On the High-Level Error Bound for Gaussian Interpolation

0.15 MB·English

by  Lin-Tian Luh

On the High-Level Error Bound for
Gaussian Interpolation
LIN-TIAN LUH
6
0 Department of Mathematics, Providence University, Shalu, Taichung, Taiwan
0
2
Abstract

n
a
J
It’swell-knownthatthereisaverypowerfulerrorboundforGaussians
9
put forward by Madych and Nelson in 1992. It’s of the formf(x) s(x)
] | − |≤
A (Cd)dc kfkh where C,c are constants, h is the Gaussian function, s is the in-
N
terpolating function, and d is called fill distance which, roughly speaking,
.
h measures the spacing of the points at which interpolation occurs. This
t
a error bound gets small very fast as d 0. The constants C and c are
m →
very sensitive. A slight change of them will result in a huge change of the
[

error bound. The number c can be calculated as shown in [9]. However,
1
v C cannot be calculated, or even approximated. This is a famous question
9
in the theory of radial basis functions. The purpose of this paper is to
5
1 answer this question.
1
0
6
0 Keyword: radial basis function, interpolation, error bound, Gaussian
/
h
t 1.INTRODUCTION
a
m
:
v
i Let h be a continuous function on Rn which is conditionally positive
X
definite of order m. Given data (x f ),j = 1,...,N, where X = x ...,x
r j, j { 1, N}
a is a subset of points in Rn and the f s are real or complex numbers, the
j′
so-called h spline interpolant of these data is the function s defined by
N
s(x) = p(x)+ c h(x x ), (1)
j j
−
j=1
P
where p(x) is a Polynomial in and c s are chosen so that
Pm−1 j ′
N
c q(x ) = 0 (2)
j j
j=1
P
Author 2
for all polynomials q in and
m 1
P −
N
p(x )+ c h(x x ) =f , i = 1,...N. (3)
i j i j i
−
j=1
P
Here denotes the class of those polynomials of Rn of degree
m 1
P −
m 1.
≤ −
It is well known that the system of equations (2) and (3) has a unique
solution when X is a determining set for and h is strictly condi-
m 1
P −
tionally positive definite. For more details see [7]. Thus, in this case, the
interpolant s(x) is well defined.
We remind the reader that X is said to be a determining set for
m 1
P −
if p is in and p vanishes on X implies that p is identically zero.
m 1
P −
1.1A Bound for Multivariate Polynomials
A key ingredient in the development of our estimates is the following
lemma which gives a bound on the size of a polynimial on a cube in Rn in
terms of its values on a discrete subset which is scattered in a sufficiently
uniform manner. For its proof, please see [9].
LEMMA1. For n = 1,2,...,define γ by the formulas γ = 2 and, if
n 1
n > 1, γ = 2n(1+γ ). Let Q be a cube in Rn that is subdivided into qn
n n 1
−
identical subcubes. Let Y be a set of qn points obtained by selecting a
point from each of those subcubes. If q γ (k+1), then for all p in
n k
≥ P
sup p(x) e2nγn(k+1)sup p(y) .
| | ≤ | |
x Q y Y
∈ ∈
