Convergence of Quintic Spline Interpolation
Y.P. Dubey
Department Of Mathematics,L.N.C.T. Jabalpur 482 003
Anil
Shukla
Department Of Mathematics Gyan Ganga College Of Technogy,Jabalpur 482 003
ABSTRACT
In this paper, we have investigate existance, uniqueness and error bound of quintic spline interpolation.
Key words : Convergence, Quintic spline,
Interpolation, Error Bound, Deficient.
1.
INTRODUCTION
Piecewise Quartic, Quintic and higher degree spline are popular for smooth and best approximation see Deboor [1]. Interpolation by lower degree spline are widely used in the method of Piecewise Polynomial approximation to represent a non analytic funciton. Interpolation by Quintic and higher degree spline, the maximum error between a function and its interpolant can be controlled by mesh spacing and such function have no corner at the joint of two pieces and therefore no more data are required than lower order method to get desire accuracy. Therefore, quintic and higher degree spline are useful than lower degree spline for best approximation See [1]. In the direction of more higher degree spline Jianzhone and Hung [9] have obtained optimal error bounds for quartic and quintic interpolatory splines (see also Howell and Verma) [2] Wong and Agrawal [5], Gmelling Meling [6] have obtained explicit error bounds for quintic and biquintic interpolatory spline Dubey [10] has obtained existence, uniqueness and error bound for higher degree spline (See also Rana and Dubey) [8]. In the present paper we shall obtain precise error estimate of deficient quintic spline interpolation matching the given function values at mesh points and its derivative at two interior points of the interval.
2. EXISTENCE
AND
UNIQUENESS
Consider a mesh P of [0, 1] given by 0 = xo <x1,..., < xn = 1
such that xi+1 - xi = hi, i = 0, 1,. . . , n - 1. For a positive integer
m, let 5 [0, 1] denote the set of all real algebraic polynomials
of degree less than or equal to m5. For a function s defined
over P. we denote the restriction of s over [xi, xi+1] by s. The
class R(5, 2, P) of deficient quintic splines defined over P is given by
R(5,2,P)={s:s C2[0,1], si5, i=0,1,2 n—1}, wherein R*(5, 2, P) denotes the class of all deficient quintic splines R(5, 2, P) which satisfy the boundary conditions
s’(x0) =f ’ (x0), s’ (xn) =f ’ (xn). (2.1)
We introduce the following interpolatory conditions: s(xi) =f(xi,), i = 0, 1, ...., n (2.2) s' (i) = f '(i), i = 1, 2, ...., n
s'(i) = f' (i), i = 1, 2, ...., n (2.3)
where i = xi-1 + (1/3)hi-1
i = xi-1 + (2/3)hi-1
In fact, we shall prove the following:
THEOREM 2.1. Let f’ exist, then there exists a unique deficient quintic spline s in R*(5, 2, P) which satisfies the interpolatory conditions (2.1) - (2.3).
Proof : Considering a quintic polynomial P(z) on [0, 1], it can be easily verified that
E(z) = E (0) Q1(z) + E '(1) . Q2(z) + E'(1/3) . Q3(z) + E '(2/3) . Q4(z) + E’(O) Q5(z) + E’(1). Q6(z) (2.4) where
2 3 4 5
1
4
120
440
540
216
4
1
)
(
z
z
z
z
z
Q
2 3 4 5
2
120
440
540
216
4
1
)
(
z
z
z
z
z
Q
]
81
189
135
27
[
4
1
)
(
2 3 4 53
z
z
z
z
z
Q
]
81
216
189
54
[
4
1
)
(
2 3 4 54
z
z
z
z
z
Q
]
27
54
23
8
4
[
4
1
)
(
2 3 4 55
z
z
z
z
z
z
Q
]
27
63
51
15
[
4
1
)
(
2 3 5 56
z
z
z
z
z
Q
' 1 '
1 '
1
2
15
)
4
15
(
2
23
h
is
ih
ih
is
is
i1
...,
,
2
,
1
'
,
F
ii
n
(2.6)where
1 1 1 1 1
1
60
60
ii i i i i
i i
i i i
i
f
h
h
f
h
h
h
h
h
h
f
F
1 1
1
)
'
(
))
(
'
(
27
f
if
ih
i))
(
'
)
(
'
(
27
h
if
i
f
i
Since
0
2
7
1
ii
h
h
forh
i
h
i1It may be seen easily that the coefficient matrix of the system of equations (2.6) is diagonally dominant and hence
invertible. This completes the proof of Theorem 2.1.
