγnγn(m+1)δ where
m, γn, and ρ are defined as in Definition1.1, 1.2 and 1.3 renspectively. o
Reason: In (9) (λ)1δ can be ignored. The only things influenced by c are cβ+14−n and ec2σ. Let
7
g(c) := cβ+14−nec2σ. Obviously g(c) as c 0+ or c . The minimum of g occurs when
→ ∞ → → ∞
g (c)=0. Now
′
g′(c) = β+1−ncβ−4n−3ec2σ +cβ+14−nσec2σ
4 2
= ec2σ β+1−ncβ−4n−3 + σcβ+14−n .
4 2
(cid:26) (cid:27)
β+1 n β n 3 σ β+1 n
g′(c)=0 iff − c −4− + c 4− =0
4 2
σ β+1 n n β 1 β n 3
iff c 4− = − − c −4−
2 4
4σ β n 3 n β 1
iff =c −4− c −4−
2(n β 1)
− −
2σ
iff =c 1
−
n β 1
− −
n β 1
iff c= − − .
2σ
In order to meet the requirement δ δ in Theorem1.4, we set δ < b0 which is the δ ob-
≤ 0 4γn(m+1) 0
tained by letting C = 2 . As c decreases, δ will also decrease. The acceptable domain of c is then
3b0 0
[12ρ√ne2nγnγn(m+1)δ, ). Thereforewechoosethe maximumofthe twovalues asthe optimal c.
∞
Remark: (a)Case2 includes the well-known cases β = 1, n 3. (b)Note that in Case2, the
≥
upper bound of f(x) s(x) goes to infinity if c or c 0+.
| − | →∞ →
In both Case1 and 2, we require n+β 1. This is a drawback. The worst thing is that the
| | ≥
frequently seen case β = 1, n = 1 is excluded. For β = 1, it’s just the inverse multiquadric
h(x) = Γ(12) . For suc−h h(x), Case1 and 2 only deal wit−h n 2. If n = 1, we need special
√c2+x2 ≥
| |
treatment which is absolutely nontrivial.
Lemma 2.1 Let h be as in (1) with β = 1, n=1. For any σ >0, if f B , then f and
σ h,m
− ∈ ∈C
1/2
1 1
kfkh ≤(2π)−n2−41 (K0(1)Z|ξ|≤1c |fˆ(ξ)|2dξ+ a0 Z1c<|ξ|≤σ|fˆ(ξ)|2pc|ξ|ec|ξ|dξ) (10)
1
if 1 <σ, where a = √π3−2, and
c 0 2Γ(1)
2
1/2
1
f h (2π)−n2−41 fˆ(ξ)2dξ (11)
k k ≤ (K0(1)Z|ξ|≤1c | | )
if 1 σ.
c ≥
Proof. By (3.9) of [6], f = fˆ and
h L2(ρ)
k k k k
1/2
fˆ = fˆ(ξ)2dρ(ξ)
L2(ρ)
k k | |
(cid:26)ZRn (cid:27)
8
1/2
= fˆ(ξ)2r(ξ)dξ
| |
(cid:26)ZRn (cid:27)
1/2
1
= fˆ(ξ)2(2π) 2n dξ where w()=hˆ() by Theorem5.2 of [6]
−
| | · w( ξ) · ·
(cid:26)ZRn − (cid:27)
1/2
1
= fˆ(ξ)2(2π)−2n2−12 dξ by (8.7) of [8](p.109)
| | · (cξ )
(cid:26)ZRn K0 | | (cid:27)
1/2
fˆ(ξ)2 fˆ(ξ)2
= [(2π)−2n2−12]12 | | dξ+ | | dξ
(cξ ) (cξ )
(Z|ξ|≤1c K0 | | Z|ξ|>1c K0 | | )
1/2
1 1
(2π)−n2−41 fˆ(ξ)2dξ+ fˆ(ξ)2 cξ ec|ξ|dξ
≤ (K0(1)Z|ξ|≤1c | | a0 Z|ξ|>1c | | p | | )
√π
by Corollary5.12 of [8] and p.374 of [1] where a = .
0 2Γ(1)√3
2
Since fˆ(ξ)=0 for ξ >σ, our conclusionfollowsimmediately. ♯
| |
Lemma 2.2 For any positive number σ and any function f B , under the conditions of Theo-
σ
∈
rem1.4 with β = 1 and n=1,
−
A+B 1/2
|f(x)−s(x)|≤√2π ∆0(λ)1δ c (12)
(cid:26) (cid:27)
p
where A = K01(1) |ξ|≤1c |fˆ(ξ)|2dξ, B = 2√√3Γπ(12) 1c<|ξ|≤σ|fˆ(ξ)|2 c|ξ|ec|ξ|dξ if 1c ≤ σ and B = 0 if
1 σ.
c ≥ R R p
Proof. This is just an immediate result of Theorem1.4 and Lemma2.1. ♯
With Lemma2.1 and 2.2 we can now begin to analyze the appropriate choice of c.
Case3 n=1,β = 1 Let f B and E be a cube in Rn with side length b as in Theorem1.4.
