# An Explicit Solution of the Cubic Spline Interpolation for Polynomials

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1 An Explicit Solution of the Cubic Spline Interpolation for Polynomials Byung Soo Moon Korea Atomic Energy Research Institute P.O. Box105, Taeduk Science Town Taejon, Korea Abstract An algorithm for computing the cubic spline interpolation coefficients for polynomials is presented in this paper. The matrix equation involved is solved analytically so that numerical inversion of the coefficient matrix is not required. For f (t) = t m, a set of constants along with the degree of polynomial m are used to compute the coefficients so that they satisfy the interpolation constraints but not necessarily the derivative constraints. Then, another matrix equation is solved analytically to take care of the derivative constraints. The results are combined linearly to obtain the unique solution of the original matrix equation. This algorithm is tested and verified numerically for various examples. 1 Introduction There are many recent reports of studies on representing functions by fuzzy systems or neural networks[1,2,3]. Futher works, however, are yet to follow in order to present the characteristics of the particular function in its representation. For the problem of representing a function using its cubic spline interpolation either by a fuzzy system or a neural network, the numerical values of the coefficients obtained from solving the matrix equation do not give any insights on how the coefficients are related to the characteristics of the function. In this paper, we present an algorithm by which one can compute the coefficients of the cubic spline interpolation for polynomials of the form f (t) = t m 75

2 76 BYUNG SOO MOON without solving the matrix equation. For an arbitrary polynomial, the coefficients can then be computed by a linear combination of those for f (t) = t m.asurveyof literature indicates that related studies on cubic splines are focused on the inverse of the coefficient matrix[4], or on monotone functions[5,6], but none seem to consider the same problem discussed here. In the following, we assume f (t) = t m and consider the cubic spline interpolation of f (t) in the interval [-1, 1]. Let t j = 1+jh, j=-3,-2,-1,0,...,2n+4 with h = 1 n and let (t t i 2 ) 3, if t [t i 2, t i 1 ] B i (t) = 1 h 3 + 3h 2 (t t i 1 ) + 3h(t t i 1 ) 2 3(t t i 1 ) 3, if t [t i 1, t i ] h 3 + 3h 2 (t i+1 t) + 3h(t i+1 t) 2 3(t i+1 t) 3, if t [t i, t i+1 ] 6 (t i+2 t) 3, if t [t i+1, t i+2 ] 0, otherwise (1) for i=-1,0,1,...,2n+1. Consider the problem of finding the coefficients c j s in S(t) = c 1 B 1 (t) + c 0 B 0 (t) + c 1 B 1 (t) +...c 2n+1 B 2n+1 (t). such that S(t i ) = f (t i ),0 i 2n,S (t 0 )=α,ands (t 2n )=β,whereb j s are cubic B-spline functions and α, β are arbitrary real numbers. Using the properties of cubic B-splines [7,p80], the problem can be written as solving the matrix equation Ac = b (2) where c = (c 1, c 0, c 1,...c 2n+1 ) T, b = (α, f (t 0 ),... f (t 2n ), β) T and A = (3) Recall that we must have use α = h 3 f (t 0 ) and β = h 3 f (t 2n ) for O(h 4 ) accuracy.

