On a mixed interpolation with integral conditions at arbitrary nodes

Abstract

In this paper, we present a symbolic algorithm for a mixed interpolation of the form $$\begin{array}{c}\hfill acoskx+bsinkx+\sum _{i=0}^{s-2}{c}_i{x}^i,s\ge 2,\end{array}$$

where $k>0$ is a given parameter and the coefficients a, b, and $c_0,\dots ,c_{s-2}$ are determined by a given set of independent integral conditions at arbitrary nodes. Implementation of the proposed algorithm in Maple is described and sample computations are provided. This algorithm will help to implement the manual calculations in commercial packages such as Mathematica, Matlab, Singular, Scilab, etc.

Keywords:

• mixed interpolation
• integral conditions
• symbolic algorithm

AMS Subject Classifications:

• 41A05
• 65D05
• 47A57

Public Interest Statement

We often come across a number of data points (obtained by sampling or experimentation) which represent the values of a function for a limited number of values of the independent variable, and the need to estimate the value of the function at other point of the independent variable. This may be achieved by interpolation. The interpolation problem naturally arises in many applications of science and engineering, for example, the orbit problems, solving differential and integral equations, quantum mechanical problems, etc. where we consider the mixed interpolation instead of the polynomial interpolation.

1. Introduction

The interpolation problem naturally arises in many applications, for example, the orbit problems, quantum mechanical problems, etc. The general form of a mixed interpolation problem is as follows (Coleman, 1998; Lorentz, 2000; Sauer, 1997): suppose we have a normed linear space $\mathcal{S}$, a finite linearly independent set $\mathrm{\Theta }\subset \mathcal{S}$ of bounded functionals and an associated values $\mathrm{\Sigma }=\{{\mathit{\alpha }}_{\mathit{\theta }}:\mathit{\theta }\in \mathrm{\Theta }\}\subset \mathbb{R}$. Then the mixed interpolation problem is to find an approximation ${\tilde{f}}_s(x)\in \mathcal{S}$ of the form(1) $$\begin{array}{c}\hfill {\tilde{f}}_s(x)=acoskx+bsinkx+\sum _{i=0}^{s-2}{c}_i{x}^i,s\ge 2,\end{array}$$(1)

such that(2) $$\begin{array}{c}\hfill \mathrm{\Theta }({\tilde{f}}_s)=\mathrm{\Sigma },\text{i.e.}\mathit{\theta }{\tilde{f}}_s={\mathit{\alpha }}_{\mathit{\theta }},\mathit{\theta }\in \mathrm{\Theta }.\end{array}$$(2)

Here s is called the order of the interpolating function ${\tilde{f}}_s(x)$. One can observe that, the interpolation problem given in Equations (1), (2) may have many solutions if there is no restriction on the dimension of the space. But our interest is to find the single interpolating function that must match with a finite number of conditions. Hence for a unique solution of the problem, one must have finite dimensional subspace $\mathrm{\Theta }$ of $\mathcal{S}$ having dimension equal to the number of functionals.

The mixed interpolation problem, its formulation, and error estimation have been studied by several engineers and scientists with general nodes at uniformly spaced and arbitrary points on a chosen interval (see e.g. de Meyer, Vanthournout, & Vanden Berghe, 1990; de Meyer, Vanthournout, Vanden Berghe, & Vanderbauwhede, 1990; Chakrabarti & Hamsapriye, 1996; Coleman, 1998). In literature survey, we observe that there is no mixed interpolation algorithm available with integral conditions at arbitrary points on a chosen interval. Therefore, in this paper, we present a symbolic algorithm for the mixed interpolation with integral conditions using the algorithm presented by the authors in Thota and Kumar (2015). Indeed, we discuss a symbolic algorithm for mixed interpolation with a linearly independent set of the integral functionals/conditions at arbitrary nodes on a chosen interval. This is the first symbolic algorithm which deals with integral conditions. The rest of paper is organized as follows: Section 1.1 gives some definitions and basic concepts of the mixed interpolation, which are required to justify our proposed algorithm. Symbolic algorithm for the mixed interpolation with a finite linearly independent set of integral conditions is discussed in Section 2, the proposed algorithm for mixed interpolation is presented in Section 2.1 and some numerical examples are given in Section 2.2. Maple implementation of the proposed algorithm is presented in Section 3 with sample computations.

1.1. Preliminaries

In this section, we present some definitions and basic concepts of the mixed interpolation, which are required to justify our proposed algorithm.

Definition 1

A mixed interpolation problem is called regular for subspace $\mathcal{M}$ of linear space $\mathcal{S}$ with respect to $\mathrm{\Theta }$ if the interpolation problem has a unique solution for each choice of values of $\mathrm{\Sigma }\subset \mathbb{R}$ such that $\mathrm{\Theta }({\tilde{f}}_s)=\mathrm{\Sigma }$. Otherwise, the interpolation problem is called singular.

