Helle Hallik Rational spline histopolation
Tartu 2015 ISSN 1024-4212 ISBN 978-9949-32-834-5
DISSERTATIONES MATHEMATICAE UNIVERSITATIS TARTUENSIS
98
Helle Hallik
Rational spline histopolation
DISSERTATIONES MATHEMATICAE UNIVERSITATIS TARTUENSIS
98
DISSERTATIONES MATHEMATICAE UNIVERSITATIS TARTUENSIS 98
HELLE HALLIK
Rational spline histopolation
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i = 1, . . . , n,
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x i
x i−1
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x 0
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S(x n ) = β > z n
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S 1
S 2
=[x i−1 , x i ]
S 1 (x) = a 1i + b 1i (x − x i−1 )
1 + d 1i (x − x i−1 )
S 2 (x) = a 2i + b 2i (x − x i−1 ) 1 + d 2i (x − x i−1 ) ,
x ∈ [x i−1 , x i ]
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c 2i = b 2i − a 2i d 2i
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(1 + d 1i (x − x i−1 )) 2 − c 2i
(1 + d 2i (x − x i−1 )) 2 .
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(1 + d 1i (x − x i−1 )) 2 = c 2i
(1 + d 2i (x − x i−1 )) 2 .
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1+d 2i (x − x i−1 ) > 0
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c 2i
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g (x) = 0
-[x i−1 , x i ]
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g (x) = 0
>
1 + d 1i (x − x i−1 ) 1 + d 2i (x − x i−1 ) =
c 1i c 2i
1/2
1 + d 1i (x − x i−1 ) = c 1i
c 2i 1/2
(1 + d 2i (x − x i−1 )).
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[x i−1 , x i ]
--" -
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-g
n
G[x 0 , x n ]
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S 2
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x i−1
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g(ξ i ) = 0
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[x i−1 , x i ]
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g
n + 1
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i = 1, . . . , n
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[x i−1 , x i ]
S (x) = c i
(1 + d i (x − x i−1 )) 2 .
."8+
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i = 0, . . . , n
h i = x i − x i−1
i = 1, . . . , n
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(1 + d i h i ) 2 = m i .
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m i
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m i < 0
m i = 0
i = 0, . . . , n
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-
d i = 1 h i
m i−1 m i
1/2
− 1
.
."10+
d i = 0
m i−1 = m i
"S(x) = b i
d i − c i
d i (1 + d i (x − x i−1 )) .
."11
x i
x i−1
S(x)dx = x i
x i−1
b i
d i − c i
d i (1 + d i (x − x i−1 ))
dx
= b i d i h i − c i
d 2 i log(1 + h i d i ),
.". -
b i d i − c i
h i d 2 i log(1 + h i d i ) = z i .
."1.>."1. -
b i
d i = z i + c i
h i d 2 i log(1 + h i d i ).
."11
S(x) = z i + m i−1
h i d 2 i log(1 + h i d i ) − m i−1
d i (1 + d i (x − x i−1 )) ,
."15) >."10
S(x) = z i + h i m i−1 m
m i−1 i
1/2
− 1 2 log
m i−1 m i
1/2
− h i m i−1
m i−1
m i
1/2
− 1 1 + t m i−1
m i
1/2
− 1 ,
."14
x ∈ [x i−1 , x i ], i = 1, . . . , n, t = (x − x i−1 )/h i .
!
d i = 0
-m i−1 = m i
b i = c i = m i
" .".a i = z i − h i m i /2
-x ∈ [x i−1 , x i ]
S(x) = z i + m i
x −
x i−1 + h i 2
.
