Research Article
Inhomogeneous Lacunary Interpolation by Splines (0, 2; 0, 1, 4) Case
Department of Mathematics, College of Science Education, University of Sulaimani, Iraq
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In approximation theory, spline functions occupy in an important position having a number of applications, in a variety of diverse fields of mathematics and engineering sciences including differential equations, optimal control and nonlinear optimization. There are a great number of techniques developed for various instances of this problem, such as polynomial regression, wavelets and from the view point of differential geometry, developable surfaces are composed of general cylinders and cones (Kvasov, 2000; De Boor, 2001; Saeed and Jwamer, 2004; Sarfraz, 2008) studied the differences types for cardinal interpolation based on a representation of the Fourier transform of the fundamental interpolation.
Meir and Sharma (1973) obtained the error bounds for lacunary interpolation of certain functions by deficient quintic splines, also Saxena and Joshi (1980) studied the inhomogeneous lacunary interpolation by quintic spline (0, 2; 0, 3) case. Jwamer (2001) constructed a non-homogeneous interpolatory by sixtic spline with the third derivatives. In this study, I shall consider the new lacunary interpolation by spline function case (0, 2; 0, 1, 4) to finding the error bounds and suitable assumptions with showed that this spline are exist and unique and also we show that this type of construction of spline functions which interpolates the lacunary data is useful in approximating complicate function and their derivatives on the given interval.
CINSTRUCTION A LACUNARY INTERPOLATION BASED ON SIXITIC SPLINE
The object of this section is to obtain the existence, uniqueness and error bounds of deficient spline interpolating the lacunary data (0, 2; 0, 1, 4).
Let S(x) ε S(5)n,6 denote the class of sixtic splines S(x) on [0,1] such that
• | S(x) ε C5 [0, 1] |
• | S(x) is a polynomial of degree six on each subinterval |
(1) |
It can be verified that if P(x) is a sixtic on [0, 1] then:
Where:
(2) |
For later references we have:
and
and
and
Further, a sixtic P(x) on [1, 2] can be written as:
Where:
(3) |
It is easy to verify that a sixtic Q(x) on [0, 1] can be expressed in the following form:
(4) |
Where:
(5) |
for later references we have:
Also a sixtic Q(x) on [1, 2] can be written as:
(6) |
Where:
(7) |
THE APPROXIMATION OF THE SPLINE FUNCTIONS
Descriptions of the method: Let (Sn, C5) be the class of spline functions with respect to the set of knots xi. The spline functions will denoted by Si(x), where i = 0, 1,..., n. We shall prove the following:
Theorem 1 (Existence and Uniqueness)
For every odd integer n and for every set of 5n+9/2 real numbers
there exists a unique S(x) ε S(5)n,6 such that:
(8) |
Theorem 2
Let fεC5[0,1] and n an odd integer. then the unique sixtic spline Sn(x) satisfying conditions of Theorem 3.1, with fv = f(v/n), v = 0,1,
,n;
we have:
Proof of Theorem 1
For a given S(x) ε S(5)n,6 set h = n-1, Mv = S(5) (vh+), v = 0,1,
..,n-1, Nv = S(5) (vh-), v = 0,1,
.,n. Since, S(5) is linear in each internal (vh, (v+1)h), it is completely determined by the (2n) constants . Also, if S(x) satisfies the requirements of Theorem 1 that for 2vh≤x ≤ (2v+1)h, v = 0,1,
, n-1/2 , it must have the following form:
(9) |
and for (2v+1)h ≤ x ≤ (2v+2)h , v=0,1, ., n-3/2, S(x) has the form:
(10) |
We shall show that it is possible to determine the (2n) parameters , such that the function S(x) given by Eq. 9 and 10 will also satisfy (5) in Theorem 1 and S' (x), S"(x) and S(4) will be continuous on [0,1]. S(x) is continuous because of the interpolating condition Eq. 8 in Theorem 1, S'(x) and S(4)(x) are continuous on [0,1] except at the points (2vh) and (2v+1)h, respectively, v = 0, 1, , n-1/2.
From Eq. 10 we see that Eq. 8 in Theorem 1 is equivalent to:
(11) |
(12) |
Simple calculations show that S"((2v+2)h-) = S"((2v+2)h+) and S(5)((2v+2)h+) are equivalent to:
(13) |
(14) |
Similarly are equivalent to:
(15) |
(16) |
Thus, the theorem will be established if we show that the system of linear Eq. 12-16 has a unique solution. This end will be achieved by showing that the homogeneous system corresponding to Eq. 12-16 has only zero solution.