3 An Explicit Solution 77 2 Interpolation Without Derivative Constraints In this section, we prove that for polynomials of the form f (t) = t m, the interpolation coefficients c j, j = 1,0,...,2n +1 can be computed directly without solving the matrix equation (2), even though the constraints α = h 3 f (t 0 ) and β = h 3 f (t 2n ) are satisfied only when m 4. These constraints will be considered in the next section. Lemma 1. Let m be a positive integer and let λ j be defined successively by λ 0 = 1, λ 1 = 1, and λ 2k = 1 1 k 1 2kC 2i λ 2i, λ 2k+1 = 1 1 k 1 2k+1C 2i+1 λ 2i i=0 i=0 If q 0 = λ m and q j = equations mc l j m l λ l for j=±1, ±2,...,then q js satisfy the set of q j 1 + 4q j + q j+1 = 6( j 1) m, j = 0, ±1, ±2, ±3,.... Proof. By a routine arithmetic, we have mc l {( j 1) m l +4 j m l +( j +1) m l }=6mC l j m l +2 Hence, the sum q j 1 + 4q j + q j+1 becomes m l k=2,4,6,.. m 2 λ l {( j 1) m l +4 j m l +( j+1) m l }mc l = 6 mc l λ l j m l +2 k+lc l mc k+l j m l k. m l λ l k=2,4,6,.. k+lc l mc k+l j m l k. We separate the double sum into one part of even k +l s and the other of odd k +l s, and reorder the sum so that the power of j is decreasing. Then we obtain 6 j m + 6mC 1 λ 1 j m k=even k=odd mc k (3λ k + mc k (3λ k + k 2 i=0,2,4,... k 2 i=1,3,5,... kc i λ i ) j m k. kc i λ i ) j m k

4 78 BYUNG SOO MOON Now, we substitute λ k s by the successive definition formula, along with λ 0 = 1, λ 1 = 1, to obtain 6 j m 6mC 1 j m 1 6 k=odd mc k j m k + 6 k=even mc k j m k which is identical to 6( j 1) m, and the proof is completed. Note that the above works for the case of j =±1, or j=0 with q 0 = λ m. Q.E.D When some of the λ j s are computed iteratively, we find that λ 2 = 2 3, λ 3 = 0, λ 4 = 2 3, λ 5 = 2 3, λ 6 = 2 3, λ 7 = 10 3, λ 8 = 34 9, λ 9 = 14, λ 10 = 66. Note that λ j s do not depend on the value of m which will be used as the degree of polynomial throughout this paper, and that q j s depend only on the value of m. Lemma 2. Let λ k s be defined as in Lemma 1 for k=0,1,2,....thenwehave λ k k k for all k 1. Proof. We apply the mathematical induction separately on even k s and odd k s. It is clear that λ j satisfies the relation for j=1,2,3 and 4. Assume the statement is true for all i < k. Then from λ 2k kC k 1 3 2i λ 2i k 1 2kC 2i (2i) 2i (1+2k 1)2k and similarly from λ 2k (2k+ i=0 1) 2k+1,wehave λ k k k for all k 1. Q.E.D Lemma 3. Let m be an even positive integer and q j s be defined as in Lemma 1. Then the q j s satisfy q j+1 = q j+1 for j=1,2,...,n+1. Proof. Let r j = q j+1 q j+1, j=1,2,...,n+1. Then we have 4r 1 + r 2 = 0 from the two equations q 1 + 4q 0 + q 1 = 6andq 1 +4q 2 +q 3 =6. Also from q j 1 + 4q j + q j+1 =6( j 1) m =6(j+1) m,andq j+1 +4q j+2 +q j+3 = 6(j+1) m,wehaver j +4r j+1 +r j+2 =0, for j=1,2,3,...,. From the first relation 4r 1 + r 2 = 0, we have r 2 = 4r 1 and hence r 2 4 r 1. From the second relation r 1 + 4r 2 + r 3 = 0, we have r 3 = r 1 4r 2 = 17r 1 from which we obtain r r 1. Continuing the same process, we have r j 4 j 1 r 1. On the other hand, we have q j mc l j m l λ l mc l j m l m l ( j +m) m and hence, r j = q j+1 q j+1 2(m+ j +1) m. Now, note that we can have 4 j 1 r 1 r j 2(m+ j +1) m for all j=1,2,..., only when r 1 = 0. Q.E.D i=0

