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Piecewise trigonometric Hermite interpolation
Applied Mathematics and Computation 268 (2015) 616–627
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Piecewise trigonometric Hermite interpolation Xuli Han∗ School of Mathematics and Statistics, Central South University, Changsha, 410083, PR China
a r t i c l e
i n f o
Keywords: Trigonometric basis Trigonometric polynomial Trigonometric interpolation Trigonometric Hermite interpolation
a b s t r a c t Based on the symmetric, nonnegative and normalized basis of the trigonometric polynomial space, the piecewise trigonometric Hermite interpolation methods are presented. The C n−1 and Cn continuous piecewise trigonometric Hermite interpolants of degree n are constructed and the interpolation methods are local. The integral and the differential representations of the errors of the trigonometric Hermite interpolants are given. Several examples are supplied to support the practical value of the given interpolation methods. © 2015 Elsevier Inc. All rights reserved.
1. Introduction Trigonometric Hermite interpolation is an important topic in approximation theory. Salzer [27] has already given Lagrangian and Barycentric formulae for arbitrary interpolation points and problems where higher order derivatives were also prescribed. Kress [18] has derived Lagrangian as well as remainder formulae and asymptotic convergence results for the most important case of an even number of equidistant points. The trigonometric Hermite interpolation on equidistant nodes have been addressed in [3], [4]. Several researchers have focused their attentions on this subject, see [6], [8], [16], [23], [26], even for arbitrary points. Several authors were mostly interested in existence questions [4], [17], convergence results [19], [25] and formulae other than Lagrange’s, see [15], [21], [24], [28]. There are a few results on the construction of trigonometric polynomial sequence approximating continuous function. Some authors were interested in the problem of constructing nonnegative trigonometric polynomials, see [2], [7], [9]. Askey and Steining’s paper [4] gave an idea of how new nonnegative trigonometric polynomials can be generated by a known sequence of nonnegative trigonometric polynomials. In [1], authors approximated continuous functions defined on a compact set E ⊂ [−π , π ] by trigonometric polynomials. A stable recurrence relation for the trigonometric B-spline was given in [20]. In [22], a general theory of quasi-interpolants based on trigonometric splines was developed which is analogous to the polynomial spline case. The aim was to construct quasiinterpolants which are local, easy to compute, and which apply to a wide class of functions. Trigonometric splines can be used to solve the problems of geometric modeling. Some types of trigonometric splines have been introduced having different features, see [10]–[12]. These trigonometric splines have been used for constrained curve design in [5]. In [13] and [14], the symmetric, nonnegative and normalized basis of trigonometric polynomial space has been presented. In [13], based on the basis, the symmetric trigonometric polynomial approximants like Bernstein polynomials are constructed. Two kinds of nodes are given to show that the trigonometric polynomial sequence is uniform convergent. The trigonometric quasi-interpolants of reproducing one degree of trigonometric polynomials are constructed. In [14], the total positivity and the B-basis property of the symmetric trigonometric polynomial basis functions are shown. Then the symmetric trigonometric
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http://dx.doi.org/10.1016/j.amc.2015.06.125 0096-3003/© 2015 Elsevier Inc. All rights reserved.
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
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polynomial curves like Bézier curves are presented and some theoretics and methods are proposed for curve representation of the trigonometric polynomial space. In the polynomial space, the piecewise Hermite interpolants are classic and important. Represented by the Bernstein basis functions, the piecewise polynomial Hermite interpolants with two points for each piece expression are useful for numerical computation and functional approximation. The piecewise polynomial Hermite interpolants of degree 2n + 1 are Cn continuous. If a problem of numerical computation or functional approximation concerns some data which relate to trigonometric functions, then a trigonometric interpolation polynomials can be considered. However, in the trigonometric polynomial space, a general piecewise trigonometric Hermite interpolant like the piecewise polynomial Hermite interpolant represented by the Bernstein basis functions had not been presented. The purpose of this paper is to present explicit piecewise trigonometric Hermite interpolants which are local and easy to compute. The main result is that the presented piecewise trigonometric Hermite interpolant of degree n achieves C n−1 or Cn continuity. Since we have the symmetric, nonnegative and normalized trigonometric basis functions like Bernstein polynomials, it is conceivable and convenient to construct the piecewise trigonometric Hermite interpolant with two points for each piece expression. The remainder of this paper is organized as follows. In Section 2, the basis functions of the trigonometric polynomial space are described and the derivative properties of the basis functions are shown. In Section 3, the C n−1 and Cn continuous piecewise trigonometric Hermite interpolants of degree n are presented. The convergence of the piecewise trigonometric Hermite interpolation method is discussed in Section 4. Conclusions are given in Section 5. 2. The symmetric trigonometric basis functions Consider the trigonometric polynomial space
Tn := span{1, sin (u), cos (u), sin (2u), cos (2u), . . . , sin (nu), cos (nu)} for u ∈ [0, π /2]. The basis functions sin (iu), cos (iu) have been used almost for all the topic concerning the trigonometric polynomial problems. However, the basis functions sin (iu), cos (iu) are not nonnegative and normalized. It is not convenient for the representation of a trigonometric Hermite interpolant by using these basis functions. Now we consider the symmetric, nonnegative and normalized basis functions of Tn presented in [13] and [14]. For u ∈ [0, π /2], n ∈ N, let s(u) := 1 − sin (u), c(u) := 1 − cos (u), w(u) := sin (u) + cos (u) − 1, we introduce the trigonometric polynomials of degree n as follows:
Ti,n (u) := where
ai,1 :=
ai,n+1 :=
i = 0, 1, . . . , n, i = n + 1, n + 2, . . . , 2n,
i = 0, 1, 2, i = 0, 1, 2,
1, 0,
and for n ≥ 1,
ai,n sn−i (u)wi (u), ai,n w2n−i (u)ci−n (u),
(2)
⎧1 ⎨ 2 ai−2,n + ai−1,n + ai,n , ⎩
1 a 2 i−2,n
+ ai−1,n + 12 ai,n ,
ai−2,n + ai−1,n +
(1)
1 a , 2 i,n
i ≤ n, i = n + 1,
(3)
i ≥ n + 2.
It has been shown that the trigonometric polynomials of the set {T0,n (u), T1,n (u), . . . , T2n,n (u)} form a basis for the trigonometric polynomial space Tn and
Tn = span{T0,n (u), T1,n (u), . . . , T2n,n (u)}. We refer to these trigonometric functions as symmetric trigonometric basis functions. Many properties of the symmetric trigonometric basis functions can be found in [13] and [14]. In order to represent the trigonometric Hermite interpolants, we need to discuss the derivative properties of the basis functions. In [14], the derivatives of (1) are given as follows (u) = Ti,n
iai,n (2n − i)ai,n Ti−1,n (u) − (n − i)Ti,n (u) − Ti+1,n (u) ai−1,n 2ai+1,n
(4)
for i = 0, 1, . . . , n − 1, (u) = Tn,n
nan,n [Ti−1,n (u) − Ti+1,n (u)], an−1,n
(5)
(u) = Ti,n
(2n − i)ai,n iai,n Ti−1,n (u) + (i − n)Ti,n (u) − Ti+1,n (u) 2ai−1,n ai+1,n
(6)
for i = n + 1, n + 2, . . . , 2n.
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X. Han / Applied Mathematics and Computation 268 (2015) 616–627
From (4) to (6), we have (u), T (u), . . . , T (u)) = (T (u), T (u), . . . , T (T0,n 0,n 1,n 2n,n (u))Bn , 1,n 2n,n
(7)
where Bn := (bij ) is a tridiagonal matrix of order 2n + 1 with
bi,i = i − n − 1,
bi,i+1 =
i = 1, 2, . . . , 2n + 1,
iai,n /ai−1,n , iai,n /(2ai−1,n ),
bi+1,i = bi, j = 0,
i = 1, 2, . . . , n, i = n + 1, n + 2, . . . , 2n,
−(2n − i + 1)ai−1,n /(2ai,n ), −(2n − i + 1)ai−1,n /ai,n ,
i = 1, 2, . . . , n, i = n + 1, n + 2, . . . , 2n,
|i − j| > 1. (r)
Lemma 1. Let bi, j be the ij entry of the rth power of the matrix Bn for i, j = 1, 2, . . . , 2n + 1 and r = 1, 2, . . . , n (For convenience, we (r)
omit a superscript or subscript n for bi, j ), then
1 r) b(1,r = − r!(2n − r + 1)ar−1,n , 2 r) b(2n+1,2n−r+1 = (−1)r r!a2n−r,n ,
r) b(1,r+1 = r!ar,n ,
(8)
1 r) b(2n+1,2n−r+2 = (−1)r+1 r!(2n − r + 1)a2n−r+1,n . 2
(9)
Proof. Obviously, the rth power Brn is a band matrix with band width 2r + 1, r = 1, 2, . . . , n, and 1) 1) 1) 1) b(1,1 = −n, b(1,2 = n, b(2n+1,2n = −n, b(2n+1,2n+1 = n.
We assume that the formulas (8) and (9) are true for r ≤ n − 1, then, when r + 1 ≤ n, by the elements of Brn and Bn we have
rar,n r) − b(1,r+1 (n − r) ar−1,n 1 1 rar,n = − r!(2n − r + 1)ar−1,n − r!ar,n (n − r) = − (r + 1)!(2n − r)ar,n , 2 ar−1,n 2
r+1) r) b(1,r+1 = b(1,r
r+1) r) = b(1,r+1 b(1,r+2
(r + 1)ar+1,n ar,n
r+1) r) b(2n+1,2n−r = −b(2n+1,2n−r+1 r+1)
= (−1) (r+1)
= r!ar,n
(r + 1)ar+1,n
(r + 1)a2n−r−1,n a2n−r,n
ar,n
= (r + 1)!ar+1,n ,
= (−1)r+1) r!a2n−r,n
(r + 1)a2n−r−1,n a2n−r,n
(r + 1)!a2n−r−1,n ,
(r)
r) b2n+1,2n−r+1 = b2n+1,2n−r+1 (n − r) − b(2n+1,2n−r+2
ra2n−r,n a2n−r+1,n
1 ra2n−r,n = (−1)r r!a2n−r,n (n − r) + (−1)r r!(2n − r + 1)a2n−r+1,n 2 a2n−r+1,n 1 = (−1)r (r + 1)!(2n − r)a2n−r,n . 2 Thus we obtain the results. The matrices Brn will be used to construct piecewise trigonometric Hermite interpolants. 3. The piecewise trigonometric Hermite interpolants The piecewise two points Hermite interpolants are classical polynomial interpolants and can be constructed conveniently based on the Bernstein polynomials. In this section, we construct the piecewise trigonometric Hermite interpolants by using the symmetric trigonometric basis functions. 3.1. The C n−1 continuous piecewise Hermite interpolants of degree n (r)
Given a partition : a = x1 < x2 < · · · < xm = b of the bounded interval [a, b] of R, and let n ≥ 1, fi (i = 1, 2, . . . , m; r = 0, 1, . . . , n − 1) be the given values at nodes x = xi . For 1 ≤ k ≤ m − 1, x ∈ [xk , xk+1 ], hk = xk+1 − xk , u = π (x − xk )/2hk , we construct a piecewise trigonometric interpolant
Tn (x) :=
2n i=0
Ti,n (u) p[k] i,n
(10)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
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to satisfy the following interpolation conditions:
Tn(r) (xi ) = fi(r) ,
i = k, k + 1;
r = 0, 1, . . . , n − 1.