The following is the homogeneous system of equations for v= 0,1, , n-3/2:
(17) |
(18) |
(19) |
(20) |
(21) |
(22) |
Form Eq. 19 and 20 we have for v = 0, 1, , n-3/2:
and
(23) |
Putting the values and Mn-1 = -13/7 Nn from Eq. 19 in 20 we have:
(24) |
Also, from Eq. 14 and 20 we have:
(25) |
and M0 = -N1 from Eq. 21
Using Eq. 19 we obtain 17Mn-3+5Nn-2 and using Eq. 21 with 25 and 24 we have Mn-3 = Nn-2 = Mn-2 = Nn-1 = 0. Also obtain the system:
By the same manner we get M0 = M1 = = Mn-1 = 0 and N1=N2= N3 = = Nn = 0, (Saxena and Joshi, 1980; Jwamer, 2001), to solution of the homogeneous system for n = 4p and n = 4p+2.
This completes the proof of the Theorem 1.
For the proof of Theorem 2, we shall need the following lemmas:
Lemma 1
Let f ε C6[0,1], n any odd integer and h = n-1, then for Sn(x) ≡ Sn(f, x) of theorem 2, we have
Proof
Since, S(x) is sixtic in 2vh≤ x≤ (2v+1)h, we easily obtain from Eq. 9:
(26) |
Similarly from Eq. 10, since S(x) is sixtic in (2v+1)h≤ x≤ (2v+2)h:
(27) |
Writing v+1 for v in Eq. 26 and subtracting from Eq. 27 we have for v = 0,1,2, ,(n-3)/2 we obtain:
Setting
(28) | |
we have from Eq. 28:
Then using Taylor's expansions, we have:
(29) |
or
(30) |
Fix k, 0≤ k≤ n-3/2. On summing both sides of Eq . 30 for v = k, k+1, , n-3/2 and using the fact that An = 0 (i.e., An = S'"(1) -f"' = 0 since we have
Thus
This completes the proof of lemma 1.
To proof of second part lemma1, since S(x) is sixtic in 2vh≤ x≤ (2v+1)h, we easily obtain from Eq. 10
(31) |
similarly from Eq. 10, since S(x) is sixtic in 2vh≤x≤(2v+1)h:
(32) |
v = 0,1,...,n-1/2.
writing (v+1) for v in (31) and subtracting from (32), we have for v = 0,1,
, n-3/2.
or
and using equation (2.2), to obtain:
(33) |
Then using Taylor's expansions, we have:
(34) |
(35) |
Fix m such that m = 0, 2, ,n-2/2. On summing both sides of (35) for v = k, k+2, , n-2/2, we have:
(36) |
This completes the proof of Lemma (second part).
Lemma 2
Let f ε C6[0,1], n an add integer and h = n-1. Then for Sn(x) = Sn(f,x) of Theorem 1, we have the following:
(37) |
(38) |
(39) |
Proof of Lemma 2
From Eq. 9 for v = 0, 1,
, n-1/2, we have:
Then using Taylor's expansions in above equation we obtain:
Now using lemma 1, we have:
This proves Eq. 38.
We now prove Eq. 39, since S(x) is a sixtic in 2vh≤ x ≤(2v+1)h, we have from Eq. 9 and 10, for v = 0,1, , n-1/2
Then using Taylor's expansions in above equation we obtain:
Now using Lemma 1, we have:
this proves Eq. 39.
Similarly from Eq. 9 and 10 arguing in the previous manner we get.
Thus the Lemma is proved.
Proof of Theorem 2
Let 2vh ≤ x ≤ (2v+1) h ,v = 0, 1,
, n-1/2 from Eq. 9 we have:
(40) |
Since,
we have:
or
(41) |
Since, form (1) for 0 ≤ x ≤1 we have:
(42) |
Since,
(43) |
where, 2vh<α, 2v<(2v+1)h and |x-2vh|≤h,
on using Lemma 2, Eq. 42 and 43, we have:
or
(44) |
Which proves the Theorem 2, when 2vh ≤ x ≤ (2v+1)h and r =5.
The rest of the argument is the same and the theorem is proved for r = 5 and for r = 0,1,2, 3,4 we proceed as follows:
If 2vh ≤ x ≤ (2v+1)h, then:
Since,
and if (2v+1)h ≤ x ≤ (2v+2)h, then:
Hence in every case, i.e., x ∈ [0,1], we have:
The Theorem is proves for r = 4.
and for r = 3, If 2vh ≤ x ≤ (2v+1)h, then
and if (2v+1)h ≤ x ≤ (2v+2)h, then
The Theorem is proves for r = 3.
and for r = 2, If 2vh ≤ x ≤ (2v+1)h, then
and if (2v+1)h ≤ x ≤ (2v+2)h, then
Hence in every case i.e., x∈[0,1] we have:
Theorem is proves for r = 2.
Similarly for r = 0, 1, we have:
and
This completes the proof of Theorem 2.
Two new constructions are present for determine the existence and uniqueness lacunary interpolation using inhomogeneous spline, theoretical background for these inhomogeneous splines proven in the class C5. Also best error bounds for certain combination formula based on the values of the derivative of this spline at all consecutive points approximating function are established.
ambrish kumar pandey Reply
this would be very useful in future .i would like to work wth this.
Faraidun K. Hamasalh
Thank you
Just let me know, are you still working in Numerical Analysis.