3. ERROR BOUNDS
In this section of the paper, we shall estimate the bounds of error function by using method of Hall and Meyer [2] and of its derivatives, i.e. e(r)(x) = s(r)(x) - f(r)(x), r =0, 1 for the spline interpolant of Theorem 2.1 which are best possible. Let s(x) be the twice continuously differentiable quintic spline function satisfying the conditions of Theorem 2.1. Now considering f C(6) [0,1] and denoting the unique quintic polynomial by Li[f, x] which agrees with given functional values and derivatives f' (i+1), f' (i+1), f (r)(xi) and f(r) (xi+1), for r= 0, 1, we have for x [xi, xi+1],
|s(x) - f(x)| |si(x) -f(x)| < |si(x) - Li[f, x]| + |Li[f, x] - f(x)|
...(3.1)
We now proceed to get pointwise bounds of both the terms on the right hand side of (3.1). The estimate of the second term can be obtained by following a well known remainder theorem for polynomial interpolation of Davis [4].
|Li[f, x] - f(x)| <
!
6
6
i
h
|t2(1-t) 2 (1-3t)(2-3t)|F/9 …..(3.2)
Where t = (x-xi)/hi and F =
1 0
max
x |f
6 (x)|
We next turn over attention to get a similar bounds for |si(x)-Li [f,x]|. From (2.4), we have
)
(
).
(
'
)
(
).
(
'
]
,
[
)
(
x
L
f
x
h
e
x
Q
5t
h
e
x
1Q
6t
s
i
i
i i
i i……(3.3)
Thus,
|
)
(
||
)
(
'
|
|
]
,
[
)
(
|
s
ix
L
if
x
h
ie
x
iQ
5t
.
|
)
(
||
)
(
'
x
1)Q
6t
e
h
i i
……(3.4)
As
|
4
19
19
|
)
1
(
|
4
1
|
)
(
|
|
)
(
|
Q
5t
Q
6t
t
t
t
2
……(3.5)
Thus, applying (3.5) in (3.4), we have
)
(
|}
)
(
'
|
|,
)
(
'
{|
max
|
]
,
[
)
(
|
s
ix
L
if
x
h
ie
x
ie
x
i1k
t
……(3.6)
where k(t) =
|
(
1
)
|
19
19
4
|
4
1
2
t
t
t
Let the max |e’(xi)| exist for i=j, therefore (3.6) may be written as
0 < i < n
|
)
(
)
(
'
|
|
]
,
[
)
(
|
s
ix
L
if
x
h
e
x
ik
t
……(3.7)Where h=max hi, we now next proceed to find the upper
bound for |e’(xi)| needed in eq. (3.7). Replacing s’(xj) by e’(xj) in (2.6), we have
' 1 1 1
' 1
2
15
)
4
15
(
2
23
j ji j j j j
j
e
h
h
e
h
e
h
' 1 1 '
1 '
1
2
15
)
4
15
(
2
23
F
jh
jf
jh
jh
jf
jh
jf
j= B(f) (say) …(3.8) In view of the fact that B(f) is a linear functional which is zero for polynomial of degree 5 or less, we can apply the Peano Theorem Davis [4] to obtain
11
.
]
)
[(
!
5
)
(
)
(
5) 6 ( j
j
x
x
B
x
y
dy
y
f
f
B
…..(3.9)Thus from (3.9), we get,
11
.
|
]
)
[(
|
!
5
|
)
(
|
j 5j
x
f
B
x
y
dy
F
f
B
…..(3.10)Further, it can be observed from (3.9) that for 1
1
jj
y
x
x
4 1
5
)
(
)
4
15
(
]
)
[(
x
y
h
h
x
y
B
j j j5 4
1
1
(
)
60
(
)
2
75
h
jx
jy
x
jy
5 1 1 1
1
)
(
60
x
y
h
h
h
h
h
h
j j j
j j
j j
4
1 5
1
)
(
)
(
135
j
y
j
y
4 4
1
135
(
y
)
(
y
)
In order to estimate the integral of r.h.s. of (3.10), we rewrite the above expression in the following symmetric form about xj to get.