σ 0
− ∈
Let h be as in (1) with n = 1, β = 1. If 0 < δ < min 1 , b0 in Theorem1.4 is fixed, under
− 24σe4 8
the conditions of Theorem1.4, the suggested choice for c is to let c= 1.
(cid:8) (cid:9) σ
Reason: Aspointedoutinthe beginningofthissection,the number(λ)1δ in(12)canbe essentially
ignored. Hencethe onlythingin(12)influencedbychangingcis A+B. Note thatifc (0, 1], then
c ∈ σ
B =0. We define H (c)= A if c (0, 1] and H (c)= A+B if c [1, ). Then
1 c ∈ σ 2 c ∈ σ ∞
1 σ σ 1
H =σA= fˆ(ξ)2dξ = f 2 =H .
1 σ (1) | | (1)k kL2(Rn) 2 σ
(cid:18) (cid:19) K0 Z|ξ|≤σ K0 (cid:18) (cid:19)
Now, for c (0, 1],
∈ σ
1
lim H (c) = lim fˆ(ξ)2dξ
c→0+ 1 c→0+ cK0(1)Z|ξ|≤1c | |
1
= lim f 2
c 0+ c 0(1)k kL2(Rn)
→ K
= .
∞
9
For c [1, ),
∈ σ ∞
1 1 2√3Γ(1)
lim H (c) = lim fˆ(ξ)2dξ+ 2 fˆ(ξ)2 cξ ecξdξ
c→∞ 2 c→∞ c (K0(1)Z|ξ|≤1c | | √π Z1c<|ξ|≤σ| | p | | | | )
lim 1 kfk2L2(Rn) + 2√3Γ(12) √cσecσ f 2
≤ c→∞ c ( K0(1) √π · k kL2(Rn))
= .
∞
By the structure of the integrands, we know that in fact lim H (c)= , not just .
c 2
Therefore, the upper bound of f(x) s(x) in (12) goes →to∞infinity as∞c 0+ or c≤∞ . This
| − | → → ∞
phenomenon together with the agreement of H and H at 1 suggests that c = 1 is a safe and
1 2 σ σ
natural choice.
Remark: (a) In Theorem1.4 we require that δ δ where δ 0+ as c 0+. If δ is fixed,
0 0
≤ → →
c cannot approach0 arbitrarily. Note that for n=1, β = 1,
−
c if c 3b e4
δ0 = b204e4 if c≤3b0e4.
(cid:26) 8 ≥ 0
The requirement δ c is equivalent to 24e4δ c. Therefore our exploration is in fact restricted
≤ 24e4 ≤
in the range c [24e4δ, 3b e4] [3b e4, ) = [24e4δ, ). This is a drawback. However the
0 0
∈ ∪ ∞ ∞
assumption δ <min 1 , b0 in Case3 guarantees that (i)δ δ for our final choice c = 1, and
24e4σ 8 ≤ 0 σ
δ δ for all c [24e4δ, ) and (ii)24e4δ 1. If c < 24e4δ, then δ δ does not hold and (12)
≤ 0 ∈ (cid:8) ∞ (cid:9) ≤ σ ≤ 0
may not be true. Therefore in Case3 c= 1 is just a suggested value, and we know that the smaller
σ
δ is, the more reliable our choice c= 1 is because 24e4δ 0 as δ 0 and lim H (c)= .
(b)Asinthecasen=1, β =1(Casσe1),thenumberλm→aynotb→everylargewc→h0e+nn1=1, β∞= 1.
−
Therefore it seems that in Case3 the influence of (λ)1δ should be taken into consideration. However
we put it into Case6 of subsection3.1 because it’s meaningful and useful to see what will happen
when (λ)1δ is ignored.
Remark: All the criteria provided in this section are based on Theorem1.4 where we require that
δ δ . Note that δ := 1 is usually a very small number. Most time C = 2ρ√ne2nγn where
≤ 0 0 6Cγn(m+1) c
ρ=1 or a bit greater than 1. For such C,
δ δ iff c [12ρ√ne2nγnγ (m+1)]δ.
0 n
≤ ≥
If δ = 0.125, n =1 and m = 1, then ρ =1 and c 6e4. The lower bound of c increases rapidly as
≥
n increases. If δ =0.125, n=2, m=1, then ρ=1 and
c 36√2e48 >36√2(210)4 >36√21012
≥
which will make the linear system (3),(4) numerically unsolvable because c is too large. Since
γ very fast as n , the minimum requirement for c will become extremely stringent for
n
→ ∞ → ∞
high dimensions. On the other hand, as is well known by RBF people, ill-conditioning will happen
whenever the fill distance is small. Therefore the restriction δ δ is a big problem. Madych’s
0
≤
experiment [7] shows that when the domain size b =8, andn=1, the conditionnumber is already
0
toolargeifδ =0.125. Ofcourseδ canbeincreasedbyincreasingc. Howeverthelinearsystem(3),(4)
0
will become numerically unsolvable when c is too large. Moreover,the conditionnumber also grows
10