1.2 A Variational Framework for Interpolation
The precise statement of our estimate concerning h splines requires a
certain amount of technical notation and terminology which is identical
to that used in [8]. For the convenience of the reader we recall several
basic notions.
Author 3
The space of complex valued functions on Rn that are compactly sup-
portedand infinitelydifferentiable is denotedby .The Fourier transform
D
of a function φ in is
D
φ∧(ξ)= e i<x,ξ>φ(x)dx.
−
R
In what follows h will always denote a continuous conditionally posi-
tive definite function of order m. The Fourier transform of such functions
uniquelydeterminesa positive Borel measure µonRn ∼ 0 andconstants
{ }
a , γ = 2m as follows: For all ψ
γ
| | ∈ D
h(x)ψ(x)dx = ψ∧(ξ) χ∧(ξ) Dγψ∧(0)ξγ dµ(ξ)
− γ!
Z ( γ<2m )
R | |P
+ Dγψ∧(0)aγ, (4)
γ!
γ 2m
| |≤
P
where for every choice of complex numbers c , α = m,
α
| |
a c c 0.
α+β α β
≥
α=mβ=m
| | | |
P P
Here χ is a function in such that 1 χ∧(ξ) has a zreo of order 2m+1
D −
at ξ =0; both of the integrals
ξ 2mdµ(ξ), dµ(ξ)
0<ξ<1| | ξ 1
| | | |≥
R R
are finite. The choice of χ affects the value of the coefficients a for
γ
γ < 2m.
| |
Our variational framework for interpolation is supplied by a space we
denote by . If
h,m
C
= φ : xαφ(x)dx = 0 for all α < m
m
D ∈D | |
(cid:8) R (cid:9)
Author 4
,then is the class of those continuous functions f which satisfy
h,m
C
1
f(x)φ(x)dx c(f) h(x y)φ(x)φ(y)dxdy 2 (5)
≤ −
(cid:12)R (cid:12) nR o
(cid:12) (cid:12)
for some constant c(f) and all φ in . If f let f denote the
Dm ∈ Ch,m k kh
smallest constant c(f) for which (4) is true. Recall that f is a semi-
k kh
norm and is a semi-Hilbert space; in the case m = 0 it is a norm and
h,m
C
a Hilbert space respectively.
Given a function f in and a subset X of Rn, there is an element
h,m
C
s of minimal norm which is equal to f on X.
h,m
C
The function space is not very easy to understand. However Luh
h,m
C
made a lucid characterization for this sapce in [3] and [4]. For an easier
understanding of this space we suggest that reader read [3] and [4] first.
2.MAIN RESULTS
Before showingour mainresults, we needsome lemmas. Thefollowing
lemma is cited directly from [9].
LEMMA2. Let Q,Y, and γ be as in LEMMA1. Then, given a point
n
x in Q, there is a measure σ supported on Y such that
p(y)dσ(y) = p(x)
R
for all p in , and
k
P
d σ (y) e2nγn(k+1).
| | ≤
R
Now we need a famous formula.
Stirling’s Formula: n!∽ √2πn(n)n.
e
Author 5
Remark The approximation is very reliable even for small n. For example,
when n = 10, the relative error is only 0.83%. The larger n is, the better the
approximation is. For further details see [1] and [2].
1
LEMMA3. Let ρ = 1 and ρ = 36 ∽ 1.2 . Then
1 e 2 e e
√2πρkkk k! √2πρkkk
1 ≤ ≤ 2
for all positive integer k.
1 1 1 1
Proof. Notethat 1,√2,√3,√4, √5,...canbeexpressedby1,(24)2,(36)3,(48)4,(510)5,....