5 An Explicit Solution 79 Lemma 4. Let m be an odd positive integer and q j s be defined as in Lemma 1. Then the q j s satisfy q 1 = 0andq j+1 = q j+1 for j=1,2,...,n+1. The above follows from an argument similar to the proof of Lemma 3 with s j = q j+1 + q j+1 replaced for r j. Note that when m is even, we have q n+2 q n = q n+2 q n. Similarly when the degree m is odd, we have q n+2 q n = q n+2 q n, which will be used in Theorem 1. Theorem 1. Let f (t) = t m and let φ(t) = 2n+1 i= 1 c i B i (t) be the cubic spline interpolation of f (t) on t j = 1+ j n in the interval [-1,1] with β = q n+2 q n and α = ( 1) m 1 β.then(c 1,c 0,...c 2n+1 ) (q n,q n+1,...,q 1,q 0,q 1,...,q n+2 )/( ) q j satisfies the equation (2), i.e., for j=-n to n+2 are the interpolation coefficients. 6nm Proof. Recall that c i s are the unique solution of the (2n+3) equations c 1 + c 1 = α c j 1 + 4c j + c j+1 = f (t j ) = ( 1 + j n )m, j = 0, 1, 2,...,2n c 2n 1 +c 2n+1 =β. First, we consider the middle equations. By Lemma 1, q j s satisfy the equations q j 1 + 4q j + q j+1 = 6( j 1) m, j = 0, ±1, ±2, ±3... and hence if we define c j = q n+ j+1 /( ),thenwehave c j 1 +4c j +c j+1 = q n+j +4q n+j+1 +q n+j+2 =( n+ j ) m =t m j n for j = 0, 1,...,2n. Hence, the set of middle equations are satisfied. The last equation is satisfied by the definition of β, and the first equation is satisfied due to Lemmas 3 and 4. Q.E.D Lemma 5. Let q j s and λ j s be defined as in Lemma 1. If m 4, then we have q n+2 q n = 2mn m 1,andq n+2 q n =( 1) m 1 2mn m 1, and hence α = h 3 f ( 1) and β = h 3 f (t 1 ) are satisfied, where α and β are as defined in Theorem 1.

6 80 BYUNG SOO MOON Proof. From the definition of q j s,wehave m 1 q n+2 q n = mc l λ l ((n + 2) m l n m l ), By directly substituting m=1,2,3, and 4 in the left hand side of the above relation, it is trivial to verify that the result is the same as 2mn m 1 for each m. The second part of Lemma follows from h 3 f ( 1) = ( 1)m 1 m and h 3n 3 f (1) = m. Q.E.D 3n 3 Corrections of the Solution for Derivative Constraints In this section, we consider the remainder r 0 = h 3 f (1) β and s 0 = h 3 f ( 1) α in the first and the last equation of (2) respectively, where β = q n+2 q n and n m α = ( 1) m 1 β.ifmiseven,thenq j s are symmetric in the sense of Lemma 4, while they are antisymmetric when m is odd as in Lemma 5. It is also easy to verify that the cubic spline interpolation coefficients c j s in (2) for an arbitrary differentiable function f (t) are symmetric when it is even, and antisymmetric when it is odd. Thus, we may consider only half of the equations in (2). For the case of even m, the middle equation q 0 + 4q 1 + q 2 = 0 becomes 4q 1 + 2q 2 = 0, the next set of equations are q j + 4q j+1 + q j+2 = 6 j m, j=1,2,...,n, and the last equation is q n + q n+2 = 0, so that we have only (n+2)-equations for the (n+2)-unknowns; q 1, q 2,...,q n+2. For the case of odd m, we start with the equation q 1 +4q 2 +q 3 = 6 which becomes 4q 2 + q 3 = 6 from q 1 = 0. The rest of the equations are the same as in the case of even m, so that there are (n+1)-equations for the (n+1)-unknowns. Using the following Lemma, one can solve the remaining part of the equation (2) explicitly, i.e., without solving a matrix equation. Lemma 6. Let A be an n n matrix whose entries are the same as in (3) except for the first row which is replaced by (4, 2, 0,...,0)and let p = (p 1, p 1,...,p n ) be the solution of Ap = e n,wheree n is the unit vector with the last entry of value 1. If α 1 = 0.5,α k+1 = 1/(4 α k ), for k=1,2,...,n, then we have p n = 1/(1 α n 2 α n 1 ) and p k = α k p k+1 =( 1) n k α k α k+1 α k+2...α n 1 p n for k=n- 1,n-2,...,1. When the first row is (4, 1, 0,...,0), the same p k s for k 1and p 1 = 1 4 p 2satisfy the equation. Proof. First, consider the case where the first row is (4, 2, 0,...,0). It is

