International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015
The Change Effect for Boundary Conditions to Minimize Error Bound in Lacunary Interpolation by Forth Spline Function Muhannad A. Mahmoud
i , i 0,2,....,2m and n=2m+1 . We define 2m the class of Spline function S P ( 4,3, n ) as follows : any element S ( x)S P (4,3, n) if the following xi
Abstract—The researcher interested in spline function because it is necessary function in interpolation process, by putting a certain condition for spline function from forth degree which is as follows
Sn( x0 ) 0, Sn( x2 m ) 0 , then the aim of this research
condition satisfied :
is to change the previous boundary conditions which is as
Sn( x0 ) f ( x0 ), Sn( x2 m ) f ( x2 m )
(ii) S ( x) is polynomial of deg ree four in each [ x2i , x2i 2 ] (1) i 0,1,..., m 1 (i ) S ( x) C 3[0,1]
to get minimum
error for spline function from forth degree.
In this paper we prove the following : Theorem 1:
Index Terms— Spline function ; Boundary Conditions.
Given arbitrary number
f ( 2) ( x ), f ( 2) ( x2 m ) there S n ( x) S p (4,3, n) such that
and
I. INTRODUCTION Spline functions appears in the late middle of twentieth century, According to Schoenbery [5] , the interest in Spline functions is due to the fact that , Spline function are good tool for the numerical approximation of functions and that they suggest new ,challenging and rewarding problems . For more information about a Spline function , one is referred to Alberg ,Milson and walsh [1] .Lacunary interpolation by Spline appears whenever observation gives scattered or irregular information about a function and its derivatives , but with out Hermite condition . Mathematically , in the problem of interpolation given data
a i, j
f ( x2 i ) , f (t 2 i )
by polynomial
p n (x )
S n ( x2 i ) f ( x2i )
i 0,1,2....., m
S n (t2i ) f (t2i )
i 0,1,2...., m 1
S
( 2) n
( x0 ) f
( 2)
( x0 ) , S
( 2) n
( x2 m ) f
( 2)
i=0,1,2…,n-1
exists a unique spline
( x2 m )
(2)
Theorem 2:
f C 4 [0,1] and S n Sp(4,3, n) be
Let
a unique
spline
satisfying the condition of (theorem 1.1) , then 1 S ( x) f ( r ) ( x) 5.6404257m r 4 w( f , ) m r 4 f ( 4) ; r 0,1,2,3 m
of
(r ) n
degree of most n satisfying :
where
p ( xi ) ai , j
, i 1,2,3,..., n , j 0,1,2 ,......., m
( j) n
w( f f
we have Hermit interpolation if for i , the order j of derivatives in form unbroken sequence . If some of the sequence are broken , we have lacunary interpolation . In another communications we shall give the applicationsof Spline functions obtaining the approximation solution of boundary value problem , for more about applications of Spline functions , see [2,4] . For description our problem , let
of
the
interval
[0,1]
,
II. TECHNICAL PRELIMINARIES: If p(x) is a polynomial of degree Four on [0,1] (because we want to construct a spline function of degree four ) then we have :
1 P( x) P(0) A0 ( x) P( ) A1 ( x) P(1) A2 ( x) P // (0) A3 ( x) P // (1) A4 ( x) 3
:0 xo x2 x3 ..... x2 m 1 by a uniform partition
( 4)
1 ) max f ( 4 ) ( x) f ( 4 ) ( y ) : x y h, x, y [0,1] m max f ( 4 ) ( x) : 0 x 1
( 4)
with
……..(3) Where