e e2 e3 e4 e5 e e e e e
1 1 1 1 1
Now, sup 1,24,36,48,510,... = 36implies that √k ρk for all k. Thusk!∽
e e e e e e ek ≤ 2
(cid:26) (cid:27)
√2π√k kk √2πρk kk.The remaining part √2πρkkk k! follows by observing
ek · ≤ 2 · 1 ≤
that √2π(1)kkk √2π(1)k √k kk ∽ k!
e ≤ e · ·
(cid:3)
LEMMA4. Let ρ = √3.Then k! √2πρkkk 1 for all k 1.
e ≤ − ≥
3 3
Proof. k! ∽ √2π(1)k √k kk = √2π k2 kk 1. Note that k2 :k = 1,2,3...
e · · · ek · − ek
(cid:26) (cid:27)
3 1 3
can be expressed by 1,(24)2,(32)3,(48)4.... Our lemma follows by noting that
e e e e
(cid:26) (cid:27)
3 1 3
sup 1, 24,32,48... = √3.
e e e e e
(cid:26) (cid:27)
(cid:3)
LEMMA5. Let h(x) = e βx2,β > 0, be the Gaussian function in Rn,
− | |
and µ be the measure defined in (4). For any positive even integer k,
|ξ|kdµ(ξ)≤ πn+21 ·n·αn·2k+n2+2 ·ρk+n2−1 ·βk2 ·(k+n−1)k+n2−3 ·(2+ 1e)
Rn
R
for add n, and
ξ kdµ(ξ) πn+21 n αn 2k+n2+3 ρk+n2−2 βk2 (k+n 2)k+n2−4
| | ≤ · · · · · · −
Rn
R
Author 6
for even n,where ρ = √3 and α is the volume of the unit ball in Rn.
e n
Proof. ξ kdµ(ξ)
| |
Rn
R
= (πβ)n2 ||ξξ||k2 dξ
Rn e 4β
R
= (πβ)n2 ·n·αn· 0∞ rk·rγn2−1dγ
e4β
R
= (πβ)n2 ·n·αn·(2√β)k+n· 0∞ rke+rn2−1dr
= πn2 ·n·αn·2n−1+k ·βRk2 ·(k+n2−2)(k+n2−2 −1)···(12) 0∞ √u1eudu if n is odd,
( πn2 ·n·αn·2n−1+k ·βk2(k+n2−2)! if n is even. R
Let b = k +n 2 . Then b! √2πρb bb 1.
2− ≤ · −
(cid:6) (cid:7)
For odd n, b = k+n 1. Thus
2−
b! ≤ √2πρk+n2−1 ·(k+n2−1)k+n2−3
= √2π 2−k−2n+3 ρk+n2−1 (k+n 1)k+n2−3,
· · · −
and
|ξ|kdµ(ξ) ≤ πn+21 ·n·αn ·(2+ 1e)·2k+n2+2 ·βk2 ·ρk+n2−1 ·(k+n−1)k+n2−3,
Rn
R
by noting that ∞ 1 du 2+ 1.
0 √ueu ≤ e
R
For even n, b = k +n 2. Thus
2−
b!≤ √2πρk+n2−2 ·(k+n2−2)k+n2−4
= √2π 2−k−2n+4 ρk+n2−2 (k+n 2)k+n2−4,
· · · −
and
Author 7
ξ kdµ(ξ) πn+21 n αn 2k+n2+3 βk2 ρk+n2−2 (k+n 2)k+n2−4.
| | ≤ · · · · · · −
Rn
R
(cid:3)
Theorem 1 Let h(x) = e βx2, β > 0, be the Gaussian function in Rn, and µ
− | |
be the measure defined in (4). Then, given a positive number b , there are positive
0
constants δ ,c,and C for which the following is true: If f and s is the h
0 h,m
∈ C
spline that interpolates f on a subset X of Rn, then
|f(x)−s(x)| ≤ △′′(Cδ)δc ·kfkh
,where △′′ = πn−41 ·(n·αn)12 ·2n+41(√e3)n−42 for even n and △′′ = πn−41 ·(n·
αn)12 ·(2+ 1e)12 ·2n4 ·(√e3)n−41 for add n, holds for all x in a cube E provided
that (i)E has side b and b b ,(ii)0 < δ δ , and (iii)every subcube of E
0 0
≥ ≤
of side δ contains a point of X. Here, α deontes the volume of the unit
n
ball in Rn.
The number c is equal to b0 where γ is defined in LEMMA1.The
8γ n
n
number C is equal to 334 ·e·√2ρβ·√n·e2nγn 4 · b03 · γn, where ρ = √e3.
Moreover, δ0 can be defi(cid:16)ned by (cid:17)
δ = min 1 ,δ
0 3 n
(cid:26)(34·e·√2ρβ·√n·e2nγn)4·b03·γn (cid:27)
where
b0 if n = 1,
2γ
n
δn =  2γnb(b0n0−1) iiff nn iiss oevdedn,,aannddnn>>12,, 
 2γn(nb−02) if n = 2. 
2γ
 n 
 