Definition 2

We call the triplet ($M,\mathrm{\Theta },\mathrm{\Sigma }$) an interpolation problem, where $M=\{coskx,sinkx,1,x,\dots ,x^{s-2}\}\subset \mathcal{M}$ a basis for a finite dimensional space $\mathcal{S}$, and $\mathrm{\Theta }\subset {\mathcal{S}}^{\ast }$ a finite linearly independent set of functionals with associated values $\mathrm{\Sigma }\subset \mathbb{R}$.

If $\mathrm{\Sigma }=\mathrm{\Theta }\mathit{\psi }$, for $\mathit{\psi }\in \mathcal{S}$, then the interpolation problem ($M,\mathrm{\Theta },\mathrm{\Sigma }$) can be stated in a different way equivalently: Let $\mathrm{\Omega }=\text{span}\{\mathit{\theta }:\mathit{\theta }\in \mathrm{\Theta }\}$. Then $\mathrm{\Omega }\subseteq {\mathcal{S}}^{\ast }$ and the interpolation problem is to find a ${\tilde{f}}_s(x)$ such that $\mathrm{\Omega }{\tilde{f}}_s=\mathrm{\Omega }\mathit{\psi }$ for given $\mathit{\psi }\in \mathcal{S}$. There is a connection between the regularity in terms of algebraic geometry and linear algebra as given in the following proposition.

Proposition 1

Let $M=\{m_0,\dots ,m_t\}$ be a basis for $\mathcal{M}$, a finite dimensional subspace of $\mathcal{S}$, and $\mathrm{\Theta }=\{{\mathit{\theta }}_0,\dots ,{\mathit{\theta }}_s\}$ be a finite linearly independent subset of ${\mathcal{S}}^{\ast }$. Then the following statements are equivalent:

(i)	The mixed interpolation problem is regular for $\mathcal{M}$ with respected to $\mathrm{\Theta }$.
(ii)	$t=s$, and the matrix, so-called evaluation matrix, (3) $$\begin{array}{c}\hfill \mathrm{\Theta }M=\left(\begin{array}{ccc}{\mathit{\theta }}_0{m}_0\hfill & \cdots \hfill & {\mathit{\theta }}_0{m}_t\hfill \\ ⋮\hfill & \ddots \hfill & ⋮\hfill \\ {\mathit{\theta }}_s{m}_0\hfill & \cdots \hfill & {\mathit{\theta }}_s{m}_t\hfill \end{array}\right)\in {\mathbb{R}}^{(s+1)\times (s+1)}\end{array}$$(3) is regular. Denote the evaluation matrix $\mathrm{\Theta }M$ by $\mathcal{E}$ for simplicity.
(iii)	$\mathcal{S}=M\oplus {\mathrm{\Theta }}^{\perp }$.

If we denote the integral condition by a symbol/operator ${\mathtt{A}}_x$ defined by ${\mathtt{A}}_x\bullet ={\int }_p^x\bullet dx$, i.e. ${\mathtt{A}}_{x}f(x)={\int }_p^{x}f(x)dx$, for a fixed $p\in \mathbb{R}$, then the set of integral conditions is(4) $$\begin{array}{c}\hfill \mathrm{\Theta }=\{{\mathtt{A}}_{x_0},\dots ,{\mathtt{A}}_{x_s}\},\end{array}$$(4)

where $x_0,\dots ,x_s$ are arbitrary nodes. Now, the symbolic representation of the mixed interpolation problem (1), (2) is to find a function of the form (1) such that(5) $$\begin{array}{c}\hfill {\mathtt{A}}_{x_i}{\tilde{f}}_s={\mathit{\alpha }}_{x_i},\text{where}{\mathtt{A}}_{x_i}\in \mathrm{\Theta }.\end{array}$$(5)

2. Symbolic algorithm for mixed interpolation

Consider the mixed interpolation problem defined in Section 1 for ($M,\mathrm{\Theta },\mathrm{\Sigma }$), where $M\subseteq \mathcal{M}\subset \mathcal{S}$, and $\mathrm{\Theta }=\{{\mathit{\theta }}_0,\dots ,{\mathit{\theta }}_s\}$ a finite set of integral conditions of the form (4). From Proposition (3), the mixed interpolation problem is regular with respect to linearly independent set $\mathrm{\Theta }$ if and only if there exists a finite linearly independent set M of $\mathcal{S}$ such that the evaluation matrix $\mathcal{E}$ in (3) is regular.

