Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
INTERPOLATION BY QUARTIC SPLINE
K.K.Nigam
1, Y.P. Dubey
2, Brajendra Tiwari
3, Anil Shukla
41
Mathematics Deptt., LNCT Jabalpur, MP, (India)
2
Sarswati Institution of Engg. & Technology, Jabalpur MP, (India)
3
R.K.D.F University, Bhopal MP, (India)
4
Mathematics Deptt., GGCT, Jabalpur, (India)
ABSTRACT
In this paper we have obtained a precise estimate of error bounds or deficient quartic spline interpolation.
Keywords: Defficient, Error Bound, Quartic Spline, Interpolation
I. INTRODUCTION
For smooth and more efficient approximation of function cubic and higher degree spline are popular (see Deboor [1]), As in the study of linear spline the maximum error between function and its interpolation can be controlled by mesh spacing such function have corner at the joint two linear pieces and therefore the usually require more data than higher order method to get desired accuracy. In the direction of same higher order method, we refer to Rana & Dubey [5,6], Hall & Mayer [7], the error bounds for quartic spline interpolation obtain by Howell and Verma [ 2 ] and the error bound of spline of degree six obtained by Dubey [ 3 ], (also see Meir and Sharma [9]). The purpose of this paper is to construct a spline method for solving an interpolation problem using piecewise quartic polynomial and obtain error bound.
II. EXISTENCE AND UNIQUENESS
Let a mesh on [0, 1] be given by P : O = x0 < x1<……….xn=1 with hi =xi-xi-1 for i=1, 2………n.
Let S(4, p) denote the set of or all algebraic polynomial of degree 4.
And si is the restriction of s over [ , ], the class s*(4, p) of deficient quantic spline is defined by S* (4, P) = {si : si𝝐 c1[0, 1] , si 𝝐 s (4, P) for i=1, 2, ……..n}
Where in S**(4, P) denotes the class of all deficient spline S * (4, P) which satisfies the boundary condition.
s(x0) = f(x0), s(xn) = f(xn) (2.1)
Problem 2.1 : suppose f exist over P then there a exit unique spline interpolation s 𝝐 s** (4, P) of f which satisfies interpolation conditions
s(i) = f(i) s(i) = f(i) s1(i) = f 1(i) (2.2) Where
Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
Let Q(z) be a quartic polynomial on [0, 1] then it is easy to verify that(2.3)
Where
We are now set to answer problem 2.1 in theorem 2.1
Theorem 2.1:
There exist a unique deficient quartic spline in S**(4, P) which satisfies interpolatory condition ( 2.1 ) – (2.2 )Proof : Let
1
Then in view of condition (2.1) - (2.2) we now express equation (2.3 ) in terms of restriction si of s in [xi,xi+1] as follows
si(x)=f(i)P1(t)+f(i)P2(t)+hi f1(i)P3(t)+si(x)P4(t)+si+1P5(t)____ (2.4) Since s𝝐 C 1[a, b] we have from equation ( 2.4 ) we get
Clearly matrix is diagonally dominant and hence invertible. Thus the system of equation has unique solution this completes the proof of theorem 2.1
III. ERROR BOUNDS
In this section of paper error bounds
e(x) = f(x) – s(x) is obtained for the spline interpolation of theorem 2.1 by following approach used by Hall and Meyar. We sell denote by hi [f, x] the unique quartic acquiring with
f(i), f(i), f’( i), f(xi) and f1 (xi+1) and let f 𝝐 C5 [0, 1]
Now, consider a first continuously differentiable quartic spline S of theorem 2.1 we have for xi < x < xi+1
| f(x) – s(x) | = | f(x) – si(x) |
< | f(x)-Li[f, x]| + |Li[f,x] - S(x)| (3.1) Thus it is clear from (3.1) that in order to get the bounds of e(x)
We have to estimate point wise bounds of both the terms on the right hand side of (3.1) by a well know remainder theorem of Cauchy (see Davis [8 ] we see that
Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
(3.2) WhereF = max | f(5) (x)|
we next proceed to obtain bounds for
| Li [f, x] – si(x) | it follows from (2.4 ) that
| Li [f, x]-si(x) | < | e(xi) P4(t) + e(xi+1) P5(t)| (3.3) Thus
| Li[f, x] – si(x)| < | e(xi)| | P4(t) | + | e(xi+1) | |P5(t) (3.4) Since
P4(t) = [1-9t + 29t2 – 39t3 + 18t4] and
0 < t < 1
= K(t) (say) (3.5) Now using (3.5) in (3.4) we have