7 An Explicit Solution 81 routine to verify that 4p 1 + 2p 2 = 0and p n 2 +p n =1. For 2 k n 1, note that p k 1 + 4p k + p k+1 = ( 1) n k+1 (α k 1 α k 4α k + 1)α k+1 α k+2...α n 1 p n which is 0 since α k is defined by α k = 1/(4 α k 1 ). A similar proof for the case where the first row is (4, 1, 0,...,0)is omitted. Q.E.D Combining the results of Theorem 1 and Lemma 6, we have proved the following Theorem. Theorem 2. Let f (t) = t m where m is an even positive integer and let q j = mc l j m l λ l, j=-n,-n+1,...,n+2, where λ l s are as defined in Lemma 1. If p j s are defined iteratively by p n+2 = 1/(1 α n α n+1 ) and p k = α k p k+1,whereα k s are as defined in Lemma 6, then (q n + ρp n+2, q n+1 + ρp n+1,...,q 0 +ρp 2,q 1 + ρp 1,q 2 +ρp 2,...,q n+2 +ρp n+2 )where ρ = m 3n q n+2 q n, define the cubic spline interpolation coefficients for f (t). If m is odd, then with p n+1 = 1/(1 α n 1 α n ), p k+1 = α k p k, for k=n-1,n-2,...,2, and p 1 = 1 4 p 2, the coefficients become (q n ρp n+1, q n+1 ρp n,...,q 0 ρp 1,q 1,q 2 +ρp 1,...,q n+2 +ρp n+1 ). Example 1. The interpolation coefficients for f (t) = t 6 at t j = 1 + j 5, j=0,1,2,...,10, computed by the above algorithm using double precision calculations are as follows; Original Proposed Algorithm Proposed Algorithm Cubic Spline Without Correction With Correction c c c c c c c The differences between the coefficients in the first and the third column are purely computational errors and the differences in the second and third column are due to ρp j,whereρ= m 3n q n+2 q n. Note that the magnitude of ρ should decrease as the number of divisions n increases since the second term in ρ is of order ( n )m ( 1 1 n )m. Therefore, the corrections contribute less as the number of divisions becomes larger.

8 82 BYUNG SOO MOON References [1] B. Kosko, Fuzzy Systems are Universal Approximators, IEEE Int. Conf. on Neural Networks and Fuzzy Systems (1992) pp [2] K. Hornik, Multilayer Feedforward Networks are Universal Approximators, Neural Networks, Vol.II (1989) pp [3] F. A. Watkins, Constructing Fuzzy Function Representations, World Congress on Neural Networks, Vol.II (1995) pp [4] S. S. Rana, Local Behaviour of the First Difference of a Discrete Cubic Spline Interpolator, Approx. Theory & its Appl., Vol.6, No.4 (1990) pp [5] S. Pruess, Shape Preserving C 2 Cubic Spline Interpolation, IMA J. of Numerical Analysis, Vol.13 (1993) pp [6] F. I. Uteras, Convergence Rates for Monotone Cubic Spline Interpolation, J. Approx. Theory Vol.36 (1982) pp [7] P. M. Prenter, Splines and Variational Methods, John Wiley and Sons (1975) pp

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