2.1. Proposed symbolic algorithm

The mixed interpolation problem ($M,\mathrm{\Theta },\mathrm{\Sigma }$), i.e. ${\tilde{f}}_s(x)=acoskx+bsinkx+\sum _{i=0}^{s-2}{c}_i{x}^i$ such that it satisfy $\mathrm{\Theta }{\tilde{f}}_s=\mathrm{\Sigma }$, can be expressed as a linear system(6) $$\begin{array}{c}\hfill \mathcal{E}u=\mathit{\sigma },\end{array}$$(6)

where $u={(a,b,c_0,\dots ,c_{s-2})}^T$, $\mathit{\sigma }={({\mathit{\alpha }}_{{\mathit{\theta }}_0}\dots ,{\mathit{\alpha }}_{{\mathit{\theta }}_s})}^T$ and $\mathcal{E}$ is the evaluation matrix of $\mathrm{\Theta }$ and M given by(7) $$\begin{array}{cc}\hfill \mathcal{E}& =\left(\begin{array}{cccccc}\frac{sink{x}_0}{k}\hfill & \frac{-cosk{x}_0}{k}\hfill & x_0\hfill & \frac{x_0^2}{2}\hfill & \cdots \hfill & \frac{x_0^{s-1}}{s-1}\hfill \\ \frac{sink{x}_1}{k}\hfill & \frac{-cosk{x}_1}{k}\hfill & x_1\hfill & \frac{x_1^2}{2}\hfill & \cdots \hfill & \frac{x_1^{s-1}}{s-1}\hfill \\ ⋮\hfill & ⋮\hfill & ⋮\hfill & ⋮\hfill & \ddots \hfill & ⋮\hfill \\ \frac{sink{x}_s}{k}\hfill & \frac{-cosk{x}_s}{k}\hfill & x_s\hfill & \frac{x_s^2}{2}\hfill & \cdots \hfill & \frac{x_s^{s-1}}{s-1}\hfill \end{array}\right)-\left(\begin{array}{cccccc}\frac{sinkp}{k}\hfill & \frac{-coskp}{k}\hfill & p\hfill & \frac{p^2}{2}\hfill & \cdots \hfill & \frac{p^{s-1}}{s-1}\hfill \\ \frac{sinkp}{k}\hfill & \frac{-coskp}{k}\hfill & p\hfill & \frac{p^2}{2}\hfill & \cdots \hfill & \frac{p^{s-1}}{s-1}\hfill \\ ⋮\hfill & ⋮\hfill & ⋮\hfill & ⋮\hfill & \ddots \hfill & ⋮\hfill \\ \frac{sinkp}{k}\hfill & \frac{-coskp}{k}\hfill & p\hfill & \frac{p^2}{2}\hfill & \cdots \hfill & \frac{p^{s-1}}{s-1}\hfill \end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cccccc}\frac{sink{x}_0}{k}-\frac{sinkp}{k}\hfill & \frac{coskp}{k}-\frac{cosk{x}_0}{k}\hfill & x_0-p\hfill & \frac{x_0^2}{2}-\frac{p^2}{2}\hfill & \cdots \hfill & \frac{x_0^{s-1}}{s-1}-\frac{p^{s-1}}{s-1}\hfill \\ \frac{sink{x}_1}{k}-\frac{sinkp}{k}\hfill & \frac{coskp}{k}-\frac{cosk{x}_1}{k}\hfill & x_1-p\hfill & \frac{x_1^2}{2}-\frac{p^2}{2}\hfill & \cdots \hfill & \frac{x_1^{s-1}}{s-1}-\frac{p^{s-1}}{s-1}\hfill \\ ⋮\hfill & ⋮\hfill & ⋮\hfill & ⋮\hfill & \ddots \hfill & ⋮\hfill \\ \frac{sink{x}_s}{k}-\frac{sinkp}{k}\hfill & \frac{coskp}{k}-\frac{cosk{x}_s}{k}\hfill & x_s-p\hfill & \frac{x_s^2}{2}-\frac{p^2}{2}\hfill & \cdots \hfill & \frac{x_s^{s-1}}{s-1}-\frac{p^{s-1}}{s-1}\hfill \end{array}\right).\hfill \end{array}$$(7)

Remark

If $p=0$, then $\mathcal{E}$ in Equation (7) is given by$$\begin{array}{cc}\hfill \mathcal{E}& =\left(\begin{array}{cccccc}\frac{sink{x}_0}{k}\hfill & \frac{1-cosk{x}_0}{k}\hfill & x_0\hfill & \frac{x_0^2}{2}\hfill & \cdots \hfill & \frac{x_0^{s-1}}{s-1}\hfill \\ \frac{sink{x}_1}{k}\hfill & \frac{1-cosk{x}_1}{k}\hfill & x_1\hfill & \frac{x_1^2}{2}\hfill & \cdots \hfill & \frac{x_1^{s-1}}{s-1}\hfill \\ ⋮\hfill & ⋮\hfill & ⋮\hfill & ⋮\hfill & \ddots \hfill & ⋮\hfill \\ \frac{sink{x}_s}{k}\hfill & \frac{1-cosk{x}_s}{k}\hfill & x_s\hfill & \frac{x_s^2}{2}\hfill & \cdots \hfill & \frac{x_s^{s-1}}{s-1}\hfill \end{array}\right)\hfill \end{array}$$