."12
m i = 0
-S(x) = z i
x ∈ [a, b]
- -."12 "
S
m i
."14 ."12S
" ! -- ?S
)x 1 , . . . , x n−1
>"
E
m i > 0
i
m i < 0
i
" +
d i = 0
"m i−1 = m i (1 + h i d i ) 2
--."15
S(x i − 0) = z i + h i m i (1 + h i d i ) 2 log(1 + h i d i ) − h i d i (1 + h i d i ) h 2 i d 2 i
= z i + h i m i (1 + h i d i ) 2 (log(1 + h i d i ) − 1) + 1 + h i d i h 2 i d 2 i
= z i + h i m i m i−1
m i
log m i−1
m i 1/2
− 1
+ m i−1
m i 1/2
m i−1 m i
1/2
− 1
2 .
."1/$ ."15 -
S(x i−1 + 0) = z i + h i m i−1 log(1 + h i d i ) − h i d i h 2 i d 2 i
= z i − h i m i−1 (1 + h i d i ) −2
log(1 + h i d i ) −1 − 1
+ (1 + h i d i ) −1
((1 + h i d i ) −1 − 1) 2
= z i − h i m i−1 m i m i−1
log
m i m i−1
1/2
− 1
+ m i
m i−1 1/2
m i m i−1
1/2
− 1
2 .
."17."12 -
d i = 0 S(x i − 0) = z i + h i
2 m i , S(x i−1 + 0) = z i − h i 2 m i .
+
ϕ(x) =
⎧ ⎪
⎨
⎪ ⎩
x 2 (log x − 1) + x
(x − 1) 2 for x > 0 , x = 1 , 1
2 for x = 1 .
/ 3 3 "
#
ϕ(x) > 0
$ϕ (x) > 0
ϕ (x) < 0
x > 0
$
lim
x→1 ϕ(x) = 1 2
$lim
x→1 ϕ (x) = 1 3
$
lim
x→0+
ϕ(x)
x = 1
$lim
x→∞
ϕ(x) log x = 1
$
1
2 < ϕ(x)
x < 1
0 < x < 1
$%
ϕ(x 1/2 ) log x
x 1.84.
1 $
x > 0
?(x 2 (log x − 1)+x)/(x − 1) 2 .
:
x→0 lim + (x 2 (log x − 1) + x) = lim
x→0 +
log x − 1 1 x 2
= lim
x→0 +
1 x
− 2 x 3
= lim
x→0 +
− x 2 2 = 0.
lim x→0 + ϕ(x) = 0
"'?
x > 0
x = 1
-@ϕ (x) =
(2x(log x − 1) + x 2 · 1
x + 1)(x − 1) 2
(x − 1) 4 − (x 2 (log x − 1) + x) · 2(x − 1) (x − 1) 4
= x 2 − 2x log x − 1 (x − 1) 3 .
!
x > 1
(x − 1) 3 > 0
" +f (x) = x 2 − 2x log x − 1
" !f(1) = 0
"'-
f (x) = 2x − 2 log x − 2 = 2(x − log x − 1)
" Ef (1) = 0
" !f (x) = 2 (1 − 1/x) > 0 x > 1
f (x) > 0 x > 1
-f (x) > 0 x > 1
" !>ϕ (x) > 0 x > 1
" E@x→1 lim + ϕ (x) = lim
x→1 +
x 2 − 2x log x − 1 (x − 1) 3 = lim
x→1 +
2 x 2
6 = 1 3 > 0.
+-
0 < x < 1
ϕ (x)
"(x − 1) 3 < 0
"f (x) = x 2 − 2x log x − 1
-lim x→0 + f (x) = − 1
lim x→0 + f (x) = ∞
"f (x) = 2 (1 − 1/x) < 0 0 < x < 1
""
f
(0, 1)
" (f (1) = 0
f (x) > 0
(0, 1)
"f
(0, 1)
"f (x) < 0 0 < x < 1
" :ϕ (x) > 0
-0 < x < 1
" ! -lim x→1 − ϕ(x) = 1/3
" ! -ϕ (x) > 0 x > 0
-ϕ
x > 0
"lim x→0 + ϕ(x) = 0,
--
ϕ(x) > 0 x > 0
"-
ϕ
". E
x→1 lim
x=1
ϕ(x) = lim
x→1 x=1
=
2x(log x − 1) + x 2 · 1 x + 1 2(x − 1)
= lim
x→1 x=1
2x log x − x + 1 2(x − 1)
= lim
x→1 x=1
2 log x + 2x · 1 x − 1 2
= lim
x→1 x=1
2 log x + 1
2 = 1
2 .
x→1 lim ϕ(x) = 1 2 .