(11)
Thus, the interpolant Tn ∈ C n−1 ([a, b]). From (1) to (3), for x ∈ [xk , xk+1 ], we have expressions
T1 (x) = s(u) p[k] + w(u) p[k] + c(u) p[k] , 0,1 1,1 2,1 + 2s(u)w(u) p[k] + 2w2 (u) p[k] + 2w(u)c(u) p[k] + c2 (u) p[k] , T2 (x) = s2 (u) p[k] 0,2 1,2 2,2 3,2 4,2 9 + 4w3 (u) p[k] s(u)w2 (u) p[k] 2,3 3,3 2
+ 3s2 (u)w(u) p[k] + T3 (x) = s3 (u) p[k] 0,3 1,3
9 2 w (u)c(u) p[k] + 3w(u)c2 (u) p[k] + c3 (u) p[k] . 4,3 5,3 6,3 2
+
Theorem 1. The pi,n (i = 0, . . . , n − 1, n + 1, . . . , 2n) in (10) can be obtained uniquely by the interpolation conditions (11). [k]
Proof. According to (7), we have (r) (r) (r) (T0,n (u), T1,n (u), . . . , T2n,n (u)) =
π r 2hk
(T0,n (u), T1,n (u), . . . , T2n,n (u))Brn
for r = 0, 1, . . . , n. Thus, the interpolation conditions (11) lead to the linear equations
⎛
p[k] 0,n
⎜ [k] ⎜ p1,n BL,n ⎜ ⎜ .. ⎝ . p[k] n−1,n and
⎛
⎞
⎛
⎞
fk
⎟ ⎜ 2hk ⎟ ⎟ ⎜ π fk ⎟ ⎟=⎜ ⎟, . ⎟ ⎝ .. ⎠ ⎠ 2hk n−1 (n−1)
p[k] n+1,n
fk
π
⎞
⎛
(12)
⎞
fk+1
⎜ p[k] ⎟ ⎜ 2hπk f ⎟ k+1 ⎜ n+2,n ⎟ ⎜ ⎟ BR,n ⎜ . ⎟ = ⎜ ⎟, .. ⎝ .. ⎠ ⎝ ⎠ . 2hk n−1 (n−1) [k] p2n,n
where
⎛ ⎜
(13)
fk+1
π
⎞
0) b(1,1
(1) BL,n := ⎜ b1,1 ⎝ ··· n−1) b(1,1
⎟ ⎟, ⎠
1) b(1,2 ··· n−1) b(1,2
···
n−1) b(1,n
⎛ ⎜ ⎝
1) b(2n+1,2n ··· n−1) b(2n+1,2n
BR,n := ⎜ n−1) b(2n+1,n+2
···
(14)
0) b(2n+1,2n+1
⎞ ⎟
1) b(2n+1,2n+1 ⎟ ⎠ ··· (n−1) b2n+1,2n+1
(r)
(15)
(r)
are triangular matrices. By Lemma 1, b1,r+1 = 0, b2n+1,2n−r+1 = 0, r = 0, 1, . . . , n − 1. Therefore, by (12) and (13), pi,n (i = 0, . . . , n − 1, n + 1, . . . , 2n) can be obtained easily. [k]
[k]
[k]
Obviously, for C n−1 continuity, pn,n can be chosen freely. Now we show the choice of the pn,n . (n)
If fk
(n)
and fk+1 are not given, then a simply method is to choose
p[k] n,n :=
1 [k] pn−1,n + p[k] . n+1,n 2
(n)
(n)
If fk
(16) (n)
(n)
and fk+1 are given, then, let Tn (xk ) = fk , we have
n+1 b(1,n)j p[k] = j−1,n j=1
2hk
π
n
fk(n) .
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X. Han / Applied Mathematics and Computation 268 (2015) 616–627
From Lemma 1, we obtain
p[k] n,n
=
p˜[k] n,n
1 := n!an,n
2hk
n − b(1,n)j p[k] . j−1,n
(n)
fk
π
(n)
n
j=1
(n)
In the same way, let Tn (xk+1 ) = fk+1 , we have 2n+1
n) b(2n+1, p[k] = j j−1,n
2hk
π
j=n+1
and then
p[k] n,n
=
p¯ [k] n,n
(−1)n
:= (n)
Therefore, when fk
n
2hk
n
(n)
fk+1 −
π
n!an,n
(n) fk+1 ,
2n+1
(n)
b2n+1, j p[k] j−1,n
.
j=n+2
(n)
and fk+1 are given, we would like to set
1 [k] p˜ n,n + p¯ [k] n,n . 2
p[k] n,n :=
(17)
Of course, we can also choose pn,n by letting Tn (xk + hk /2) = fk+1/2 if the value fk+1/2 is given at node x = xk+1/2 . [k]
Theorem 2. With u = π (x − xk )/(2hk ), the interpolation operator Tn reproduces all trigonometric polynomials of degree ≤ n in Tn by (12), (13) and (17). Proof. Take u = π (x − xk )/(2hk ) as the trigonometric function variable, we can know that any trigonometric polynomial f(x) of [k] [k] degree ≤ n in Tn is determined uniquely by (12), (13) and p˜ n,n . Therefore, Tn (x) = f (x) by (12), (13) and p˜ n,n . Simultaneously, [k] Tn (x) = f (x) by (12), (13) and p¯ n,n . Then Tn (x) = f (x) by (12), (13) and (17).
Now we give some explicit expressions of the trigonometric interpolants of lower degree for x ∈ [xk , xk+1 ]. Since
B1 =
−1 −1 0
1 0 −1
0 1 , 1
it is easy to know that
T1 (x) = s(u) fk + w(u)
1 ( f + fk+1 ) + c(u) fk+1 2 k
(18)
by (12), (13) and (16), and
hk 1 T1 (x) = s(u) fk + w(u) ( f + fk+1 ) + ( fk − fk+1 ) + c(u) fk+1 2 k π
by (12), (13) and (17). Since
⎛−2
⎜−1 B2 = ⎜ 0 ⎝ 0 0
2 −1 −3/2 0 0
0 2 0 −2 0
⎞
0 0 3/2 1 −2
0 0⎟ 0⎟, ⎠ 1 2
⎛ 2 ⎜ 3 B22 = ⎜3/2 ⎝ 0 0
−6 −4 3/2 3 0
4 −2 −6 −2 4
(19)
0 3 3/2 −4 −6
⎞
0 0 ⎟ 3/2⎟, ⎠ 3 2
we can obtain
p[k] = fk , 0,2
p[k] = fk + 1,2
hk fk ,
π
p[k] = fk+1 − 3,2
hk fk+1 ,
π
p[k] = fk+1 4,2
(20)
by (12), (13), and
p[k] = 2,2
1 h ) ( f + fk ) + k ( fk − fk+1 2 k 2π
(21)
2 1 3h ) + hk ( f + f ) ( fk + fk+1 ) + k ( fk − fk+1 k+1 2 4π 2π 2 k
(22)
by (16), and
p[k] = 2,2 by (17).
X. Han / Applied Mathematics and Computation 268 (2015) 616–627 2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1 −4
−3
−2
−1
0
1
2
3
4
−1 −4
−3
(a) m = 5, hk = π/2
−2
−1
0
621
1
2
3
4
2
3
4
(b) m = 9, hk = π/4
Fig. 1. The curves of T2 (x) and the cubic Hermite interpolants to (27). 2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1 −4
−3
−2
−1
0
1
2
3
4
−1 −4
−3
(a) m = 5, hk = π/2
−2
−1
0
1
(b) m = 9, hk = π/4
Fig. 2. The curves of T3 (x) and the quintic Hermite interpolants to (27).
By (12), (13) and (16), we have
p[k] = fk , 0,3
p[k] = fk + 1,3
2hk f , 3π k
p[k] = fk + 2,3
4h2k 10hk fk + f , 9π 9π 2 k
2 1 5h ) + 2hk ( f + f ), ( fk + fk+1 ) + k ( fk − fk+1 k+1 2 9π 9π 2 k 2 4hk 10hk 2hk = fk+1 − f + f , p[k] = fk+1 − f , 5,3 9π k+1 9π 2 k+1 3π k+1
(23)
p[k] = 3,3 p[k] 4,3
(24)
p[k] = fk+1 , 6,3
(25)
and by (17) we have
p[k] = 3,3
3 2 1 19hk ) + hk ( f + f ) + hk ( f − f ). ( fk + fk+1 ) + ( fk − fk+1 k k+1 k+1 2 2 24π 2π 6π 3 k
(26)
Example 1. Consider the function
f (x) = sin (4x) + (r)
and use the data fi
1 , 1 + x2
x ∈ [−π , π ],
(27)
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n).
Fig. 1 shows the graphs of the function (27) (dotted lines) and the C1 continuous piecewise trigonometric Hermite interpolant T2 (x). On the left, we take m = 5, hk = π /2 and use the expression (21) (solid lines). For comparison, the cubic polynomial Hermite interpolation curves are given (dashed line). On the right, we take m = 9, hk = π /4 and use the expression (21) (dashed lines) and (22) (solid lines). When the expression (22) is used, the curve merges with the curve of the function (27). Fig. 2 shows the graphs of the function (27) (dotted lines) and the C2 continuous piecewise trigonometric Hermite interpolant T3 (x). We show the curves by using the expression (24), On the left, we take m = 5, hk = π /2 (solid lines). For comparison, the quintic polynomial Hermite interpolation curves are given (dashed line). On the right, we take m = 9, hk = π /4. Example 2. Consider the function
f (x) = cos (3x) exp (−x) + cos (2x) exp (x − 2π ), (r)
and use the data fi
x ∈ [0, 2π ],
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n).