Uniqueness of the solution is possible if and only if the evaluation matrix (7) is regular (non-singular). The simple form of the determinant of $\mathcal{E}$ is given by(8) $$\begin{array}{c}\hfill det\mathcal{E}=\frac{{\prod }_{i=0}^s{x}_i}{k^2(s-1)!}\prod _{k<j}(x_j-x_k)\left|\begin{array}{cc}{\sum }_{i=0}^{s-1}\frac{\frac{cosk{x}_i}{x_i}}{{\prod }_{j=0,j\ne i}^{s-1}(x_i-x_j)}\hfill & {\sum }_{i=0}^{s-1}\frac{\frac{sink{x}_i}{x_i}}{{\prod }_{j=0,j\ne i}^{s-1}(x_i-x_j)}\hfill \\ {\sum }_{i=0}^s\frac{\frac{cosk{x}_i}{x_i}}{{\prod }_{j=0,j\ne i}^s(x_i-x_j)}\hfill & {\sum }_{i=0}^s\frac{\frac{sink{x}_i}{x_i}}{{\prod }_{j=0,j\ne i}^s(x_i-x_j)}\hfill \end{array}\right|.\end{array}$$(8)

This simple form is obtained by performing the following steps:

I.	Dividing i-th row by $x_i$ in the determinant of $\mathcal{E}$ in Equation (7), we get (9) $$\begin{array}{c}\hfill det\mathcal{E}=\frac{{\prod }_{i=0}^s{x}_i}{k^2(s-1)!}\left|\begin{array}{cccccc}\frac{cosk{x}_0}{x_0}& \frac{sink{x}_0}{x_0}& 1& x_0& \dots & x_0^{s-2}\\ \frac{cosk{x}_1}{x_1}& \frac{sink{x}_1}{x_1}& 1& x_1& \dots & x_1^{s-2}\\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ \frac{cosk{x}_s}{x_s}& \frac{sink{x}_s}{x_s}& 1& x_s& \dots & x_s^{s-2}\end{array}\right|.\end{array}$$(9)
II.	Subtract first row from the other $(i+1)$-th row, for $i=1,2,\dots s$, and divide $(i+1)$-th row by $x_i-x_0$ for $i=1,2,\dots s$, we get $$\begin{array}{c}\hfill det\mathcal{E}=\frac{{\prod }_{i=0}^s{x}_i}{k^2(s-1)!}\prod _{k=1}^s(x_k-x_0)\left|\begin{array}{cccccc}\frac{cosk{x}_0}{x_0}& \frac{sink{x}_0}{x_0}& 1& x_0& \dots & x_0^{s-2}\\ \frac{\frac{cosk{x}_1}{x_1}-\frac{cosk{x}_0}{x_0}}{x_1-x_0}& \frac{\frac{sink{x}_1}{x_1}-\frac{sink{x}_0}{x_0}}{x_1-x_0}& 0& 1& \dots & x_1^{s-3}\\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ \frac{\frac{cosk{x}_s}{x_s}-\frac{cosk{x}_0}{x_0}}{x_s-x_0}& \frac{\frac{sink{x}_s}{x_s}-\frac{sink{x}_0}{x_0}}{x_s-x_0}& 0& 1& \dots & x_s^{s-3}\end{array}\right|.\end{array}$$ This reduces to a determinant of a matrix of order s similar to (9).
III.	Repeating the step II finite number of times, we arrive at the simple form of $det\mathcal{E}$ as in (8).

From the procedure given for simplification of the determinant of $\mathcal{E}$, we can construct the interpolating function ${\tilde{f}}_s(x)$ for ($M,\mathrm{\Theta },\mathrm{\Sigma }$) in terms of evaluation matrix. The following theorem presents an algorithm to construct ${\tilde{f}}_s(x)$. Denote $\mathcal{D}=det(\mathcal{E})$ for simplicity.

Theorem 1

Let $\mathrm{\Theta }$ be a finite set of integral conditions of the form $\mathrm{\Theta }=\{{\mathtt{A}}_{x_0},\dots ,{\mathtt{A}}_{x_s}\}$ with associated values $\mathrm{\Sigma }$ and $M=\{coskx,sinkx,1,\dots ,x^{s-2}\}\subset \mathcal{S}$ be a finite linearly independent set such that the evaluation matrix $\mathcal{E}$ is regular. Then there exists unique interpolating function ${\tilde{f}}_s(x)$ of the form (1), such that $\mathrm{\Theta }{\tilde{f}}_s=\mathrm{\Sigma }$ as(10) $$\begin{array}{c}\hfill {\tilde{f}}_s(x)=\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^1{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}coskx+\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^2{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}sinkx+\sum _{l=0}^{s-2}\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^{l+3}{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}{x}^l,\end{array}$$(10)

where ${\mathcal{D}}_i^j$ is the determinant of ${\mathcal{E}}_i^j$ obtained from $\mathcal{E}$ by replacing j-th column by the i-th unit vector.