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The Change Effect for Boundary Conditions to Minimize Error Bound in Lacunary Interpolation by Forth Spline Function
....( 4)
1 [27 x 4 54x 3 38x 11] 11 1 A1 ( x) [81x 4 162x 3 81x] 22 1 A2 ( x) [27 x 4 54x 3 5 x] 22 1 A3 ( x) [15x 4 41x 3 33x 2 7 x] 66 1 A4 ( x) [12x 4 13x 3 x] 66 A0 ( x)
f C 4 [0,1] we have the following expansions: 4 2 f ( x2i 2 ) f ( x2i ) 2hf / ( x2i ) 2h 2 f // ( x2i ) h 3 f /// ( x2i ) h 4 f ( 4 ) (1, 2i ) 3 3 x2i 1, 2i x2i 2 4 2 f ( x2i 2 ) f ( x2i ) 2hf / ( x2i ) 2h 2 f // ( x2i ) h 3 f /// ( x2i ) h 4 f ( 4 ) ( 2 , 2i ) 3 3 x2 i 2 2 , 2 i x2 i 2 2 4 2 4 ( 4) f (t 2i ) f ( x2i ) hf / ( x2 i ) h 2 f // ( x2i ) h 3 f /// ( x2i ) h f ( 3 , 2 i ) 3 9 81 243 x2 i 3 , 2 i t 2 i
In the subsequent section we need the following values for
4 8 32 32 4 ( 4) f (t2i 2 ) f ( x2i ) hf / ( x2i ) h 2 f // ( x2i ) h3 f /// ( x2i ) h f (4 , 2i ) 3 9 81 243 t2i 2 4 , 2i x2i
2 // 2 4 hf ( x2i ) h 2 f /// ( x2i ) h 3 f ( 4 ) (5, 2i ) 3 9 81 x2 i 5 , 2 i t 2 i 4 8 f // (t2i 2 ) f // ( x2i ) hf /// ( x2i ) h 2 f ( 4 ) (6, 2i ), t2i 2 6, 2i x2i 3 9 2 2 f // (t2i ) f // ( x2i ) hf /// ( x2i ) h 2 f ( 4) (7 , 2i ), x2i 7 , 2i t2i 3 9 // // /// f ( x2i 2 ) f ( x2i ) 2hf ( x2i ) 2h2 f ( 4) (8, 2i ) , x2i 2 8, 2i x2i
f / (t 2i ) f / ( x2i )
f // ( x2i 2 ) f // ( x2i ) 2hf /// ( x2i ) 2h2 f ( 4) (9, 2i ), x2i 9, 2i x2i 2 f /// (t2i ) f /// ( x2i )
2 ( 4) hf (10, 2i ) 3
, x2i 10 , 2i t2i ....( 5) III. PROOF OF THEOREM 1:
The proof depends on the following representation of
Sn ( x) for 2ih x (2i 2)h , i 0,1,...., m 1
we
have
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International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015
x 2ih x 2ih x 2ih ) f (t2i ) A1 ( ) f ( x2i 2 ) A2 ( ) 2h 2h 2h x 2ih x 2ih // // 4h 2 S n ( x2i ) A3 ( ) 4h 2 S n ( x2i 2 ) A4 ( ) .......... (6) 2h h
S n ( x) f ( x2i ) A0 (
On using (3.0) and the conditions
Sn (0) f // (0) , Sn (1) f // (1) //
//
we say that
S n (x )
(7)
as given by (6) satisfies (1) is fourth in
[ x2 i , x2 i 2 ] , i 0,1,2,...., m 1.we also need to show that
S // n ( x2i ) where i 1,2,...., m 1 uniquely .For this purpose we use the fact that S /// n ( x2i 2 ) S /// n ( x2i ) , i 1,2,..., m 1 with the help of (6) and (7) reduce to
whether it is possible to determine
41 2 // 32 35 h S n ( x2i 1 ) h 2 S // n ( x2i ) h 2 S // n ( x2i 2 ) 22 22 22 243 243 81 243 81 f ( x2i ) f (t2i ) f ( x2i 2 ) f (t2i 2 ) f ( x2i 2 ) 44 44 44 44 22 ........ (8)
but (7) is a strictly tri
diagonal dominant system has a unique solution[3] . Thus
S // n ( x2i ) , i 1,2,..., m 1 can be obtained uniquely by the system (8)
This theorem is completed. IV. ESITIMATES: In order to prove theorem 1.2 , we need the following: Lemma 4.0 : Let
E2i S // n ( x2i ) f // ( x2i )
max E2i
f C 4 [0,1] we have
then for
81 2 1 h w( f ( 4) ; ) , i 0,1,2,..., m 1 108 m
(9)
Proof: From (8) we have
32 2 // 41 2 // // h ( S h ( S n ( x2i ) f // ( x2i )) n ( x 2 i 2 ) f ( x 2 i 2 )) 22 22
35 2 // 243 81 243 h ( S n ( x2i 2 ) f // ( x2i 2 )) f ( x2i ) f (t 2i ) f ( x2i 2 ) 22 44 44 44
243 81 41 32 35 f (t 2i 2 ) f ( x2i 2 ) h 2 f // ( x2i 2 ) h 2 f // ( x2i ) h 2 f // ( x2i 2 ) 44 22 22 22 22 2 4 h f 44
( 4)
( 3, 2 i )
82 4 h f 22