(28)
622
X. Han / Applied Mathematics and Computation 268 (2015) 616–627 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) use expression(22)
2
3
4
5
6
7
5
6
7
(b) use expression(21)
Fig. 3. The curves T2 (x) interpolate to (28). 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) m = 5, hk = π/2
2
3
4
(b) m = 9, hk = π/4
Fig. 4. The curves T3 (x) interpolate to (28).
Fig. 3 shows the graphs of the function (28) (dotted lines) and the C1 continuous piecewise trigonometric Hermite interpolant T2 (x) (solid lines). On the left, we take m = 5, hk = π /2 and use the expression (22). On the right, we take m = 9, hk = π /4 and use the expression (21). Fig. 4 shows the graphs of the function (28) (dotted lines) and the C2 continuous piecewise trigonometric Hermite interpolant T3 (x). We show the curves by using the expression (24), taking m = 5, hk = π /2 on the left and m = 9, hk = π /4 on the right respectively. 3.2. The Cn continuous piecewise Hermite interpolants of degree n It is well known that the B-splines and the spline interpolation functions of degree n have C n−1 continuity. The C n−1 continuous spline interpolation method is not local interpolation method. In [11], the C2 continuous piecewise quadratic trigonometric polynomials were presented. Now we give some Cn continuous piecewise trigonometric interpolants of degree n and the given interpolation methods will be local interpolation methods. From (12) and (13) we have
p[k] n−1,n
1 = (n − 1)!an−1,n
(−1) = p[k] n+1,n (n − 1)!an+1,n n−1
(n−1)
where fk (n−1)
dk+1
for
(n−1)
and fk+1
[k] pn−1,n
p¯ [k] n−1,n p¯ [k] n+1,n
and
2hk
n−1
fk
π
2hk
(n−1)
n−1
π
n−1 ) [k] − b(1,n−1 p j−1,n , j j=1
(n−1) fk+1 −
2n+1
n−1) [k] b(2n+1, p , j j−1,n
j=n+3
(n−1)
are the given values. Now we change them (except for f1
[k] pn+1,n
(not for
1 = (n − 1)!an−1,n
(−1)n−1 = (n − 1)!an+1,n
[k] pn,n )
2hk
respectively, let
n−1
dk
π
2hk
π
(n−1)
n−1
dk+1 −
2n+1 j=n+3
(n−1)
) to undetermined dk
and
n−1 ) [k] − b(1,n−1 p j−1,n , j j=1
(n−1)
(n−1)
and fm
(29)
(n−1)
b2n+1, j p[k] j−1,n
,
(30)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
623
and then take the piecewise trigonometric interpolant as 2n
Tn (x) =
Ti,n (u) p[k] + Tn−1,n (u) p¯ [k] + Tn+1,n (u) p¯ [k] n−1,n n+1,n i,n
(31)
i=0, j=n−1,n+1
(n−1)
(n−1)
for 1 ≤ k ≤ m − 1, x ∈ [xk , xk+1 ], u = π (x − xk )/(2hk ). The dk used to achieve Cn continuity. Since (n)
Tn
(xk +) =
π n n−1 2hk
=− Tn(n) (xk+1 −) =
(n) [k]
b1, j p j−1,n + b1,n p¯ n−1,n +
j=1
n(n + 1)π (n−1) dk + 4hk
π n
2hk
2n+1
π n 2hk
[k]
are not included in the term pn,n and they will be
(n)
b1,n+1 p[k] n,n
+ αk,n ,
n) n) n) b(2n+1, p[k] + b(2n+1,n+2 + b(2n+1,n+1 p[k] p¯ [k] n,n n+1,n j j−1,n
j=n+3
n(n + 1)π (n−1) dk+1 + 4hk
=
(n) [k]
and dk+1
π n 2hk
− αk+1,n ,
where + αk,n =
n−1 1 ) b(1,n)j + n(n + 1)b(1,n−1 p[k] + n!an,n p[k] n,n , j−1,n j 2 j=1
− αk+1,n =
2n+1
n) b(2n+1, − j
j=n+3
(n)
1 n−1) n(n + 1)b(2n+1, p[k] + (−1)n n!an,n p[k] n,n , j−1,n j 2
(n)
let Tn (xk +) = Tn (xk −), we obtain the Cn continuous conditions: (n−1)
dk
1 π n−1 hk−1 hk 1 + − = n−2 α − α 2 n(n + 1)(hk−1 + hk ) hnk k,n hnk−1 k,n
(32)
for k = 2, 3, . . . , m − 1. (n−1) (n−1) (n−1) (n−1) (n−1) (n−1) and fk+1 , then, by (32), fk−1 , fk and fk+1 are used to represent dk for Tn (x). If we use (16) which involves fk (r)
(r)
(r)
(r)
Obviously, fk−1 , fk , fk+1 and fk+2 are used for the kth segment of the trigonometric interpolants, r = 0, 1, . . . , n − 1. The values (r)
of fk
affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. In this case, we obtain
(1)
dk
1
dk(0) =
hk fk−1 + (hk−1 + hk ) fk + hk−1 fk+1 ,
2(hk−1 + hk ) 1 )−h = [π (hk fk−1 + hk−1 fk ) + hk ( fk − fk−1 k−1 ( f k+1 − f k )], 3(hk−1 + hk )
dk(2) =
π2
2(hk−1 + hk ) +
2 9(hk−1 + hk )
h hk−1 fk − k fk−1 hk hk−1
+
π
36(hk−1 + hk )
hk + 37 f ) − hk−1 (37 f + 20 f ) (20 fk−1 k k k+1 hk−1 hk
+ f ) + h [hk ( fk−1 k−1 ( f k + f k+1 )], k
(0)
where fi = ( fi+1 − fi )/hi . Note that T1 (x) interpolates dk (k = 2, 3, . . . , m − 1) in this case. (n−1)
If we use (17) which involves fk (r)
(n−1)
(n)
, fk+1 , fk
(n)
(n−1)
and fk+1 , then, by (32), some fi
(n)
and fi
(n−1)
are used to represent dk
. The
values of fk , r = 0, 1, . . . , n, affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. For example, in this case,
dk(0) =
1 hk fk−1 + (hk−1 + hk ) fk + hk−1 fk+1 2(hk−1 + hk ) +
hk−1 hk
π (hk−1 + hk )
− f ). ( fk−1 k+1
(n−1)
If we do not want to use fi
p[k] n,n =
1 [k] (p + p[k] ) n+2,n 2 n−2,n
(n)
or fi
(n ≥ 2) to represent dk(n−1) , then instead of (16) or (17), we can use (33)
624
X. Han / Applied Mathematics and Computation 268 (2015) 616–627 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) use expression (33)
2
3
4
5
6
7
(b) use expression (16) 2
Fig. 5. The C continuous interpolation curves T2 (x).
(r)
to represent (32). In this case, the values of fk , r = 0, 1, . . . , n − 2, affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. We obtain
dk(1) = (2)
dk
=
π
3(hk−1 + hk )
π2
2(hk−1 + hk )
(hk fk−1 + hk−1 fk ), hk hk−1 fk − fk−1 + hk
π
12(hk−1 + hk )
hk−1
hk + 15 f ) − hk−1 (15 f + 4 f ) . (4 fk−1 k k k+1 hk−1 hk
Fig. 5 shows the graphs of the function (28) (dotted lines) and the C2 continuous trigonometric Hermite interpolant T2 (x) (solid lines). The expression (33) is used on the left and the expression (16) is used on the right. 4. The convergence of the trigonometric interpolation It has been shown that the interpolation operator Tn can reproduce all trigonometric polynomials of degree ≤ n in Tn . Numerical examples in Section 3 show that the presented piecewise trigonometric Hermite interpolants approximate the given functions well. Further, we show the errors of the piecewise trigonometric Hermite interpolation in this section. A trigonometric Taylor expansion will be presented to show the integral representation of the error. This expansion is different from the trigonometric Taylor expansion in [20] for which the recurrence relations were based on even and odd order trigonometric polynomials, respectively. 4.1. The trigonometric Taylor expansion Let Ln be the differential operators defined by L1 = D( ≡ d/dx), and
Ln+1 =
D2 +
4n2 q2
Ln
(34)
for integer n ≥ 1 and real number q = 0. Theorem 3. Let f ∈ C 2n+1 [a, b], then
f (x) = sn (x) +
q2n (2n)!
x
sin a
x−y q
2n
where sn is determined by s0 (x) = f (a), and
q2n−1 x−a sin q (2n − 1)!
sn (x) = sn−1 (x) +
Ln+1 f (y)dy,
2n−1
cos
x−a q2n x−a Ln f (a) + sin q q (2n)!
for n ≥ 1. Proof. Obviously,
f (x) = s0 (x) +
a
x
L1 f (y)dy.
Therefore, (35) holds for n = 1. Assume (35) holds for sn−1 , that is
f (x) = sn−1 (x) +
q2n−2 (2n − 2)!
x
sin a
x−y q
(35)
2n−2
Ln f (y)dy.
2n
DLn f (a)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
625
By integration-by-parts argument, we have
x
In := a
=
x−y q
sin 4n2 q2
x
a
x−a = − sin q and then
In−1 =
Ln+1 f (y)dy
2n
x−y q
sin
2n
2n
Ln f (y)dy − sin
x−a q
2n x−a DLn f (a) − sin q q
1 x−a sin 2n − 1 q
2n
2n−1
DLn f (a) +
x
sin a
x−a q2 x−a Ln f (a) + sin q q 2n(2n − 1)
cos
x−y q
2n−1
x−a 2n(2n − 1) Ln f (a) + cos q q2
2n−1
2n q
2n
cos
x
sin
a
DLn f (a) +
x−y DLn f (y)dy q x−y q
2n−2
Ln f (y)dy,
q2 In . 2n(2n − 1)
From this and
f (x) = sn−1 (x) +
q2n−2 I , (2n − 2)! n−1
we obtain (35). The proof is complete by induction.