60
(
)
2
45
]
)
[(
1 1 5y
x
h
h
h
y
x
B
j jj j
41
)
(
x
j
y
h
jx
j1
y
j
1 3 4 1)
(
90
)
(
2
255
j j j j j jh
y
x
y
x
h
h
h
5 1 4 1 3 1 225
)
(
10
)
(
15
x
jy
h
jx
jy
h
jh
jj
j
y
3 2 1 4 1 1)
(
90
)
(
2
15
y
x
h
y
x
h
h
h
j j j j j j 3 2 1 3 1 2)
(
30
)
(
75
x
j
y
h
j
h
jx
j
y
3 13
80
)
(
x
j
y
h
jj j
y
x
5 1
)
(
60
[
x
y
h
h
j j j
3 2 4)
(
90
)
(
2
15
y
x
h
y
x
h
j j
j j
]
3
80
)
(
30
)
(
75
x
j
y
2h
3j
h
j4x
j
y
h
j5
1
j jy
x
4 5 1)
(
2
255
)
(
60
[
x
y
h
x
y
h
h
j j j j j
3 2 3)
(
15
)
(
90
x
j
y
h
j
x
j
y
h
j
]
25
)
(
10
x
j
y
h
4j
h
5j
1
1
jj
y
41
)
(
60
2
45
j j j j j jh
y
x
y
x
h
h
h
1 1
jj
y
x
From the above expression, it follows that
1 1 4 1 4 1 572
.
256
|
]
)
[(
|
j j xx
B
x
y
dy
h
jh
jh
jh
j….(3.11) Combining (3.10) with (3.11), we have
h
h
F
h
h
f
B
j j 1 4j j41!
5
.
72
.
256
|
)
(
|
….(3.12) Thus from (3.8) and (3.12), it follows that
1 4 1 4 12
23
2
7
!
5
.
72
.
256
|
|
|
)
(
'
|
j j j j j j j jh
h
h
h
h
h
e
x
e
….(3.13) Now using (3.2), (3.7) along with (3.13) in (3.1), we have9
/
)
3
2
)(
3
1
(
)
1
(
|
)
!
5
(
|
)
(
|
2 26
F
t
t
t
t
h
x
e
|
)
4
19
9
)(
1
(
|
)
(
'
|
2
h
e
t
jt
t
t
t
|
)
(
|
)
!
5
(
6t
C
F
h
….(3.14) Where
(
1
3
)(
2
3
)
(
9
19
4
)
/
15
)
1
(
)
(
|
C
t
F
t
t
t
t
t
2
f
)
(
*
t
K
(say) .(3.15)THEOREM 3.1. Let s(x) be the quintic spline interpolant of Theorem 2.1 interpolating a given function f(x) and
]
1
,
0
[
6C
f
then,|
)
(
|
max
!
5
/
|
)
(
|
(6)1 0 6
0
h
f
x
K
x
e
x
....(3.16)Where
max
|
(
)
|,
1 0
0
C
t
K
x
defined by the equation (3.15)Also, we have
|,
)
(
|
max
!
5
/
|
)
(
'
|
(6)1 0 5 1
1
K
h
f
x
x
e
x
....(3.17)Where K1 = 412/15
Furthermore, it can be seen easily that K in (3.16) be improved for an equally spaced partition. Inequality (3.17) is also best possible. Also, we have
|,
)
(
|
max
)
!
5
/
(
|
)
(
'
|
(6)1 0 5
1
h
f
x
K
x
e
x
....(3.18)where K2 is positive constant. Equation (3.14) proves (3.16)
Now, we turn to see that the inequality (3.16) is best possible in the limit. Considering f(x) = x6/6! and using the Cauchy formula Davis [4], we have
2 2 6 6 6
)
1
(
!
6
/
]
,
!
6
/
[
t
x
x
h
t
t
L
!
6
/
)
3
/
2
)(
3
/
1
(
t
t
....(3.19)Moreover, for the function under consideration (3.8) the following relation holds for equally spaced knots,
1 1 1 1 1 5 6
2
15
19
2
23
!
5
72
.
256
!
6
j jj
e
e
e
h
x
B
....(3.20) Considering for a moment
)
(
'
)
(
'
)
15
)(
!
5
(
72
.
256
)
(
'
1 15
i ii
e
x
e
x
h
x
e
....(3.21) we have from (3.4),
))
(
)
(
(
)
!
5
(
15
72
.