ϕ
1
- 1 "5 E
x→0 lim +
ϕ(x) x = lim
x→0 +
x(log x − 1) + 1 (x − 1) 2 = 1
lim x→0 + x log x = 0
"E )-
ϕ(x) > 0 x > 0
" $x→∞ lim ϕ(x) log x = lim
x→∞
x 2 (log x − 1) (x − 1) 2 log x + lim
x→∞
x (x − 1) 2 log x .
G
x → ∞
" :x→∞ lim ϕ(x) log x = lim
x→∞
x 2
(x − 1) 2 · lim
x→∞
log x − 1 log x
- >" E
x→∞ lim ϕ(x) log x = 1.
4
ψ(x) = ϕ(x)/x = (x(log x − 1) + 1)/(x − 1) 2
"ψ (x) =
((log x − 1) + x · 1
x )(x − 1) 2 − (x(log x − 1) + 1)2(x − 1) (x − 1) 4
= − (x + 1) log x + 2(x − 1) (x − 1) 3 .
+
ψ (x) x ∈ (0, 1)
"E
χ(x) = − (x + 1) log x + 2(x − 1)
"
lim x→0 + χ(x) = ∞
" !χ(1) = 0
" Eχ (x) = − 1/x + 1/x 2 = ( − x + 1)/x 2 > 0 0 < x < 1
"χ
0 < x < 1
" >χ (1) = 0
-χ (x) < 0 0 < x < 1
-
χ
0 < x < 1
"χ(1) = 0
-χ(x) > 0 0 < x < 1
" E--ψ (x) = χ(x)/(x − 1) 3 < 0 0 < x < 1
"
ϕ(x)/x
0 < x < 1
" 5. -
lim x→0 + ϕ(x)/x = 1
lim x→1 ϕ(x)/x = 1/2
4"
2 5 -
x
ϕ(x)/ log x 2
ϕ(x) 2 log x
" + @ ?x ∗
>ϕ(x) = 2 log x
" '
ϕ(1) = 1/2
log 1 = 0
-lim x→1 + ϕ(x)/ log x = ∞
x ∗
?" !x x ∗
ϕ(x)/ log x 2
" $x ∗ < √
1.84
x √
1.84
x x ∗
ϕ(x)/ log x 2
"E+."1 "
ϕ
- - >."1/ ."17S(x i − 0) = z i + h i m i ϕ
m i−1 m i
1/2
."18
S (x i−1 + 0) = z i − h i m i−1 ϕ
m i m i−1
1/2
.
."16
S
""S(x i − 0) = S(x i + 0)
i = 1, . . . , n − 1
-m i
h i ϕ
m i−1 m i
1/2
+ h i+1 ϕ
m i+1 m i
1/2
= δ i ,
.".0-
δ i = z i+1 − z i
" $m i > 0
S
-
δ i > 0
"!
d i = 0
d i+1 = 0
>S(x i − 0) = z i + h i
2 m i
.".1
S(x i−1 + 0) = z i − h i 2 m i
-
m i (h i + h i+1 ) = 2δ i .
."..!
d i = 0
d i+1 = 0
d i = 0
d i+1 = 0
."16.".1 -
m i h i
2 + h i+1 ϕ
m i+1 m i
1/2
= δ i .
.".5'.".. .".5 .".0 "
d i = 0
m i−1 = m i
ϕ(1) = 1/2
"( ."5 @?
m 0 = α
m n = β
" (."4
d 0
d n
-z 1 − h 1 m 0 ϕ
m 1 m 0
1/2
= α , z n + h n m n ϕ
m n−1 m n
1/2
= β.