4.2. The error of the piecewise trigonometric Hermite interpolation 4h
Take a = xk and q = πk in (35), we have
2n−1 (4hk )2n−1 π (x − xk ) π (x − xk ) sn (x) = sn−1 (x) + sin cos Ln f (xk ) 4hk 4hk (2n − 1)!π 2n−1 2n (4hk )2n π (x − xk ) + sin DLn f (xk ), 4hk (2n)!π 2n 2n (4hk )2n x π (x − y) f (x) = sn (x) + sin Ln+1 f (y)dy. 4hk (2n)!π 2n xk
(36)
(37)
Obviously, sn is a trigonometric polynomial of degree n regarding variable u = π (x − xk )/(2hk ). (r)
Theorem 4. Let f ∈ C 2n+1 [a, b], fi = f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n), Tn ( f ; x) be the piecewise trigonometric Hermite interpolant corresponding to f with (17), then
f (x) − Tn ( f ; x) =
(4hk )2n (2n)!π 2n
xk+1
K (x, y)Ln+1 f (y)dy,
xk
where
(38)
T (v− ; x), y ∈ [xk , x], v (x, y) − Tn (v ) = n n + −Tn (vn ; x), y ∈ [x, xk+1 ], 0, y ∈ [xk , x], vn (x, y), y ∈ [xk , x], + − vn (x, y) = vn (x, y) = 0, y ∈ [x, xk+1 ], vn (x, y), y ∈ [x, xk+1 ]. 2n π (x − y) vn (x, y) = sin . K (x, y) =
+ n
+ n;x
4hk
Proof. Let
en (x) =
(4hk )2n (2n)!π 2n
x
sin
xk
π (x − y) 4hk
2n Ln+1 f (y)dy.
Since Tn is a linear operator and sn is a trigonometric polynomial of degree n regarding variable u = π (x − xk )/(2hk ), we have
Tn ( f ; x) = Tn (sn + en ; x) = sn (x) + Tn (en ; x). A straightforward computing gives that
e(nr) (x) =
(4hk )2n (2n)!π 2n
x xk
dr vn (x, y)Ln+1 f (y)dy, dxr
r = 0, 1, . . . , n,
626
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
and then
(4hk )2n xk (r) v (x , y)Ln+1 f (y)dy = 0, (2n)!π 2n xk n k (4hk )2n xk+1 (r) vn (xk+1 , y)Ln+1 f (y)dy. e(nr) (xk+1 ) = (2n)!π 2n xk
e(nr) (xk ) =
Therefore,
Tn (en ; x) =
(4hk )2n (2n)!π 2n
xk+1 xk
Tn (v+ n ; x)Ln+1 f (y)dy.
Since vn (x, y) = Tn (vn , x), for y < x we have
vn (x, y) − Tn (v+n ; x) = Tn (v−n ; x). The proof is complete. Note that Tn = Ker(Ln+1 ) regarding variable u = π (x − xk )/(2hk ). From (38) we can know that Tn reproduces all trigonometric polynomials of degree ≤ n in Tn . (r)
Theorem 5. Let f ∈ C 2n [a, b], fi
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n), Tn ( f ; x) be the piecewise trigonometric Hermite in-
terpolant corresponding to f with any pn,n , then to each x in [xk , xk+1 ] there corresponds a point ξ in (xk , xk+1 ) such that
f (x) − Tn ( f ; x) =
[k]
(x − xk )n (x − xk+1 )n (2n) [f (ξ ) − Tn(2n) ( f ; ξ )], (2n)!
(39)
for 1 ≤ k ≤ m − 1. Proof. If x is the node xk or xk+1 , the assertion is obviously true since both sides of Eq. (41) reduce to 0. So, let x be a fixed point in (xk , xk+1 ). Put
f (x) − Tn ( f ; x) (t − xk )n (t − xk+1 )n . (x − xk )n (x − xk+1 )n
ϕ(t ) := f (t ) − Tn ( f ; t ) −
Notice that ϕ has at least 2n + 1 zeros in [xk , xk+1 ], namely, x, xk ( n multiple zeros), xk+1 (n multiple zeros). By Rolle’s Theorem, ϕ has at least 2n zeros in [xk , xk+1 ]. Similarly, ϕ has at least 2n − 1 zeros in [xk , xk+1 ]. If this argument is repeated, we conclude eventually that ϕ (2n) has at least one zero, say ξ , in (xk , xk+1 ). Thus,
0 = ϕ(ξ ) = f (2n) (ξ ) − Tn(2n) ( f ; ξ ) −
f (x) − Tn ( f ; x) (2n)! (x − xk )n (x − xk+1 )n
which is another form of the equation to be proved. Let
Tnv (u) = (T0,n (u), T1,n (u), . . . , T2n,n (u)),
P2n = ( p[k] , p[k] , . . . , p[k] )T . 0,n 1,n 2n,n
The piecewise interpolant (10) can be written as Tn ( f ; x) = Tnv (u)P2n and then
Tn(2n) ( f ; x) =
π 2n 2hk
Tnv (u)B2n n P2n .
From (39), we obtain
1 | f (x) − Tn ( f ; x)| ≤ (2n)!
hk 2
where M2n = maxxk ≤x≤xk+1 | f (2n) (x)|. When n = 2, by (21) we have
⎛
2n M2n +
π 2n 4
⎞
) 8( fk − fk+1 ) + hπk (16 fk + 14 fk+1
⎜8( f − f ) + hk (17 f + 14 f )⎟ ⎜ k k+1 π k k+1 ⎟ ⎜ ⎟ 3hk ⎟, B42 P4 = ⎜ ( fk − fk+1 ) 2π ⎜ ⎟ ⎜ ⎟ hk 8 ( f − f ) − ( 14 f + 17 f ) ⎝ k+1 k π k k+1 ⎠ ) 8( fk+1 − fk ) − hπk (14 fk + 16 fk+1
and then
31 ||B42 P4 ||∞ ≤ 8 + h M , π k 1
where M1 = maxxk ≤x≤xk+1 | f (x)|.
||
B2n n P2n
||∞ ,
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
627
5. Conclusion The new symmetric, nonnegative and normalized basis of the trigonometric polynomial space is useful for constructing a piecewise trigonometric Hermite interpolant, analogous to the piecewise polynomial Hermite interpolant with two points for each piece. The constructed piecewise trigonometric Hermite interpolants of degree n can reproduce any trigonometric polynomials of degree ≤ n. Furthermore, the given interpolation methods are local and the interpolants of degree n are C n−1 or Cn continuous. The presented general trigonometric Taylor expansion provides a simple error expression of the trigonometric interpolant with an integral kernel. Numerical examples show the efficiency of the given interpolation methods. Acknowledgment The author thanks the referees for their careful review and valuable comments. This research is supported by the National Natural Science Foundation of China (no. 11271376). References [1] V. Andrievskii, H.-P. Blatt, On approximation of continuous functions by trigonometric polynomials, J. Approx. Theory 163 (2) (2011) 249–266. [2] R. Askey, J. Steinig, Some positive trigonometric sums, Trans. Amer. Math. Soc. 187 (1974) 295–307. [3] J.-P. Berrut, A. Welscher, Fourier and barycentric formulae for equidistant Hermite trigonometric interpolation, Appl. Comput. Harmonoic Anal. 23 (2007) 307–320. [4] A.S. Cavaretta, A. Sharma, R.S. Varga, Lacunary trigonometric interpolation on equidistant nodes, in: R.A. DeVore, K. Scherer (Eds.), Quantitative approximation, Academic Press, New York, 1980, pp. pp.63–80. [5] N. Choubey, A. Ojha, Constrained curve drawing using trigonometric splines having shape parameters, Comput. Aid. Des. 39 (2007) 1058–1064. [6] F.J. Delvos, Hermite interpolation with trigometric polynomials, BIT, 1993, 33, 113–123. [7] D.K. Dimitrov, C.A. Merlo, Nonnegative trigonometric polynomials, Constr. Approx. 18 (1) (2002) 117–143. [8] J. Du, H. Han, G. Jin, On trigonometric and paratrigonometric Hermite interpolation, J. Approx. Theory 131 (2004) 74–99. [9] A. Gluchoff, F. Hartmann, Univalent polynomials and non-negative trigonometric sums, Amer. Math. Monthly 105 (1998) 508–522. [10] X. Han, Quadratic trigonometric polynomial curves with a shape parameter, Comput. Aid. Geom. Des. 19 (2002) 503–512. [11] X. Han, Piecewise quadratic trigonometric polynomial curves, Math. Comput. 72 (243) (2003) 1369–1377. [12] X. Han, Cubic trigonometric polynomial curves with a shape parameter, Comput. Aid. Geom. Des. 21 (2004) 535–548. [13] X. Han, The trigonometric polynomial like bernstein polynomial, The Scientific World J. 2014 (2014) 17. Article ID 174716. [14] X. Han, Normalized b-basis of the space of trigonometric polynomials and curve design, Appl. Math. Comput. 251 (2015) 336–348. [15] J. He, On a linear combination of S.N. Bernstein trigonometric interpolation polynomial, Appl. Math. Comput. 106 (2-3) (1999) 197–203. [16] P. Henrici, Barycentric formulas for interpolating trigonometric polynomials and their conjugates, Numer. Math. 33 (1979) 225–234. [17] D.J. Johnson, The trigonometric Hermite-Birkhoff interpolation, Trans. Amer. Math. Soc. 212 (1975) 365–374. [18] R. Kress, On general Hermite trigonometric interpolation, Numer. Math. 20 (1972) 125–138. [19] P.E. Koch, T. Lyche, Bounds for the error in Trigonometric Hermite interpolaton, in: R.A. DeVore, K. Scherer (Eds.), Quantitative approximation, Academic Press, New York, 1980, pp. 185–196. [20] T. Lyche, R. Winther, A stable recurrence relation for trigonometric B-spline, J. Approx. Theory 25 (3) (1979) 266–279. [21] T. Lyche, A Newton form for trigonometric Hermite interpolation, BIT Numer. Math. 19 (2) (1979) 229–235. [22] T. Lyche, L.L. Schumaker, Quasi-interpolants based on trigonometric splines, J. Approx. Theory 95 (2) (1998) 280–309. [23] E. Quak, Trigonometric wavelets for Hermite interpolation, Math. Comput. 65 (214) (1996) 683–722. [24] L. Reichel, G. Ammar, W. Gragg, Discrete least squares approximation by trigonometric polynomials, Math. Comp. 57 (1991) 273–289. [25] S.D. Riemenschneider, Lacunary trigonometric interpolation, Convergence, in: E.W. Cheney (Ed.), Quantitative approximation, Academic Press, New York, 1980, pp. 741–746. [26] K.P. Sahakyan, Hermite trigonometric interpolation, East J. Approx. 12 (4) (2006) 441–449. [27] H.E. Salzer, New fomulas for trigonometric interpolation, J. Math. Phys. 39 (1960) 85–96. [28] J. Szabados, The exact error of trigonometric interpolation for differentiable functions, Construc. Approx. 8 (2) (1992) 203–210.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Piecewise trigonometric Hermite interpolation Xuli Han∗ School of Mathematics and Statistics, Central South University, Changsha, 410083, PR China
a r t i c l e
i n f o
Keywords: Trigonometric basis Trigonometric polynomial Trigonometric interpolation Trigonometric Hermite interpolation
a b s t r a c t Based on the symmetric, nonnegative and normalized basis of the trigonometric polynomial space, the piecewise trigonometric Hermite interpolation methods are presented. The C n−1 and Cn continuous piecewise trigonometric Hermite interpolants of degree n are constructed and the interpolation methods are local. The integral and the differential representations of the errors of the trigonometric Hermite interpolants are given. Several examples are supplied to support the practical value of the given interpolation methods. © 2015 Elsevier Inc. All rights reserved.