256
]
,
[
)
(
5 66
t
Q
t
Q
h
x
f
L
x
s
i
}
4
/
)
4
19
9
)(
1
(
[
)
!
5
(
15
72
.
256
2 6
h
t
t
t
t
....(3.22) Now combining (3.19) and (3.22), we get for
x
i
x
x
i1
4
4
19
9
)
1
(
15
72
.
256
)
!
5
(
)
(
)
(
2 6t
t
t
t
h
x
f
x
s
(
1
)(
1
/
3
)(
2
/
3
)
6
1
t
t
t
t
....(3.23)From (3.23), it is clearly observed that (3.16) is best possible, provided we could prove that
)
15
)(
!
5
(
)
72
.
256
(
)
(
'
)
(
'
)
(
'
5 1 1h
x
e
x
e
x
e
i
i
i
....(3.24) In fact (3.24) is attained only in the limit. The difficulty will appear in the case of boundary conditions, i.e.,
0
)
(
'
)
(
'
x
0
e
x
n
e
. However, it can be shown that as wemove many subintervals away from the boundaries,
).
15
)(
!
5
/(
72
.
256
)
(
'
x
h
5e
i
For that, we shall apply(3.20) inductively to move away from the end conditions
.
0
)
(
'
)
(
'
x
0
e
x
n
e
The first step in this direction is to establish that
e
'
(
x
i)
0
for some i, i=1, ...., n which can be shown by contradictory result, let e'(xi) < 0 for some i = 1, 2....n-1. Now making use of (3.17), we get1 1 1 1 1 5
2
15
19
2
23
15
)
!
5
(
72
.
256
e
je
je
jh
for some j L.H.S. positive and r.h.s. is negative. Thus we have a contradication. Thus,
e
'
(
x
i)
0
for i=0, ...., n.Now from (3.20) we can write
11 1 1 5
2
23
2
15
)
!
5
(
72
.
256
)
(
'
19
e
x
ih
e
je
j!
5
72
.
256
)
(
'
19
5h
x
e
j
Since
e
'
(
x
i)
0
,
we get,
)
19
)(
!
5
(
72
.
256
)
(
'
5h
x
e
i
....(3.25)Now again using (3.25) in (3.20), we have
19
4
1
)
!
5
(
19
72
.
256
)
(
'
5h
x
e
i i = 2, 3,...., n-2Thus, it is clear that a repeated use of (3.20) leads us to
...]
)
19
/
4
(
)
19
/
4
1
[
)
!
5
(
19
72
.
256
)
(
'
2 5
h
x
e
i ...(3.26) Now, it can be seen easily that the right hand side of (3.26))
!
6
(
15
72
.
256
h
5
and hence in the limiting case.)
!
5
(
15
72
.
256
)
(
'
5h
x
e
i
....(3.27)which verifies the proof of (3.17).
Thus, corresponding to the function
,
)
!
6
(
)
(
6x
x
f
(3.26)and (3.27) imply
)
!
5
(
15
72
.
256
)
(
'
5h
x
e
i
in the limit forequally spaced knots. This completes the proof of Theorem 3.1
4.
REFERENCES
[1] Deboor, C., A Practical Guide to splines, Applied Mathematical Science, Vol. 27 Springer - Verlag, New York, 1979.
[2] Hall, C.A. and Meyer, W.W., J. Approximation Theory, 16(1976), pp. 105-122.
[4] Davis, P.J., Interpolation and approximation. New Yrok, 1961.
[5] Agrawal, R.P. and Wong, P.J.Y. Error Inqualities in Polynomial Interpolation and their application, Kumar Academic Publisher's (1993).
[6] Gemlling, R.H.J. and Meyling, G. In Interpolation by Bivariate Quintic splines of class ...(Construction in theory of functions 87), (Ed) Sendov et.al.) (1987), 152-61.
[7] Rana, S.S., Dubey, Y.P. and Gupta, P. Convergence of Deficient Quintic Spline Interpolation, Journal of Indian Math Soc. Vol. No. 70 No.-1-4 (2003), pp. 41-28. [8] Rana, S.S. and Dubey, Y.P. Best Error Bound of Quintic
Spline Interpolation, J. Pure and app. Math 28 (10) 1937-44, 1997.
[9] Jianzhong, W. and Hung, D. Scientia Sinica 24 (1982), 1130-41.