 
 
 
Proof. Let δ be as in the theorem. If δ δ , thenδ b0 andγnδ 1. There is an
0 ≤ 0 ≤ 2γn b0 ≤
2
γ δ
integer k 1 such that 1 ( n )k 2. This implies
≥ ≤ b0 ≤
2
b0 γ δk b .
2 ≤ n ≤ 0
Author 8
Now, let x be any point of the cube E and recall that Theorem4.2 of
[8] implies that
f(x) s(x) c f y xkd σ (y) (6)
| − |≤ k ·k kh· | − | | |
Rn
R
whenever k > 0, where σ is any measure supported on X such that
p(y)dσ(y) = p(x) (7)
R
for all polynomials p in . Here
k 1
P −
1
ξ2k 2
ck = (|k|!)2dµ(ξ)
(cid:26)Rn (cid:27)
R
whenever k > 0.
LEMMA5. now applies.
For odd n,
1
2
c = 1 ξ 2kdµ(ξ)
k k! | |
(cid:26)Rn (cid:27)
R
1
≤ k1! πn+21 ·n·αn·22k+2n+2 ·ρ2k+2n−1 ·βk ·(2k+n−1)2k+2n−3 ·(2+ 1e) 2
n o
≤ k1! πn+41 ·(n·αn)12 ·2n+42 ·ρn−41 ·(√2ρβ)k ·(2k+n−1)2k+4n−3 · 2+ 1e
n q o
πn+41·(n·αn)21·2n+42·ρn−41·(√2ρβ)k·(2k+n−1)2k+4n−3· 2+1e by LEMMA 3
≤ √2πρ k kk q
· 1 ·
Author 9
= πn−41 ·(n·αn)21 ·2n4 ·ρn−41 ·ρ3k ·(k1k)·(2k+n−1)2k+4n−3 ·( 2+ 1e)
q
where ρ = √2ρβ
3 ρ
1
= ′′ ρ k 1 (2k+n 1)2k+4n−3
△ · 3 · kk · −
where △′′ = 2+ 1e ·πn−41 ·(n·αn)21 ·2n4 ·ρn−41.
q
To obtain the desired bound on f(x) s(x) , it suffices to find a suit-
| − |
able bound for
I = c y x kd σ (y).
k
| − | | |
Rn
R
Let Q be any cube which contains x, has side γ kδ, and is contained
n
in E. Subdivide Q into (γ k)n congruent subcubes of side δ. Since each
n
of these subcubes must contain a point of X, select a point of X from
each subcube and call the resulting discrete set Y. By Lemma1 we may
conclude that there is a measure σ supported on Y which satis fies (6)
and enjoys the estimate
d σ (y) e2nγnk. (8)
| | ≤
Rn
R
We use this measure in (6) to obtain an estimate on I.
Using (8) and the fact that support of σ is contained in Q whose
diameter is √nγ kδ we may write
n
I ≤ △′′ ·ρ3k · k1k ·(2k+n−1)2k+4n−3 ·(√nγnkδ)k ·e2nγnk
= △′′ ·(B′γnδ)k ·(2k+n−1)2k+4n−3 where B′ = ρ3√ne2nγn
′′ (B′γnδ)k (3k)34k (if k n 1)
≤ △ · · ≥ −
Author 10
= ′′ (B′γnδ334k43)k
△ ·
k
≤ △′′ · 334B′γnδ(γbn0δ)34 (since k ≤ γbn0δ)
h i
k
= △′′ · B′′γn41b043δ14 where B′′ = 343B′
(cid:20) (cid:21)
b0
= △′′ · B′′b034γn41δ14 2γnδ (since 2γb0nδ ≤ k)
(cid:20) (cid:21)
where B′′b034γn41δ14 ≤ 1 if and only if B′′4b30γnδ ≤ 1 and is guaranteed by
δ δ
0
≤
b0
= ′′ B′′4b3γ δ 8γnδ
△ · 0 n
(cid:20) (cid:21)
4
= △′′ ·[Cδ]δc where C = B′′ b30γn and c= 8bγ0n.
Note that k n 1 is guaranteed by n 1 b0 k.
≥ − − ≤ 2γnδ ≤
For even n,
1
2
c = 1 ξ 2kdµ(ξ)
k k! | |
(cid:26)Rn (cid:27)
R
1
≤ k1! πn+21 ·n·αn ·22k+2n+3 ·ρ2k+2n−2 ·βk ·(2k+n−2)2k+2n−4 2
n o
= k1! πn+41 ·(nαn)12 ·2n+43 ·ρn−42 ·(√2ρβ)k ·(2k+n−2)2k+4n−4
n o