1. Introduction Trigonometric Hermite interpolation is an important topic in approximation theory. Salzer [27] has already given Lagrangian and Barycentric formulae for arbitrary interpolation points and problems where higher order derivatives were also prescribed. Kress [18] has derived Lagrangian as well as remainder formulae and asymptotic convergence results for the most important case of an even number of equidistant points. The trigonometric Hermite interpolation on equidistant nodes have been addressed in [3], [4]. Several researchers have focused their attentions on this subject, see [6], [8], [16], [23], [26], even for arbitrary points. Several authors were mostly interested in existence questions [4], [17], convergence results [19], [25] and formulae other than Lagrange’s, see [15], [21], [24], [28]. There are a few results on the construction of trigonometric polynomial sequence approximating continuous function. Some authors were interested in the problem of constructing nonnegative trigonometric polynomials, see [2], [7], [9]. Askey and Steining’s paper [4] gave an idea of how new nonnegative trigonometric polynomials can be generated by a known sequence of nonnegative trigonometric polynomials. In [1], authors approximated continuous functions defined on a compact set E ⊂ [−π , π ] by trigonometric polynomials. A stable recurrence relation for the trigonometric B-spline was given in [20]. In [22], a general theory of quasi-interpolants based on trigonometric splines was developed which is analogous to the polynomial spline case. The aim was to construct quasiinterpolants which are local, easy to compute, and which apply to a wide class of functions. Trigonometric splines can be used to solve the problems of geometric modeling. Some types of trigonometric splines have been introduced having different features, see [10]–[12]. These trigonometric splines have been used for constrained curve design in [5]. In [13] and [14], the symmetric, nonnegative and normalized basis of trigonometric polynomial space has been presented. In [13], based on the basis, the symmetric trigonometric polynomial approximants like Bernstein polynomials are constructed. Two kinds of nodes are given to show that the trigonometric polynomial sequence is uniform convergent. The trigonometric quasi-interpolants of reproducing one degree of trigonometric polynomials are constructed. In [14], the total positivity and the B-basis property of the symmetric trigonometric polynomial basis functions are shown. Then the symmetric trigonometric
∗
Tel.: +8613973116065. E-mail address: [email protected], [email protected]
http://dx.doi.org/10.1016/j.amc.2015.06.125 0096-3003/© 2015 Elsevier Inc. All rights reserved.
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
617
polynomial curves like Bézier curves are presented and some theoretics and methods are proposed for curve representation of the trigonometric polynomial space. In the polynomial space, the piecewise Hermite interpolants are classic and important. Represented by the Bernstein basis functions, the piecewise polynomial Hermite interpolants with two points for each piece expression are useful for numerical computation and functional approximation. The piecewise polynomial Hermite interpolants of degree 2n + 1 are Cn continuous. If a problem of numerical computation or functional approximation concerns some data which relate to trigonometric functions, then a trigonometric interpolation polynomials can be considered. However, in the trigonometric polynomial space, a general piecewise trigonometric Hermite interpolant like the piecewise polynomial Hermite interpolant represented by the Bernstein basis functions had not been presented. The purpose of this paper is to present explicit piecewise trigonometric Hermite interpolants which are local and easy to compute. The main result is that the presented piecewise trigonometric Hermite interpolant of degree n achieves C n−1 or Cn continuity. Since we have the symmetric, nonnegative and normalized trigonometric basis functions like Bernstein polynomials, it is conceivable and convenient to construct the piecewise trigonometric Hermite interpolant with two points for each piece expression. The remainder of this paper is organized as follows. In Section 2, the basis functions of the trigonometric polynomial space are described and the derivative properties of the basis functions are shown. In Section 3, the C n−1 and Cn continuous piecewise trigonometric Hermite interpolants of degree n are presented. The convergence of the piecewise trigonometric Hermite interpolation method is discussed in Section 4. Conclusions are given in Section 5. 2. The symmetric trigonometric basis functions Consider the trigonometric polynomial space
Tn := span{1, sin (u), cos (u), sin (2u), cos (2u), . . . , sin (nu), cos (nu)} for u ∈ [0, π /2]. The basis functions sin (iu), cos (iu) have been used almost for all the topic concerning the trigonometric polynomial problems. However, the basis functions sin (iu), cos (iu) are not nonnegative and normalized. It is not convenient for the representation of a trigonometric Hermite interpolant by using these basis functions. Now we consider the symmetric, nonnegative and normalized basis functions of Tn presented in [13] and [14]. For u ∈ [0, π /2], n ∈ N, let s(u) := 1 − sin (u), c(u) := 1 − cos (u), w(u) := sin (u) + cos (u) − 1, we introduce the trigonometric polynomials of degree n as follows:
Ti,n (u) := where
ai,1 :=
ai,n+1 :=
i = 0, 1, . . . , n, i = n + 1, n + 2, . . . , 2n,
i = 0, 1, 2, i = 0, 1, 2,
1, 0,
and for n ≥ 1,
ai,n sn−i (u)wi (u), ai,n w2n−i (u)ci−n (u),
(2)
⎧1 ⎨ 2 ai−2,n + ai−1,n + ai,n , ⎩
1 a 2 i−2,n
+ ai−1,n + 12 ai,n ,
ai−2,n + ai−1,n +
(1)
1 a , 2 i,n
i ≤ n, i = n + 1,
(3)
i ≥ n + 2.
It has been shown that the trigonometric polynomials of the set {T0,n (u), T1,n (u), . . . , T2n,n (u)} form a basis for the trigonometric polynomial space Tn and
Tn = span{T0,n (u), T1,n (u), . . . , T2n,n (u)}. We refer to these trigonometric functions as symmetric trigonometric basis functions. Many properties of the symmetric trigonometric basis functions can be found in [13] and [14]. In order to represent the trigonometric Hermite interpolants, we need to discuss the derivative properties of the basis functions. In [14], the derivatives of (1) are given as follows (u) = Ti,n
iai,n (2n − i)ai,n Ti−1,n (u) − (n − i)Ti,n (u) − Ti+1,n (u) ai−1,n 2ai+1,n
(4)
for i = 0, 1, . . . , n − 1, (u) = Tn,n
nan,n [Ti−1,n (u) − Ti+1,n (u)], an−1,n
(5)
(u) = Ti,n
(2n − i)ai,n iai,n Ti−1,n (u) + (i − n)Ti,n (u) − Ti+1,n (u) 2ai−1,n ai+1,n
(6)
for i = n + 1, n + 2, . . . , 2n.
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From (4) to (6), we have (u), T (u), . . . , T (u)) = (T (u), T (u), . . . , T (T0,n 0,n 1,n 2n,n (u))Bn , 1,n 2n,n
(7)
where Bn := (bij ) is a tridiagonal matrix of order 2n + 1 with
bi,i = i − n − 1,
bi,i+1 =
i = 1, 2, . . . , 2n + 1,
iai,n /ai−1,n , iai,n /(2ai−1,n ),
bi+1,i = bi, j = 0,
i = 1, 2, . . . , n, i = n + 1, n + 2, . . . , 2n,
−(2n − i + 1)ai−1,n /(2ai,n ), −(2n − i + 1)ai−1,n /ai,n ,
i = 1, 2, . . . , n, i = n + 1, n + 2, . . . , 2n,
|i − j| > 1. (r)
Lemma 1. Let bi, j be the ij entry of the rth power of the matrix Bn for i, j = 1, 2, . . . , 2n + 1 and r = 1, 2, . . . , n (For convenience, we (r)
omit a superscript or subscript n for bi, j ), then
1 r) b(1,r = − r!(2n − r + 1)ar−1,n , 2 r) b(2n+1,2n−r+1 = (−1)r r!a2n−r,n ,
r) b(1,r+1 = r!ar,n ,
(8)
1 r) b(2n+1,2n−r+2 = (−1)r+1 r!(2n − r + 1)a2n−r+1,n . 2
(9)
Proof. Obviously, the rth power Brn is a band matrix with band width 2r + 1, r = 1, 2, . . . , n, and 1) 1) 1) 1) b(1,1 = −n, b(1,2 = n, b(2n+1,2n = −n, b(2n+1,2n+1 = n.
We assume that the formulas (8) and (9) are true for r ≤ n − 1, then, when r + 1 ≤ n, by the elements of Brn and Bn we have
rar,n r) − b(1,r+1 (n − r) ar−1,n 1 1 rar,n = − r!(2n − r + 1)ar−1,n − r!ar,n (n − r) = − (r + 1)!(2n − r)ar,n , 2 ar−1,n 2
r+1) r) b(1,r+1 = b(1,r
r+1) r) = b(1,r+1 b(1,r+2
(r + 1)ar+1,n ar,n
r+1) r) b(2n+1,2n−r = −b(2n+1,2n−r+1 r+1)
= (−1) (r+1)
= r!ar,n
(r + 1)ar+1,n
(r + 1)a2n−r−1,n a2n−r,n
ar,n
= (r + 1)!ar+1,n ,
= (−1)r+1) r!a2n−r,n
(r + 1)a2n−r−1,n a2n−r,n
(r + 1)!a2n−r−1,n ,
(r)
r) b2n+1,2n−r+1 = b2n+1,2n−r+1 (n − r) − b(2n+1,2n−r+2
ra2n−r,n a2n−r+1,n
1 ra2n−r,n = (−1)r r!a2n−r,n (n − r) + (−1)r r!(2n − r + 1)a2n−r+1,n 2 a2n−r+1,n 1 = (−1)r (r + 1)!(2n − r)a2n−r,n . 2 Thus we obtain the results. The matrices Brn will be used to construct piecewise trigonometric Hermite interpolants. 3. The piecewise trigonometric Hermite interpolants The piecewise two points Hermite interpolants are classical polynomial interpolants and can be constructed conveniently based on the Bernstein polynomials. In this section, we construct the piecewise trigonometric Hermite interpolants by using the symmetric trigonometric basis functions. 3.1. The C n−1 continuous piecewise Hermite interpolants of degree n (r)
Given a partition : a = x1 < x2 < · · · < xm = b of the bounded interval [a, b] of R, and let n ≥ 1, fi (i = 1, 2, . . . , m; r = 0, 1, . . . , n − 1) be the given values at nodes x = xi . For 1 ≤ k ≤ m − 1, x ∈ [xk , xk+1 ], hk = xk+1 − xk , u = π (x − xk )/2hk , we construct a piecewise trigonometric interpolant
Tn (x) :=
2n i=0
Ti,n (u) p[k] i,n
(10)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
619
to satisfy the following interpolation conditions:
Tn(r) (xi ) = fi(r) ,
i = k, k + 1;
r = 0, 1, . . . , n − 1.