| Li[f, x] – si(x) | < max {|e(xi)|, |e(xi+1)|} k(t)
Setting |e(xj)| = max |e(xi)| (3.6)
j=1, 2, ……….n-1 and
h = max (hi) (3.7) i =1, 2, …………n-1
we see that (3.6) may be written as
|Li[f , x]-s(x)| < |e(xj)| k(t) (3.8)
First we have to find upper bound of |e(xj)|
Replacing s(xj) by e(xj) in (3.8) we get
(3.9)
Now
(3.10)
Since E(f) is a linear functional which is zero for polynomials of degree 4 or less and applying the peano theorem (see Davis [8 ]) we have
(3.11)
Now from (3.11) it follows that
(3.12)
Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
We rewrite the expression (3.9) in the following symmetric form of xj Thusj < y < xj+1
j < y < j
xj < y < j
xj-1 < y < j-1 = yj-1
j-1 < y < j-1
j-1 < y < xj From above expression it is follows that
E [(x-y)+4] is non negative in xj-1< y < xj+1 Thus we see that
(3.13) Thus we have following from (3.12) when we apply to (3.13)
(3.14) Combining (3.7) (3.9) (3.12) with (3.14) we have
max | e(xj) | = | ej | j=1, 2,…………..n-1
(3.15) Now making use of equation (3.2) and (3.7) in (3.1) and then using (3.15) along with (3.8) we see that
(3.16)
(3.17) Where
Thus we prove the following
Theorem 3.1 :
Suppose S(x) is the quartic spline interpolation of theorem 2.1 and f 𝝐 c5 [0, 1] then(3.18)
Where k = max |c(t)| given by (3.17) also we have
0 < x < 1 (3.19)
Equation on (3.15) and (3.17) resp. prove the inequality (3.18) and (3.19) of Theorem 3.1 we shall prove the inequality (3.18) is best possible in the limit.
we can easily seen that by cauchy formula given in [4 ] that
Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
(3.20)More over for equally spaced knots, we have from (3.9) that
(3.21) Consider for a moment
(3.22) We have from ( 3.6 )
(3.23)
Combine (3.20) and (3.23) we have
(3.24)
From (3.24) it is clearly observed that (3.18) is best possible provide that. We could prove that
(3.25)
In fact (3.25) is attained only in the limit. The difficulty will take place in the boundary condition e(x0) = e(xn) = 0
However it can be shown is that as we move many submitted away from the boundary
for that we shall apply (3.21) in inductively to move away from the end condition
e(x0) = e(xn)=0
first step in this direction is to show that e(xj)> for j=0,………..n Which can be prove by contradiction assumption.
Let e(xj)<0 for some i=1, 2, ……n-1. Now a making use of (3.19)
4>77 this is a contradiction. Hence e(xj)>0 for j=0,……….n Now from equation (3.21)
Similarly ej > 0
from j=1, 2, ………..n-1 (3.26)
Volume No.03, Issue No. 12, December 2014 ISSN (online): 2394-1537
Again using ( 3.26) in (3.21)We have,
Repeated use of (3.21) following that
(3.27)
Now it can be easily see that r, h, s of
And hence in the limiting case
(3.28) Which verifies (3.18) inequality thus corresponding to the function
(3.29)
in the limit for equally spaced knot this complete the proof of theorem (3.1)
REFERENCES Journal Papers
[1] Deboor, C.A. Practical Guide to splines,Applied Mathematical Science, vol.27 Springer,Verlage, Newyork,1979.
[2] Howell,G and Verma,A.K Best Error Bound of Quartic Spline Interpolation, J.Approximation Theory [3] 58(1989), 58-67.
[4] Dubey, Y.P. Best error bounds of spline of degree six Int.Journal of Mathematical Analysis vol5 (2011),pp 21-24
[5] Gemling,R.H.J.and Meyling,G. Interpolation by Bivariate Quintic Spline of class construction of theory of function 87(ed) Sendor et al (1987) 152-161.
[6] Rana,S.S and Dubey,Y.P.Dubey. Best Error bounds of Quinticspline interpolation .J.Pure and Applied mathematics 28(10) 1937-44(1997).
[7] Rana,S.S and Dubey, Y.P. Best error bounds of quartic spline interpolation Indian Journal of pure and applied mathematics30(4)(1999)385-393.
Books
[1] Hall, C.A. and Meyar,W.W, J.Approximation Theory 16(1976),pp105- 122.
[2] Davis,P.J. Interpolation and approximation,New York,1969.
[3] Meir, A and Sharma,A. .Convergence of a class of interpolatory spline J.Approx. Theory (1968) pp 243- 250.