Proof

It is given that the evaluation matrix associated with $\mathrm{\Theta }$ and M is regular, therefore there exists unique mixed interpolation. Suppose ${\mathcal{D}}_k^1,{\mathcal{D}}_k^2$ and ${\mathcal{D}}_k^{l+3}$ denote the determinants of the resultant matrix $\mathcal{E}$ after replacing $1^{\mathit{st}}$, $2^{\mathit{nd}}$, and lth columns by k-unit vector, respectively, for $l=0,1,\dots ,s-2$, then the coefficients $a,b,c_0,\dots ,c_{s-2}$ are determined uniquely using the Cramer’s rule, as follows(11) $$\begin{array}{cc}\hfill a& =\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^1{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}},\hfill \\ \hfill b& =\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^2{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}},\hfill \\ \hfill c_l& =\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^{l+3}{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}},l=0,1,\dots ,s-2.\hfill \end{array}$$(11)

Now, the required interpolating function ${\tilde{f}}_s(x)$ is the linear combination of elements of M with the coefficients $a,b,c_0,\dots ,c_{s-2}$. Hence, we have(12) $$\begin{array}{c}\hfill {\tilde{f}}_s(x)=\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^1{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}coskx+\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^2{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}sinkx+\sum _{l=0}^{s-2}\sum _{k=1}^{s+1}{\mathcal{D}}^{-1}{\mathcal{D}}_k^{l+3}{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}{x}^l,\end{array}$$(12)

as stated. $\square$

In general, it is very difficult to solve explicitly the linear system (6) for the coefficients $a,b,c_0,\dots ,c_{s-2}$ in terms of $\mathrm{\Sigma }$ at the interpolation points. However, we can express the coefficients of ${\tilde{f}}_s(x)$ in terms of evaluation matrix as given Theorem 4.

2.2. Examples

Now to verify the proposed algorithm in Theorem 4, we present some examples. We use Maple, the computer algebra software, for numerical computations in the following examples.

Example 2.1

Consider the integral conditions $\underset{0}{\overset{0.1}{\int }}f(x)dx=k_1$, $\underset{0}{\overset{0.3}{\int }}f(x)dx=k_2$, $\underset{0}{\overset{0.5}{\int }}f(x)dx=k_3$, $\underset{0}{\overset{0.8}{\int }}f(x)dx=k_4$ and $\underset{0}{\overset{1.0}{\int }}f(x)dx=k_5$, where $k_i$ is constant, for $i=1,2,3,4,5$. Now we construct ${\tilde{f}}_4(x)=acoskx+bsinkx+c_0+c_{1}x+c_2{x}^2$ such that ${\tilde{f}}_4(x)$ satisfies the given conditions. For simplicity, take $k=1$. Here, $\mathrm{\Theta }=\{{\mathtt{A}}_{0.1},{\mathtt{A}}_{0.3},{\mathtt{A}}_{0.5},{\mathtt{A}}_{0.8},{\mathtt{A}}_{1.0}\}$, $M=\{cosx,sinx,1,x,x^2\}$, and $\mathrm{\Sigma }=\{k_1,k_2,k_3,k_4,k_5\}$. Following Theorem 4, we have the evaluation matrix$$\begin{array}{cc}\hfill \mathcal{E}& =\left(\begin{array}{ccccc}0.099833& 0.004996& 0.1& 0.005& 0.000333\\ 0.295520& 0.044664& 0.3& 0.045& 0.009000\\ 0.479426& 0.122417& 0.5& 0.125& 0.041667\\ 0.717356& 0.303293& 0.8& 0.320& 0.170667\\ 0.841471& 0.459698& 1.0& 0.500& 0.333333\end{array}\right),\hfill \\ \hfill \mathcal{D}& =det\mathcal{E}=7.19735\times {10}^{-11};{\mathcal{D}}^{-1}=1.3894\times {10}^{10},\hfill \\ \hfill {\mathcal{D}}_1^1& =det\left(\begin{array}{ccccc}1\hfill & 0.004996\hfill & 0.1\hfill & 0.005\hfill & 0.000333\hfill \\ 0\hfill & 0.044664\hfill & 0.3\hfill & 0.045\hfill & 0.009000\hfill \\ 0\hfill & 0.122417\hfill & 0.5\hfill & 0.125\hfill & 0.041667\hfill \\ 0\hfill & 0.303293\hfill & 0.8\hfill & 0.320\hfill & 0.170667\hfill \\ 0\hfill & 0.459698\hfill & 1.0\hfill & 0.500\hfill & 0.333333\hfill \end{array}\right)=0.15028\times {10}^{-5}\hfill \\ \hfill {\mathcal{D}}_2^1& =det\left(\begin{array}{ccccc}0.099833\hfill & 0\hfill & 0.1\hfill & 0.005\hfill & 0.000333\hfill \\ 0.295520\hfill & 1\hfill & 0.3\hfill & 0.045\hfill & 0.009000\hfill \\ 0.479426\hfill & 0\hfill & 0.5\hfill & 0.125\hfill & 0.041667\hfill \\ 0.717356\hfill & 0\hfill & 0.8\hfill & 0.320\hfill & 0.170667\hfill \\ 0.841471\hfill & 0\hfill & 1.0\hfill & 0.500\hfill & 0.333333\hfill \end{array}\right)=-0.18396\times {10}^{-5}\hfill \\ \hfill {\mathcal{D}}_3^1& =det\left(\begin{array}{ccccc}0.099833\hfill & 0.004996\hfill & 0\hfill & 0.005\hfill & 0.000333\hfill \\ 0.295520\hfill & 0.044664\hfill & 0\hfill & 0.045\hfill & 0.009000\hfill \\ 0.479426\hfill & 0.122417\hfill & 1\hfill & 0.125\hfill & 0.041667\hfill \\ 0.717356\hfill & 0.303293\hfill & 0\hfill & 0.320\hfill & 0.170667\hfill \\ 0.841471\hfill & 0.459698\hfill & 0\hfill & 0.500\hfill & 0.333333\hfill \end{array}\right)=0.13131\times {10}^{-5}\hfill \\ \hfill {\mathcal{D}}_4^1& =det\left(\begin{array}{ccccc}0.099833\hfill & 0.004996\hfill & 0.1\hfill & 0\hfill & 0.000333\hfill \\ 0.295520\hfill & 0.044664\hfill & 0.3\hfill & 0\hfill & 0.009000\hfill \\ 0.479426\hfill & 0.122417\hfill & 0.5\hfill & 0\hfill & 0.041667\hfill \\ 0.717356\hfill & 0.303293\hfill & 0.8\hfill & 1\hfill & 0.170667\hfill \\ 0.841471\hfill & 0.459698\hfill & 1.0\hfill & 0\hfill & 0.333333\hfill \end{array}\right)=-4.82523\times {10}^{-7}\hfill \\ \hfill {\mathcal{D}}_5^1& =det\left(\begin{array}{ccccc}0.099833\hfill & 0.004996\hfill & 0.1\hfill & 0.005\hfill & 0\hfill \\ 0.295520\hfill & 0.044664\hfill & 0.3\hfill & 0.045\hfill & 0\hfill \\ 0.479426\hfill & 0.122417\hfill & 0.5\hfill & 0.125\hfill & 0\hfill \\ 0.717356\hfill & 0.303293\hfill & 0.8\hfill & 0.320\hfill & 0\hfill \\ 0.841471\hfill & 0.459698\hfill & 1.0\hfill & 0.500\hfill & 1\hfill \end{array}\right)=1.31086\times {10}^{-7},\hfill \\ \hfill \end{array}$$