(11)
Thus, the interpolant Tn ∈ C n−1 ([a, b]). From (1) to (3), for x ∈ [xk , xk+1 ], we have expressions
T1 (x) = s(u) p[k] + w(u) p[k] + c(u) p[k] , 0,1 1,1 2,1 + 2s(u)w(u) p[k] + 2w2 (u) p[k] + 2w(u)c(u) p[k] + c2 (u) p[k] , T2 (x) = s2 (u) p[k] 0,2 1,2 2,2 3,2 4,2 9 + 4w3 (u) p[k] s(u)w2 (u) p[k] 2,3 3,3 2
+ 3s2 (u)w(u) p[k] + T3 (x) = s3 (u) p[k] 0,3 1,3
9 2 w (u)c(u) p[k] + 3w(u)c2 (u) p[k] + c3 (u) p[k] . 4,3 5,3 6,3 2
+
Theorem 1. The pi,n (i = 0, . . . , n − 1, n + 1, . . . , 2n) in (10) can be obtained uniquely by the interpolation conditions (11). [k]
Proof. According to (7), we have (r) (r) (r) (T0,n (u), T1,n (u), . . . , T2n,n (u)) =
π r 2hk
(T0,n (u), T1,n (u), . . . , T2n,n (u))Brn
for r = 0, 1, . . . , n. Thus, the interpolation conditions (11) lead to the linear equations
⎛
p[k] 0,n
⎜ [k] ⎜ p1,n BL,n ⎜ ⎜ .. ⎝ . p[k] n−1,n and
⎛
⎞
⎛
⎞
fk
⎟ ⎜ 2hk ⎟ ⎟ ⎜ π fk ⎟ ⎟=⎜ ⎟, . ⎟ ⎝ .. ⎠ ⎠ 2hk n−1 (n−1)
p[k] n+1,n
fk
π
⎞
⎛
(12)
⎞
fk+1
⎜ p[k] ⎟ ⎜ 2hπk f ⎟ k+1 ⎜ n+2,n ⎟ ⎜ ⎟ BR,n ⎜ . ⎟ = ⎜ ⎟, .. ⎝ .. ⎠ ⎝ ⎠ . 2hk n−1 (n−1) [k] p2n,n
where
⎛ ⎜
(13)
fk+1
π
⎞
0) b(1,1
(1) BL,n := ⎜ b1,1 ⎝ ··· n−1) b(1,1
⎟ ⎟, ⎠
1) b(1,2 ··· n−1) b(1,2
···
n−1) b(1,n
⎛ ⎜ ⎝
1) b(2n+1,2n ··· n−1) b(2n+1,2n
BR,n := ⎜ n−1) b(2n+1,n+2
···
(14)
0) b(2n+1,2n+1
⎞ ⎟
1) b(2n+1,2n+1 ⎟ ⎠ ··· (n−1) b2n+1,2n+1
(r)
(15)
(r)
are triangular matrices. By Lemma 1, b1,r+1 = 0, b2n+1,2n−r+1 = 0, r = 0, 1, . . . , n − 1. Therefore, by (12) and (13), pi,n (i = 0, . . . , n − 1, n + 1, . . . , 2n) can be obtained easily. [k]
[k]
[k]
Obviously, for C n−1 continuity, pn,n can be chosen freely. Now we show the choice of the pn,n . (n)
If fk
(n)
and fk+1 are not given, then a simply method is to choose
p[k] n,n :=
1 [k] pn−1,n + p[k] . n+1,n 2
(n)
(n)
If fk
(16) (n)
(n)
and fk+1 are given, then, let Tn (xk ) = fk , we have
n+1 b(1,n)j p[k] = j−1,n j=1
2hk
π
n
fk(n) .
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X. Han / Applied Mathematics and Computation 268 (2015) 616–627
From Lemma 1, we obtain
p[k] n,n
=
p˜[k] n,n
1 := n!an,n
2hk
n − b(1,n)j p[k] . j−1,n
(n)
fk
π
(n)
n
j=1
(n)
In the same way, let Tn (xk+1 ) = fk+1 , we have 2n+1
n) b(2n+1, p[k] = j j−1,n
2hk
π
j=n+1
and then
p[k] n,n
=
p¯ [k] n,n
(−1)n
:= (n)
Therefore, when fk
n
2hk
n
(n)
fk+1 −
π
n!an,n
(n) fk+1 ,
2n+1
(n)
b2n+1, j p[k] j−1,n
.
j=n+2
(n)
and fk+1 are given, we would like to set
1 [k] p˜ n,n + p¯ [k] n,n . 2
p[k] n,n :=
(17)
Of course, we can also choose pn,n by letting Tn (xk + hk /2) = fk+1/2 if the value fk+1/2 is given at node x = xk+1/2 . [k]
Theorem 2. With u = π (x − xk )/(2hk ), the interpolation operator Tn reproduces all trigonometric polynomials of degree ≤ n in Tn by (12), (13) and (17). Proof. Take u = π (x − xk )/(2hk ) as the trigonometric function variable, we can know that any trigonometric polynomial f(x) of [k] [k] degree ≤ n in Tn is determined uniquely by (12), (13) and p˜ n,n . Therefore, Tn (x) = f (x) by (12), (13) and p˜ n,n . Simultaneously, [k] Tn (x) = f (x) by (12), (13) and p¯ n,n . Then Tn (x) = f (x) by (12), (13) and (17).
Now we give some explicit expressions of the trigonometric interpolants of lower degree for x ∈ [xk , xk+1 ]. Since
B1 =
−1 −1 0
1 0 −1
0 1 , 1
it is easy to know that
T1 (x) = s(u) fk + w(u)
1 ( f + fk+1 ) + c(u) fk+1 2 k
(18)
by (12), (13) and (16), and
hk 1 T1 (x) = s(u) fk + w(u) ( f + fk+1 ) + ( fk − fk+1 ) + c(u) fk+1 2 k π
by (12), (13) and (17). Since
⎛−2
⎜−1 B2 = ⎜ 0 ⎝ 0 0
2 −1 −3/2 0 0
0 2 0 −2 0
⎞
0 0 3/2 1 −2
0 0⎟ 0⎟, ⎠ 1 2
⎛ 2 ⎜ 3 B22 = ⎜3/2 ⎝ 0 0
−6 −4 3/2 3 0
4 −2 −6 −2 4
(19)
0 3 3/2 −4 −6
⎞
0 0 ⎟ 3/2⎟, ⎠ 3 2
we can obtain
p[k] = fk , 0,2
p[k] = fk + 1,2
hk fk ,
π
p[k] = fk+1 − 3,2
hk fk+1 ,
π
p[k] = fk+1 4,2
(20)
by (12), (13), and
p[k] = 2,2
1 h ) ( f + fk ) + k ( fk − fk+1 2 k 2π
(21)
2 1 3h ) + hk ( f + f ) ( fk + fk+1 ) + k ( fk − fk+1 k+1 2 4π 2π 2 k
(22)
by (16), and
p[k] = 2,2 by (17).
X. Han / Applied Mathematics and Computation 268 (2015) 616–627 2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1 −4
−3
−2
−1
0
1
2
3
4
−1 −4
−3
(a) m = 5, hk = π/2
−2
−1
0
621
1
2
3
4
2
3
4
(b) m = 9, hk = π/4
Fig. 1. The curves of T2 (x) and the cubic Hermite interpolants to (27). 2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1 −4
−3
−2
−1
0
1
2
3
4
−1 −4
−3
(a) m = 5, hk = π/2
−2
−1
0
1
(b) m = 9, hk = π/4
Fig. 2. The curves of T3 (x) and the quintic Hermite interpolants to (27).
By (12), (13) and (16), we have
p[k] = fk , 0,3
p[k] = fk + 1,3
2hk f , 3π k
p[k] = fk + 2,3
4h2k 10hk fk + f , 9π 9π 2 k
2 1 5h ) + 2hk ( f + f ), ( fk + fk+1 ) + k ( fk − fk+1 k+1 2 9π 9π 2 k 2 4hk 10hk 2hk = fk+1 − f + f , p[k] = fk+1 − f , 5,3 9π k+1 9π 2 k+1 3π k+1
(23)
p[k] = 3,3 p[k] 4,3
(24)
p[k] = fk+1 , 6,3
(25)
and by (17) we have
p[k] = 3,3
3 2 1 19hk ) + hk ( f + f ) + hk ( f − f ). ( fk + fk+1 ) + ( fk − fk+1 k k+1 k+1 2 2 24π 2π 6π 3 k
(26)
Example 1. Consider the function
f (x) = sin (4x) + (r)
and use the data fi
1 , 1 + x2
x ∈ [−π , π ],
(27)
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n).
Fig. 1 shows the graphs of the function (27) (dotted lines) and the C1 continuous piecewise trigonometric Hermite interpolant T2 (x). On the left, we take m = 5, hk = π /2 and use the expression (21) (solid lines). For comparison, the cubic polynomial Hermite interpolation curves are given (dashed line). On the right, we take m = 9, hk = π /4 and use the expression (21) (dashed lines) and (22) (solid lines). When the expression (22) is used, the curve merges with the curve of the function (27). Fig. 2 shows the graphs of the function (27) (dotted lines) and the C2 continuous piecewise trigonometric Hermite interpolant T3 (x). We show the curves by using the expression (24), On the left, we take m = 5, hk = π /2 (solid lines). For comparison, the quintic polynomial Hermite interpolation curves are given (dashed line). On the right, we take m = 9, hk = π /4. Example 2. Consider the function
f (x) = cos (3x) exp (−x) + cos (2x) exp (x − 2π ), (r)
and use the data fi
x ∈ [0, 2π ],
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n).
(28)
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X. Han / Applied Mathematics and Computation 268 (2015) 616–627 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) use expression(22)
2
3
4
5
6
7
5
6
7
(b) use expression(21)
Fig. 3. The curves T2 (x) interpolate to (28). 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) m = 5, hk = π/2
2
3
4
(b) m = 9, hk = π/4
Fig. 4. The curves T3 (x) interpolate to (28).