( 4)
54 4 h f 44
( 8 , 2 i )
( 4)
(1, 2i )
70 4 h f 22
( 4)
32 4 h f 44
( 4)
( 4 , 2 i )
54 4 h f 22
81 (9, 2i ) 1 h 4 w( f 22
( 4)
,
( 4)
( 2 , 2 i )
1 ) , 1 1 m
The result (9) follows on using the property of diagonal dominant [3]. Lemma 4.1: Let f C [0,1] then 4
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The Change Effect for Boundary Conditions to Minimize Error Bound in Lacunary Interpolation by Forth Spline Function
135 1 hw( f ( 4 ) ; ) 44 m 378 1 (ii) S lll n ( x 2i ) f lll ( x 2i ) hw( f ( 4) ; ) 88 m 33 1 (iii) S lll n (t 2i ) f lll (t 2i ) hw( f ( 4) ; ) 22 m 31 2 1 (iv) S ll n (t 2i ) f ll (t 2i ) h w( f ( 4) ; ) 44 m 2546 3 1 (v) S l n (t 2i ) f l (t 2i ) h w( f ( 4 ) ; ) 9801 m (i ) S lll n ( x 2i ) f lll ( x 2i )
Proof: From (6) we have
h 3 S lll n ( x 2i 2 )
81 243 81 123 2 ll 39 f ( x 2i ) f (t 2i ) f ( x 2i 2 ) h S n ( x 2i ) h 2 S ll n ( x 2i 2 ) 22 44 44 66 66
Hence
81 243 81 123 2 ll f ( x 2i ) f (t 2i ) f ( x 2i 2 ) h ( S n ( x 2i ) f ll ( x 2i )) 22 44 44 66 39 123 2 ll 39 h 2 ( S ll n ( x 2i 2 ) f ll ( x 2i 2 )) h 3 f lll ( x 2i ) h f ( x 2i ) h 2 f ll ( x 2i 2 ) 66 66 66 1 27 4 ( 4 ) 39 123 2 ll h 4 f ( 4 ) ( 3, 2 i ) h f (1, 2i ) h 4 f ( 4 ) ( 4, 2i ) h ( S n ( x 2i ) f ll ( x 2i )) 22 22 33 66 39 h 2 ( S ll n ( x 2i 2 ) f ll ( x 2i 2 )) 66 h 3 ( S lll n ( x 2i ) f lll ( x 2i ))
27 4 1 123 ll 39 h 2 w( f ( 4 ) ; ) ( S n ( x 2i ) f ll ( x 2i )) h 2 ( S ll n ( x 2i 2 ) f ll ( x 2i 2 )) ; 2 1 22 m 66 66
By using (9) the lemma (4.1)(i) follows ,the proof of lemma (4.1)(ii-v) are similar ,we have only mention that
h 3 S lll n ( x 2i )
81 243 81 57 2 ll 105 2 ll f ( x 2i ) f (t 2i 2 ) f ( x 2i 2 ) h S n ( x 2i 2 ) h S n ( x 2i ) 44 44 22 66 66
h 3 S lll n (t 2i )
27 81 27 63 9 2 ll f ( x 2i ) f (t 2i ) f ( x 2i 2 ) h 2 S ll n ( x 2i ) h S n ( x 2i 2 ) 22 44 44 66 66
h 2 S ll n (t 2i )
18 27 9 2 5 f ( x 2i ) f (t 2i ) f ( x 2i 2 ) h 2 S ll n ( x 2i ) h 2 S ll n ( x 2i 2 ) 11 11 11 33 33
and hS l n (t 2i )
12 39 37 34 2 ll 14 2 ll f ( x 2i ) f (t 2i ) f ( x 2i 2 ) h S n ( x 2i ) h S n ( x 2i 2 ) 11 44 44 297 297
Proof of theorem (2): For 0 Z 1, we have
A0 ( Z ) A1 ( Z ) A2 ( Z ) 1
....(10)
let x 2i Z x 2i 2 . On using (10) and (7) , we obtain
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International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-1, Issue-6, December 2015
x 2ih x 2ih ) ( S lll n ( x2i 2 ) f lll ( x)) A1 ( ) 2h 2h x 2ih ( S lll n (t2i ) f lll ( x)) A2 ( ) L1 L2 L3 2h .......... .......(11)
S lll n ( x) f lll ( x) ( S lll n ( x2i ) f lll ( x)) A0 (
A0 ( x) 1, A1 ( x) 1 and A2 ( x) 1
from (4) it follows that Since
f lll ( x) f lll ( x2i ) ( x x2i ) f ( 4) ( ) , x2i x
Therefore
x 2ih x 2ih ) ( S lll n ( x2i ) f lll ( x2i ) ( x x2i ) f ( 4) ( )) A0 ( ) On using lemma 2h 2h 2h , we obtain
L1 ( S lll n ( x2i ) f lll ( x)) A0 (
x x 2i
(4.1)(i) and
L1 S lll n ( x2i ) f lll ( x2i ) ( x x2i ) f ( 4 ) ( ) A0 (
x 2ih ) 2h
Therefore
L1
135 1 hw( f ( 4 ) ; ) 2h f ( 4 ) 44 m
(12)
Similarly
378 1 hw( f ( 4 ) ; ) 2h f ( 4) 88 m x 2ih L3 ( S lll n (t2i ) f lll ( x)) A2 ( ) ( S lll n (t2i ) f lll (t2i ) f lll (t2i ) f lll ( x2i ) 2h x 2ih ( x x2i ) f ( 4 ) ( )) A2 ( ) (13) 2h L2
.