hence,$$\begin{array}{cc}\hfill \sum _{k=1}^5{\mathcal{D}}_k^1{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}& =0.15028\times {10}^{-5}{k}_1-0.18396\times {10}^{-5}{k}_2+0.13131\times {10}^{-5}{k}_3\hfill \\ \hfill & -4.82523\times {10}^{-7}{k}_4+1.31086\times {10}^{-7}{k}_5,\hfill \end{array}$$

similarly,$$\begin{array}{cc}\hfill \sum _{k=1}^5{\mathcal{D}}_k^2{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}& =8.60525\times {10}^{-7}{k}_1-9.57686\times {10}^{-7}{k}_2+6.17919\times {10}^{-7}{k}_3\hfill \\ \hfill & -1.92581\times {10}^{-7}{k}_4+4.63585\times {10}^{-8}{k}_5,\hfill \\ \hfill \sum _{k=1}^5{\mathcal{D}}_k^3{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}& =-0.15011\times {10}^{-5}{k}_1+0.18389\times {10}^{-5}{k}_2-0.13128\times {10}^{-5}{k}_3\hfill \\ \hfill & +4.82459\times {10}^{-7}{k}_4-1.31072\times {10}^{-7}{k}_5,\hfill \\ \hfill \sum _{k=1}^5{\mathcal{D}}_k^4{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}& =-8.86346\times {10}^{-7}{k}_1+9.77076\times {10}^{-7}{k}_2-6.26828\times {10}^{-7}{k}_3\hfill \\ \hfill & +1.94676\times {10}^{-7}{k}_4-4.68155\times {10}^{-8}{k}_5\hfill \\ \hfill \sum _{k=1}^5{\mathcal{D}}_k^5{\mathit{\alpha }}_{{\mathit{\theta }}_{k-1}}& =8.52374\times {10}^{-7}{k}_1-0.10177\times {10}^{-5}{k}_2+7.11690\times {10}^{-7}{k}_3\hfill \\ \hfill & -2.55717\times {10}^{-7}{k}_4+6.88081\times {10}^{-8}{k}_5,\hfill \end{array}$$

and the coefficients are given by$$\begin{array}{cc}& a=20904.99k_1-25589.48k_2+18265.09k_3-6712.06k_4+1823.45k_5\hfill \\ \hfill & b=11970.22k_1-13321.75k_2+8595.47k_3-2678.87k_{4}644.86k_5,\hfill \\ \hfill & c_0=-20881.23k_1+25580.00k_2-18261.10k_3+6711.16k_4-1823.26k_5,\hfill \\ \hfill & c_1=-12329.34k_1+13591.47k_2-8719.40k_3+2708.02k_4-651.22k_5,\hfill \\ \hfill & c_2=11856.84k_1-14156.85k_2+9899.85k_3-3557.12k_4+957.14k_5.\hfill \end{array}$$