.".4
' >.".4 .".0 -
i = 0
i = n
h 0 = 0
z 0 = α
h n+1 = 0
z n+1 = β
"
? =
z i
α
β
"2 33 (
z i
) *α
$β
$C 1
δ i = 0
" .".0 >."...".5 .".4 .".0
m i = 0
S
.". -."5 ."4 " (
S
" -δ i > 0
i = 1, . . . , n − 1
-α > 0
β > 0
."5α < z 1
β > z n
."4 "E - > .".0 D.".5 - - 3
m i = ϕ i (m)
-m = (m 0 , . . . , m n )
- -
[c, M ]
ϕ i : [c, M] n+1 → [c, M]
ϕ i
" (3(- @?
m i = ϕ i (m)
i = 0, . . . , n
">.".0
m i = ϕ i (m) = δ i
h i ϕ m
m i−1 i
1/2
+ h i+1 ϕ m
m i+1 i
1/2 .
.".2
m i−1
m i
m i+1 ∈ [c, M ]
-0 < c < M
"ϕ
-
ϕ i (m) δ i
(h i + h i+1 )ϕ c
M
1/2 .
( + ."1 4
ϕ((c/M) 1/2 ) (c/M) 1/2 /2
" --
ϕ i (m) 2δ i h i + h i+1
M c
1/2
M
2δ i
h i + h i+1 (M c) 1/2 .
."./% H
c
M
H)
M/c 1.84
ϕ((M/c) 1/2 ) 2 log(M/c) 1/2
-ϕ i (m) δ i
2(h i + h i+1 ) log M c
1/2 c
δ i
h i + h i+1 c log M
c .
.".7E > .".. .".5
.".0 -
> ."./ .".7 " .".4
."5 - ) >
α, β ∈ [c, M]
c
M
"-
h 0 = 0
h n+1 = 0
.".4A = max
0in
2δ i
h i + h i+1 , B = min
0in
δ i h i + h i+1 .
<
A = (M c) 1/2
c
M
-B c log(M/c)
" "!"
- )
ξ i
(x i−1 , x i )
i = 1, . . . , n
3
S(ξ i ) = f (ξ i ) , i = 1, . . . , n ,
.".8
f
@[a, b]
" !B56 C -0 3 3 + * ,
f f ∈ Lip 1
$
S
C 1
-, ,
x i = a + ih
$h = (b − a)/n
$i = 0, . . . , n
*
O(h 3 )
,,#% 3 3
O(h 3 )
' ./0,1 $, ,
, 2 $ ',$
S (x 0 ) = f 0 = f (x 0 )
S i = S(x i )
$i = 0, 1
h 2 f 0 (S 1 − f (ξ 1 )) − h(f (ξ 1 ) − S 0 )(S 1 − S 0 ) = 0
2 ,# ./0
=
S
x i
x i−1
S(x)dx =
x i
x i−1
f (x)dx , i = 1, . . . , n ,
.".6
f
" (ξ i ∈ (x i−1 , x i )
x i
x i−1
S(x) − f (x))dx = (S(ξ i ) − f (ξ i ))(x i − x i−1
= 0
-
S(ξ i ) = f(ξ i )
"S
"
S (x 0 ) = f (x 0 ) , S (x n ) = f (x n )
."50
S(x 0 ) = f (x 0 ) , S(x n ) = f(x n ) .
."51( ."1- -
2 33 ( * ,
f f ∈ Lip 1
5$
f ∈ C 3 [a, b]
$S
C 1
/ , , 6 #*
O(h 3 )
,,E- 5"
# $
."14 ."12 > )-
m i
" > > .".0i = 1, . . . , n − 1
.".. .".5 -- .".4 .".0
."4
m 0 = α
m n = β
" -ψ 0 (m) ≡ m 0 − α = 0 ,
ψ i (m) ≡ m i
h i ϕ
m i−1 m i
1/2
+ h i+1 ϕ
m i+1 m i
1/2
− δ i = 0 , i = 1, . . . , n − 1 , ψ n (m) ≡ m n − β = 0
."5.