Fig. 3 shows the graphs of the function (28) (dotted lines) and the C1 continuous piecewise trigonometric Hermite interpolant T2 (x) (solid lines). On the left, we take m = 5, hk = π /2 and use the expression (22). On the right, we take m = 9, hk = π /4 and use the expression (21). Fig. 4 shows the graphs of the function (28) (dotted lines) and the C2 continuous piecewise trigonometric Hermite interpolant T3 (x). We show the curves by using the expression (24), taking m = 5, hk = π /2 on the left and m = 9, hk = π /4 on the right respectively. 3.2. The Cn continuous piecewise Hermite interpolants of degree n It is well known that the B-splines and the spline interpolation functions of degree n have C n−1 continuity. The C n−1 continuous spline interpolation method is not local interpolation method. In [11], the C2 continuous piecewise quadratic trigonometric polynomials were presented. Now we give some Cn continuous piecewise trigonometric interpolants of degree n and the given interpolation methods will be local interpolation methods. From (12) and (13) we have
p[k] n−1,n
1 = (n − 1)!an−1,n
(−1) = p[k] n+1,n (n − 1)!an+1,n n−1
(n−1)
where fk (n−1)
dk+1
for
(n−1)
and fk+1
[k] pn−1,n
p¯ [k] n−1,n p¯ [k] n+1,n
and
2hk
n−1
fk
π
2hk
(n−1)
n−1
π
n−1 ) [k] − b(1,n−1 p j−1,n , j j=1
(n−1) fk+1 −
2n+1
n−1) [k] b(2n+1, p , j j−1,n
j=n+3
(n−1)
are the given values. Now we change them (except for f1
[k] pn+1,n
(not for
1 = (n − 1)!an−1,n
(−1)n−1 = (n − 1)!an+1,n
[k] pn,n )
2hk
respectively, let
n−1
dk
π
2hk
π
(n−1)
n−1
dk+1 −
2n+1 j=n+3
(n−1)
) to undetermined dk
and
n−1 ) [k] − b(1,n−1 p j−1,n , j j=1
(n−1)
(n−1)
and fm
(29)
(n−1)
b2n+1, j p[k] j−1,n
,
(30)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
623
and then take the piecewise trigonometric interpolant as 2n
Tn (x) =
Ti,n (u) p[k] + Tn−1,n (u) p¯ [k] + Tn+1,n (u) p¯ [k] n−1,n n+1,n i,n
(31)
i=0, j=n−1,n+1
(n−1)
(n−1)
for 1 ≤ k ≤ m − 1, x ∈ [xk , xk+1 ], u = π (x − xk )/(2hk ). The dk used to achieve Cn continuity. Since (n)
Tn
(xk +) =
π n n−1 2hk
=− Tn(n) (xk+1 −) =
(n) [k]
b1, j p j−1,n + b1,n p¯ n−1,n +
j=1
n(n + 1)π (n−1) dk + 4hk
π n
2hk
2n+1
π n 2hk
[k]
are not included in the term pn,n and they will be
(n)
b1,n+1 p[k] n,n
+ αk,n ,
n) n) n) b(2n+1, p[k] + b(2n+1,n+2 + b(2n+1,n+1 p[k] p¯ [k] n,n n+1,n j j−1,n
j=n+3
n(n + 1)π (n−1) dk+1 + 4hk
=
(n) [k]
and dk+1
π n 2hk
− αk+1,n ,
where + αk,n =
n−1 1 ) b(1,n)j + n(n + 1)b(1,n−1 p[k] + n!an,n p[k] n,n , j−1,n j 2 j=1
− αk+1,n =
2n+1
n) b(2n+1, − j
j=n+3
(n)
1 n−1) n(n + 1)b(2n+1, p[k] + (−1)n n!an,n p[k] n,n , j−1,n j 2
(n)
let Tn (xk +) = Tn (xk −), we obtain the Cn continuous conditions: (n−1)
dk
1 π n−1 hk−1 hk 1 + − = n−2 α − α 2 n(n + 1)(hk−1 + hk ) hnk k,n hnk−1 k,n
(32)
for k = 2, 3, . . . , m − 1. (n−1) (n−1) (n−1) (n−1) (n−1) (n−1) and fk+1 , then, by (32), fk−1 , fk and fk+1 are used to represent dk for Tn (x). If we use (16) which involves fk (r)
(r)
(r)
(r)
Obviously, fk−1 , fk , fk+1 and fk+2 are used for the kth segment of the trigonometric interpolants, r = 0, 1, . . . , n − 1. The values (r)
of fk
affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. In this case, we obtain
(1)
dk
1
dk(0) =
hk fk−1 + (hk−1 + hk ) fk + hk−1 fk+1 ,
2(hk−1 + hk ) 1 )−h = [π (hk fk−1 + hk−1 fk ) + hk ( fk − fk−1 k−1 ( f k+1 − f k )], 3(hk−1 + hk )
dk(2) =
π2
2(hk−1 + hk ) +
2 9(hk−1 + hk )
h hk−1 fk − k fk−1 hk hk−1
+
π
36(hk−1 + hk )
hk + 37 f ) − hk−1 (37 f + 20 f ) (20 fk−1 k k k+1 hk−1 hk
+ f ) + h [hk ( fk−1 k−1 ( f k + f k+1 )], k
(0)
where fi = ( fi+1 − fi )/hi . Note that T1 (x) interpolates dk (k = 2, 3, . . . , m − 1) in this case. (n−1)
If we use (17) which involves fk (r)
(n−1)
(n)
, fk+1 , fk
(n)
(n−1)
and fk+1 , then, by (32), some fi
(n)
and fi
(n−1)
are used to represent dk
. The
values of fk , r = 0, 1, . . . , n, affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. For example, in this case,
dk(0) =
1 hk fk−1 + (hk−1 + hk ) fk + hk−1 fk+1 2(hk−1 + hk ) +
hk−1 hk
π (hk−1 + hk )
− f ). ( fk−1 k+1
(n−1)
If we do not want to use fi
p[k] n,n =
1 [k] (p + p[k] ) n+2,n 2 n−2,n
(n)
or fi
(n ≥ 2) to represent dk(n−1) , then instead of (16) or (17), we can use (33)
624
X. Han / Applied Mathematics and Computation 268 (2015) 616–627 1.2
1.2
0.8
0.8
0.4
0.4
0
0
−0.4 0
1
2
3
4
5
6
7
−0.4 0
1
(a) use expression (33)
2
3
4
5
6
7
(b) use expression (16) 2
Fig. 5. The C continuous interpolation curves T2 (x).
(r)
to represent (32). In this case, the values of fk , r = 0, 1, . . . , n − 2, affect locally 4 segments of the interpolants for x ∈ [xk−2 , xk+2 ]. We obtain
dk(1) = (2)
dk
=
π
3(hk−1 + hk )
π2
2(hk−1 + hk )
(hk fk−1 + hk−1 fk ), hk hk−1 fk − fk−1 + hk
π
12(hk−1 + hk )
hk−1
hk + 15 f ) − hk−1 (15 f + 4 f ) . (4 fk−1 k k k+1 hk−1 hk
Fig. 5 shows the graphs of the function (28) (dotted lines) and the C2 continuous trigonometric Hermite interpolant T2 (x) (solid lines). The expression (33) is used on the left and the expression (16) is used on the right. 4. The convergence of the trigonometric interpolation It has been shown that the interpolation operator Tn can reproduce all trigonometric polynomials of degree ≤ n in Tn . Numerical examples in Section 3 show that the presented piecewise trigonometric Hermite interpolants approximate the given functions well. Further, we show the errors of the piecewise trigonometric Hermite interpolation in this section. A trigonometric Taylor expansion will be presented to show the integral representation of the error. This expansion is different from the trigonometric Taylor expansion in [20] for which the recurrence relations were based on even and odd order trigonometric polynomials, respectively. 4.1. The trigonometric Taylor expansion Let Ln be the differential operators defined by L1 = D( ≡ d/dx), and
Ln+1 =
D2 +
4n2 q2
Ln
(34)
for integer n ≥ 1 and real number q = 0. Theorem 3. Let f ∈ C 2n+1 [a, b], then
f (x) = sn (x) +
q2n (2n)!
x
sin a
x−y q
2n
where sn is determined by s0 (x) = f (a), and
q2n−1 x−a sin q (2n − 1)!
sn (x) = sn−1 (x) +
Ln+1 f (y)dy,
2n−1
cos
x−a q2n x−a Ln f (a) + sin q q (2n)!
for n ≥ 1. Proof. Obviously,
f (x) = s0 (x) +
a
x
L1 f (y)dy.
Therefore, (35) holds for n = 1. Assume (35) holds for sn−1 , that is
f (x) = sn−1 (x) +
q2n−2 (2n − 2)!
x
sin a
x−y q
(35)
2n−2
Ln f (y)dy.
2n
DLn f (a)
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
625
By integration-by-parts argument, we have
x
In := a
=
x−y q
sin 4n2 q2
x
a
x−a = − sin q and then
In−1 =
Ln+1 f (y)dy
2n
x−y q
sin
2n
2n
Ln f (y)dy − sin
x−a q
2n x−a DLn f (a) − sin q q
1 x−a sin 2n − 1 q
2n
2n−1
DLn f (a) +
x
sin a
x−a q2 x−a Ln f (a) + sin q q 2n(2n − 1)
cos
x−y q
2n−1
x−a 2n(2n − 1) Ln f (a) + cos q q2
2n−1
2n q
2n
cos
x
sin
a
DLn f (a) +
x−y DLn f (y)dy q x−y q
2n−2
Ln f (y)dy,
q2 In . 2n(2n − 1)
From this and
f (x) = sn−1 (x) +
q2n−2 I , (2n − 2)! n−1
we obtain (35). The proof is complete by induction.