Now the solution of the mixed interpolation $(M,\mathrm{\Theta },\mathrm{\Sigma })$ is$$\begin{array}{cc}\hfill {\tilde{f}}_4(x)& =(20904.99k_1-25589.48k_2+18265.09k_3-6712.06k_4+1823.45k_5)cosx\hfill \\ \hfill & +(11970.22k_1-13321.75k_2+8595.47k_3-2678.87k_{4}644.86k_5)sinx\hfill \\ \hfill & -20881.23k_1+25580.00k_2-18261.10k_3+6711.16k_4-1823.26k_5\hfill \\ \hfill & +(-12329.34k_1+13591.47k_2-8719.40k_3+2708.02k_4-651.22k_5)x\hfill \\ \hfill & +(11856.84k_1-14156.85k_2+9899.85k_3-3557.12k_4+957.14k_5)x^2\hfill \end{array}$$

In particular, if we choose $k_i=i$, for $i=1,2,3,4,5$, then$$\begin{array}{cc}\hfill a=& 6790.30,b=3621.97,c_0=-6776.17,c_1=-3728.64,c_2=3799.92\hfill \end{array}$$

and the solution of the interpolation $(M,\mathrm{\Theta },\mathrm{\Sigma })$ is$$\begin{array}{cc}\hfill {\tilde{f}}_4(x)& =6790.30cosx+3621.97sinx-6776.17-3728.64x+3799.92x^2.\hfill \end{array}$$

One can easily check in both the cases that $\mathrm{\Theta }({\tilde{f}}_4)=\mathrm{\Sigma }$.

Example 2.2

Suppose we have integral conditions $\underset{0}{\overset{0.1}{\int }}f(x)dx=1$, $\underset{0}{\overset{0.2}{\int }}f(x)dx=3$, $\underset{0}{\overset{0.3}{\int }}f(x)dx=4$, $\underset{0}{\overset{0.4}{\int }}f(x)dx=5$, $\underset{0}{\overset{0.5}{\int }}f(x)dx=6$, $\underset{0}{\overset{0.6}{\int }}f(x)dx=7$, $\underset{0}{\overset{0.7}{\int }}f(x)dx=9$, $\underset{0}{\overset{0.8}{\int }}f(x)dx=13$, $\underset{0}{\overset{0.9}{\int }}f(x)dx=15$ and $\underset{0}{\overset{1.0}{\int }}f(x)dx=16$. Now we construct ${\tilde{f}}_9(x)=acoskx+bsinkx+c_0+c_{1}x+c_2{x}^2+\cdots +c_7{x}^7$ such that ${\tilde{f}}_9(x)$ satisfies the given conditions. For simplicity, take $k=0.5$. In symbolic notations, we have $\mathrm{\Theta }=\{{\mathtt{A}}_{0.1},{\mathtt{A}}_{0.2},{\mathtt{A}}_{0.3},{\mathtt{A}}_{0.4},{\mathtt{A}}_{0.5},{\mathtt{A}}_{0.6},{\mathtt{A}}_{0.7},{\mathtt{A}}_{0.8},{\mathtt{A}}_{0.9},{\mathtt{A}}_{1.0}\}$, $M=\{cos(0.5x),sin(0.5x),1,x,\dots ,x^7\}$ and $\mathrm{\Sigma }=\{1,3,4,5,6,7,9,13,15,16\}$. From Theorem 4, the coefficients are computed similar to Example 2.1 as follows$$\begin{array}{cc}& a=-1.496313140\times {10}^9,b=3.349193220\times {10}^{10},\hfill \\ \hfill & c_0=1.496313072\times {10}^9,c_1=-1.674596380\times {10}^{10},\hfill \\ \hfill & c_2=-1.87067734\times {10}^8,c_3=6.978924923\times {10}^8,\hfill \\ \hfill & c_4=3.516587822\times {10}^6,c_5=-8.177983600\times {10}^6,\hfill \\ \hfill & c_6=-4.298813138\times {10}^5,c_7=1.677402086\times {10}^5.\hfill \end{array}$$

Now, the interpolating function ${\tilde{f}}_9(x)$ is given by$$\begin{array}{cc}\hfill {\tilde{f}}_9(x)& =-1.496313140\times {10}^{9}cos(0.5x)+3.349193220\times {10}^{10}sin(0.5x)\hfill \\ \hfill & +1.496313072\times {10}^9-1.674596380\times {10}^{10}x-1.87067734\times {10}^8{x}^2\hfill \\ \hfill & +6.978924923\times {10}^8{x}^3+3.516587822\times {10}^6{x}^4-8.177983600\times {10}^6{x}^5\hfill \\ \hfill & -4.298813138\times {10}^5{x}^6+1.677402086\times {10}^5{x}^7.\hfill \end{array}$$

If we choose $k=2$, then$$\begin{array}{cc}\hfill {\tilde{f}}_9(x)& =-3.85899899\times {10}^{8}cos(2x)+2.91356330\times {10}^{8}sin(2x)\hfill \\ \hfill & +3.858998408\times {10}^8-5.82709230\times {10}^{8}x-7.71845620\times {10}^8{x}^2\hfill \\ \hfill & +3.88781409\times {10}^8{x}^3+2.560552184\times {10}^8{x}^4-7.4670907\times {10}^7{x}^5\hfill \\ \hfill & -3.9114135\times {10}^7{x}^6+1.208298237\times {10}^7{x}^7.\hfill \end{array}$$

One can easily verify in both cases that $\mathrm{\Theta }({\tilde{f}}_9)=\mathrm{\Sigma }$.