-@ ->.".4 " E-
Ψ(m) = (ψ 0 (m), . . . , ψ m (m))
Ψ(m) = 0
."5. "% - @ ."5. '-K " $
'-K
Ψ (m k )m k+1 = Ψ (m k )m k − Ψ(m k )
."55- ?" ?
Ψ
i
3-∂ψ i
∂m i−1 = 1 2 h i 1
u i ϕ (u i ) ,
∂ψ i
∂m i = h i ϕ(u i ) − 1
2 h i u i ϕ (u i ) + h i+1 ϕ(v i ) − 1
2 h i+1 v i ϕ (v i ) ,
∂ψ i
∂m i+1 = 1 2 h i+1 1
v i ϕ (v i )
-
u i = (m i−1 /m i ) 1/2
v i = (m i+1 /m i ) 1/2
"0 33 (
u i > 0
$v i > 0
∂ψ i /∂m i > 0
χ(x) = ϕ(x) − xϕ (x)/2
" - @χ (x) = (ϕ (x) − xϕ (x))/2
" + ."1 1 -χ (x) > 0 x > 0
"
χ(0) = 0
lim χ(x) = 0
x → 0 +
-"
!
∂ψ i /∂m i−1 > 0
∂ψ i /∂m i+1 > 0
?.".4 ->G " A
i
3-∂ψ i
∂m i − ∂ψ i
∂m i−1 − ∂ψ i
∂m i+1
= h i
ϕ(u i ) − 1 2
u i + 1
u i
ϕ (u i )
+ h i+1
ϕ(v i ) − 1 2
v i + 1 v i
ϕ (v i )
.
+
δ(x) = ϕ(x) − 1 2
x + 1
x
ϕ (x) .
δ (x) = 1 2
1 + 1 x 2
ϕ (x) − 1 2
x + 1
x
ϕ (x)
+ ."1
δ (x) > 0 x > 0
" >δ(x) = 0
>>
2x 4 log x − 3x 4 + 4x 3 − 2x 2 + 2x log x + 1 = 0
x ∗ ≈ 0.734
" > - -#0 33 , ' 78 , ,
m i−1 /m i (x ∗ ) 2 ≈ 0.54
$m i+1 /m i (x ∗ ) 2
+ )
n → ∞
-m i−1
m i
m i+1
-"
! - - )- -
."14 ."12 " ."..
-- -
d i = 0
>m i−1 = m i
"E- -"" B.6C
'-K "
033 9
B(m 0 , R)
$ ,,$
Ψ (x) − Ψ (y) L x − y ∀ x, y ∈ B(m 0 , R) ,
Ψ (m 0 ) −1 b 0 ,
."54Ψ (m 0 ) −1
Ψ(m 0 ) b 1
."52b 0 b 1 L 1/2
R (1 − √
1 − 2b 0 )b 1 /b 0
,*
m ∗ ∈ B(m 0 , R)
,+ +G
Ψ
>""
m k+1 − m ∗ = O( m k − m ∗ 2 )
"! +G
Ψ
ϕ
" ."54 @""& ."5"
m 0
m k
-> ."52 @
b 1
"-
f ∈ C 3 [a, b]
-
z i = 1 h i
x i
x i−1
f (x)dx , i = 1, . . . , n .
)
m i = 2(z i+1 − z i )
h i + h i+1 , i = 1, . . . , n − 1 .
."5/K
m 0 i = f (x i ) + O(h)
h = max h i
m 0 i = f (x i ) + O(h 2 )
"-
m ∗ i = f (x i ) + O(h)
>m 0 i − m ∗ i = O(h)
"h
."4'-K - "