4.2. The error of the piecewise trigonometric Hermite interpolation 4h
Take a = xk and q = πk in (35), we have
2n−1 (4hk )2n−1 π (x − xk ) π (x − xk ) sn (x) = sn−1 (x) + sin cos Ln f (xk ) 4hk 4hk (2n − 1)!π 2n−1 2n (4hk )2n π (x − xk ) + sin DLn f (xk ), 4hk (2n)!π 2n 2n (4hk )2n x π (x − y) f (x) = sn (x) + sin Ln+1 f (y)dy. 4hk (2n)!π 2n xk
(36)
(37)
Obviously, sn is a trigonometric polynomial of degree n regarding variable u = π (x − xk )/(2hk ). (r)
Theorem 4. Let f ∈ C 2n+1 [a, b], fi = f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n), Tn ( f ; x) be the piecewise trigonometric Hermite interpolant corresponding to f with (17), then
f (x) − Tn ( f ; x) =
(4hk )2n (2n)!π 2n
xk+1
K (x, y)Ln+1 f (y)dy,
xk
where
(38)
T (v− ; x), y ∈ [xk , x], v (x, y) − Tn (v ) = n n + −Tn (vn ; x), y ∈ [x, xk+1 ], 0, y ∈ [xk , x], vn (x, y), y ∈ [xk , x], + − vn (x, y) = vn (x, y) = 0, y ∈ [x, xk+1 ], vn (x, y), y ∈ [x, xk+1 ]. 2n π (x − y) vn (x, y) = sin . K (x, y) =
+ n
+ n;x
4hk
Proof. Let
en (x) =
(4hk )2n (2n)!π 2n
x
sin
xk
π (x − y) 4hk
2n Ln+1 f (y)dy.
Since Tn is a linear operator and sn is a trigonometric polynomial of degree n regarding variable u = π (x − xk )/(2hk ), we have
Tn ( f ; x) = Tn (sn + en ; x) = sn (x) + Tn (en ; x). A straightforward computing gives that
e(nr) (x) =
(4hk )2n (2n)!π 2n
x xk
dr vn (x, y)Ln+1 f (y)dy, dxr
r = 0, 1, . . . , n,
626
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
and then
(4hk )2n xk (r) v (x , y)Ln+1 f (y)dy = 0, (2n)!π 2n xk n k (4hk )2n xk+1 (r) vn (xk+1 , y)Ln+1 f (y)dy. e(nr) (xk+1 ) = (2n)!π 2n xk
e(nr) (xk ) =
Therefore,
Tn (en ; x) =
(4hk )2n (2n)!π 2n
xk+1 xk
Tn (v+ n ; x)Ln+1 f (y)dy.
Since vn (x, y) = Tn (vn , x), for y < x we have
vn (x, y) − Tn (v+n ; x) = Tn (v−n ; x). The proof is complete. Note that Tn = Ker(Ln+1 ) regarding variable u = π (x − xk )/(2hk ). From (38) we can know that Tn reproduces all trigonometric polynomials of degree ≤ n in Tn . (r)
Theorem 5. Let f ∈ C 2n [a, b], fi
= f (r) (xi ) (i = 1, 2, . . . , m; r = 0, 1, . . . , n), Tn ( f ; x) be the piecewise trigonometric Hermite in-
terpolant corresponding to f with any pn,n , then to each x in [xk , xk+1 ] there corresponds a point ξ in (xk , xk+1 ) such that
f (x) − Tn ( f ; x) =
[k]
(x − xk )n (x − xk+1 )n (2n) [f (ξ ) − Tn(2n) ( f ; ξ )], (2n)!
(39)
for 1 ≤ k ≤ m − 1. Proof. If x is the node xk or xk+1 , the assertion is obviously true since both sides of Eq. (41) reduce to 0. So, let x be a fixed point in (xk , xk+1 ). Put
f (x) − Tn ( f ; x) (t − xk )n (t − xk+1 )n . (x − xk )n (x − xk+1 )n
ϕ(t ) := f (t ) − Tn ( f ; t ) −
Notice that ϕ has at least 2n + 1 zeros in [xk , xk+1 ], namely, x, xk ( n multiple zeros), xk+1 (n multiple zeros). By Rolle’s Theorem, ϕ has at least 2n zeros in [xk , xk+1 ]. Similarly, ϕ has at least 2n − 1 zeros in [xk , xk+1 ]. If this argument is repeated, we conclude eventually that ϕ (2n) has at least one zero, say ξ , in (xk , xk+1 ). Thus,
0 = ϕ(ξ ) = f (2n) (ξ ) − Tn(2n) ( f ; ξ ) −
f (x) − Tn ( f ; x) (2n)! (x − xk )n (x − xk+1 )n
which is another form of the equation to be proved. Let
Tnv (u) = (T0,n (u), T1,n (u), . . . , T2n,n (u)),
P2n = ( p[k] , p[k] , . . . , p[k] )T . 0,n 1,n 2n,n
The piecewise interpolant (10) can be written as Tn ( f ; x) = Tnv (u)P2n and then
Tn(2n) ( f ; x) =
π 2n 2hk
Tnv (u)B2n n P2n .
From (39), we obtain
1 | f (x) − Tn ( f ; x)| ≤ (2n)!
hk 2
where M2n = maxxk ≤x≤xk+1 | f (2n) (x)|. When n = 2, by (21) we have
⎛
2n M2n +
π 2n 4
⎞
) 8( fk − fk+1 ) + hπk (16 fk + 14 fk+1
⎜8( f − f ) + hk (17 f + 14 f )⎟ ⎜ k k+1 π k k+1 ⎟ ⎜ ⎟ 3hk ⎟, B42 P4 = ⎜ ( fk − fk+1 ) 2π ⎜ ⎟ ⎜ ⎟ hk 8 ( f − f ) − ( 14 f + 17 f ) ⎝ k+1 k π k k+1 ⎠ ) 8( fk+1 − fk ) − hπk (14 fk + 16 fk+1
and then
31 ||B42 P4 ||∞ ≤ 8 + h M , π k 1
where M1 = maxxk ≤x≤xk+1 | f (x)|.
||
B2n n P2n
||∞ ,
X. Han / Applied Mathematics and Computation 268 (2015) 616–627
627
5. Conclusion The new symmetric, nonnegative and normalized basis of the trigonometric polynomial space is useful for constructing a piecewise trigonometric Hermite interpolant, analogous to the piecewise polynomial Hermite interpolant with two points for each piece. The constructed piecewise trigonometric Hermite interpolants of degree n can reproduce any trigonometric polynomials of degree ≤ n. Furthermore, the given interpolation methods are local and the interpolants of degree n are C n−1 or Cn continuous. The presented general trigonometric Taylor expansion provides a simple error expression of the trigonometric interpolant with an integral kernel. Numerical examples show the efficiency of the given interpolation methods. Acknowledgment The author thanks the referees for their careful review and valuable comments. This research is supported by the National Natural Science Foundation of China (no. 11271376). References [1] V. Andrievskii, H.-P. Blatt, On approximation of continuous functions by trigonometric polynomials, J. Approx. Theory 163 (2) (2011) 249–266. [2] R. Askey, J. Steinig, Some positive trigonometric sums, Trans. Amer. Math. Soc. 187 (1974) 295–307. [3] J.-P. Berrut, A. Welscher, Fourier and barycentric formulae for equidistant Hermite trigonometric interpolation, Appl. Comput. Harmonoic Anal. 23 (2007) 307–320. [4] A.S. Cavaretta, A. Sharma, R.S. Varga, Lacunary trigonometric interpolation on equidistant nodes, in: R.A. DeVore, K. Scherer (Eds.), Quantitative approximation, Academic Press, New York, 1980, pp. pp.63–80. [5] N. Choubey, A. Ojha, Constrained curve drawing using trigonometric splines having shape parameters, Comput. Aid. Des. 39 (2007) 1058–1064. [6] F.J. Delvos, Hermite interpolation with trigometric polynomials, BIT, 1993, 33, 113–123. [7] D.K. Dimitrov, C.A. Merlo, Nonnegative trigonometric polynomials, Constr. Approx. 18 (1) (2002) 117–143. [8] J. Du, H. Han, G. Jin, On trigonometric and paratrigonometric Hermite interpolation, J. Approx. Theory 131 (2004) 74–99. [9] A. Gluchoff, F. Hartmann, Univalent polynomials and non-negative trigonometric sums, Amer. Math. Monthly 105 (1998) 508–522. [10] X. Han, Quadratic trigonometric polynomial curves with a shape parameter, Comput. Aid. Geom. Des. 19 (2002) 503–512. [11] X. Han, Piecewise quadratic trigonometric polynomial curves, Math. Comput. 72 (243) (2003) 1369–1377. [12] X. Han, Cubic trigonometric polynomial curves with a shape parameter, Comput. Aid. Geom. Des. 21 (2004) 535–548. [13] X. Han, The trigonometric polynomial like bernstein polynomial, The Scientific World J. 2014 (2014) 17. Article ID 174716. [14] X. Han, Normalized b-basis of the space of trigonometric polynomials and curve design, Appl. Math. Comput. 251 (2015) 336–348. [15] J. He, On a linear combination of S.N. Bernstein trigonometric interpolation polynomial, Appl. Math. Comput. 106 (2-3) (1999) 197–203. [16] P. Henrici, Barycentric formulas for interpolating trigonometric polynomials and their conjugates, Numer. Math. 33 (1979) 225–234. [17] D.J. Johnson, The trigonometric Hermite-Birkhoff interpolation, Trans. Amer. Math. Soc. 212 (1975) 365–374. [18] R. Kress, On general Hermite trigonometric interpolation, Numer. Math. 20 (1972) 125–138. [19] P.E. Koch, T. Lyche, Bounds for the error in Trigonometric Hermite interpolaton, in: R.A. DeVore, K. Scherer (Eds.), Quantitative approximation, Academic Press, New York, 1980, pp. 185–196. [20] T. Lyche, R. Winther, A stable recurrence relation for trigonometric B-spline, J. Approx. Theory 25 (3) (1979) 266–279. [21] T. Lyche, A Newton form for trigonometric Hermite interpolation, BIT Numer. Math. 19 (2) (1979) 229–235. [22] T. Lyche, L.L. Schumaker, Quasi-interpolants based on trigonometric splines, J. Approx. Theory 95 (2) (1998) 280–309. [23] E. Quak, Trigonometric wavelets for Hermite interpolation, Math. Comput. 65 (214) (1996) 683–722. [24] L. Reichel, G. Ammar, W. Gragg, Discrete least squares approximation by trigonometric polynomials, Math. Comp. 57 (1991) 273–289. [25] S.D. Riemenschneider, Lacunary trigonometric interpolation, Convergence, in: E.W. Cheney (Ed.), Quantitative approximation, Academic Press, New York, 1980, pp. 741–746. [26] K.P. Sahakyan, Hermite trigonometric interpolation, East J. Approx. 12 (4) (2006) 441–449. [27] H.E. Salzer, New fomulas for trigonometric interpolation, J. Math. Phys. 39 (1960) 85–96. [28] J. Szabados, The exact error of trigonometric interpolation for differentiable functions, Construc. Approx. 8 (2) (1992) 203–210.