From (5) we have
2 ( 4) hf (10, 2i ) 3 t2i , but (3t2i x2i ) 2h
f lll (t 2i ) f lll ( x 2i ) where x2 i
10 , 2i
L3 (S lll n (t2i ) f lll (t2i ) 2h f ( 4) (0 ) f ( 4) ( ) , x2i , 0 x
Hence
On using the above result with lemma (4.1)(iii), we obtain
L3
77 1 hw( f ( 4 ) ; ) 22 m
......(14)
putting (12) ,(14) in (11) ,we obtain
956 1 hw( f ( 4) , ) 4h f ( 4 ) 88 m
S lll n ( x) f lll ( x)
.......(15)
this prove (2) for r=3 to prove (2), when r=2 x
Since S
ll
n
( x) f ( x) ( S lll n (t ) f lll (t ))dt ( S ll n (t 2i ) f ll (t 2i )) ll
t2 i
on using lemma (4.1)(iv) and (15) we obtain
S ll n ( x) f ll ( x)
987 2 1 h w( f ( 4 ) , ) 8h 2 f ( 4 ) 44 m
.......(16)
which is proof (2) for r=2 the proof (2) for r=1 since x
S l n ( x) f l ( x) ( S ll n (t ) f ll (t ))dt S l n (t 2i ) f l (t 2i ) t2 i
On using lemma (4.1)(v) and (16),we get
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The Change Effect for Boundary Conditions to Minimize Error Bound in Lacunary Interpolation by Forth Spline Function
S l n ( x) f l ( x)
884509 3 1 h w( f ( 4 ) , ) 16h3 f ( 4 ) 19602 m
.......(17)
Which proof (2) for r=1, if r=0, we have x
S n ( x) f ( x) ( S l n (t ) f l (t ))dt S n (t 2i ) f (t 2i ) t2 i
sin ce S n (t 2i ) f (t 2i ) 0 Thus
884509 4 1 h w( f ( 4 ) , ) 32h 4 f ( 4 ) 9801 m 1 Since 2mh=1 , then h put it in above result this completes the proof of theorem (1.2) . 2m Sn ( x) f ( x)
V. CONCLUSIONS: Throughout this paper we mentioned that some time the changes of boundary conditions effect on minimizing error bound in the subject of lacunary interpolation by Spline function .
REFERENCES [1] Ahlberg,J.H., E.N.Nilson and J.L.walsh ,(1967) .The theory of Spline and their applications ,Academic press ,new york and London. [2] Bokhary,M.A.,H.P. Dikshit and A.A. Sharma .(2000),Birkhoff Interpolation on some perturbed roots of unity. Numerical Algorithms. Vol.25,No.(1-4),(2000),pp.(47-54). [3] Kincaid, D and W. Cheney,(2002). Nuumerical analysis , Mathematics of scientific computing third edition, printed in USA. [4] Lenard , M., (1999), Modified (0,2) interpolation on the roots of Jacobi polynomial ,I(Explicit formulae), Acta Math.,Hungar ,Vol.84, No.(4), pp.(317-327). [5] Schoenberg, I.J. , (1986),On Ahlberg – Nilson extention of Spline interpolation : the g-Spline and their optimal properties .J math. Anal .Appl.,Vol.4,No.(2),pp.(207-231).
Muhannad A. Mahmoud, Department of Mathematics, Science , University of Kirkuk, Iraq
College of
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