The following section presents the implementation of the proposed algorithm in Maple.

3. Maple implementation

Maple implementation of the proposed algorithm is presented by creating different data types using the Maple package IntDiffOp implemented by Korporal, Regensburger, and Rosenkranz (2010). For displaying the operators, we have A for integral operator and E for evaluation operator as defined in IntDiffOp package, i.e. ${\mathtt{A}}_x=\mathtt{E}[x].\mathtt{A}$.

The data type IntegralCondition(np) is created to represent the integral condition, where np is the node point.

with(IntDiffOp):

IntegralCondition := proc(np)

return BOUNDOP(EVOP(np,EVDIFFOP(0),EVINTOP(EVINTTERM(1,1))));

end proc:

The following producer EvaluationMatrix(IC) gives the evaluation matrix of the given M and $\mathrm{\Theta }$, where IC is the column matrix of the integral conditions.

EvaluationMatrix := proc (IC::Matrix)

local r,c,elts,fs;

r,c:=LinearAlgebra[Dimension](IC);

fs:=Matrix(1,r,[cos(k*x),sin(k*x),seq(x${}^{\wedge }$(i-1),i=1..r-2)]);

elts:=seq(seq(ApplyOperator(IC[t,1],fs[1,j]),j=1..r), t=1..r);

return Matrix(r,r,[elts]);

end proc:

The procedure MixedInterpolation(IC,CM) is created to find the mixed interpolating function for ($M,\mathrm{\Theta },\mathrm{\Sigma }$), where CM is column matrix of the associated values of $\mathrm{\Sigma }$.

MixedInterpolation := proc (IC::Matrix, CM::Matrix)

local r,c,fs,evm,invevm,approx;

r,c:=LinearAlgebra[Dimension](IC);

fs: Matrix(1,r,[cos(k*x),sin(k*x),seq(x${}^{\wedge }$(i-1),i=1..r-2)]);

evm:=EvaluationMat(IC);

invevm:=1/evm; approx:=fs.invevm.CM;

return simplify(approx[1,1]);

end proc:

Example 3.1

Recall Example 2.1 for sample computations using Maple implementation.

> with(IntDiffOp):

> C1:=IntegralCondition(0.1); c1:=1;

C2:=IntegralCondition(0.3); c2:=2;

C3:=IntegralCondition(0.5); c3:=3;

C4:=IntegralCondition(0.8); c4:=4;

C5:=IntegralCondition(1.0); c5:=5; $$\begin{array}{cc}\hfill C1:=& \mathtt{E}[.1].\mathtt{A}\hfill \\ \hfill c1:=& 1\hfill \\ \hfill C2:=& \mathtt{E}[.3].\mathtt{A}\hfill \\ \hfill c2:=& 2\hfill \\ \hfill C3:=& \mathtt{E}[.5].\mathtt{A}\hfill \\ \hfill c3:=& 3\hfill \\ \hfill C4:=& \mathtt{E}[.8].\mathtt{A}\hfill \\ \hfill c4:=& 4\hfill \\ \hfill C5:=& \mathtt{E}[1.0].\mathtt{A}\hfill \\ \hfill c4:=& 5\hfill \end{array}$$> k:=1; $$\begin{array}{c}\hfill 1\end{array}$$> C:=Matrix([[C1],[C2],[C3],[C4],[C5]]); $$\left[\begin{array}{c}\mathtt{E}[.1].\mathtt{A}\\ \mathtt{E}[.3].\mathtt{A}\\ \mathtt{E}[.5].\mathtt{A}\\ \mathtt{E}[.8].\mathtt{A}\\ \mathtt{E}[1.0].\mathtt{A}\end{array}\right]$$CM:=Matrix([[c1],[c2],[c3],[c4],[c5]]); $$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\end{array}\right]$$EvaluationMat(C); $$\left[\begin{array}{ccccc}0.09983341665& 0.004995834722& 0.1& 0.005& 0.0003333\\ 0.2955202067& 0.04466351087& 0.3& 0.045& 0.009000\\ 0.4794255386& 0.1224174381& 0.5& 0.125& 0.041667\\ 0.7173560909& 0.3032932907& 0.8& 0.320& 0.170667\\ 0.8414709848& 0.4596976941& 1.0& 0.500& 0.333333\end{array}\right]$$> MixedInterpolation(C,CM); $$\begin{array}{cc}& -6776.165565-3728.640530\ast x+6790.295165\ast cos(x)\hfill \\ \hfill & +3621.967910\ast sin(x)+3799.917488\ast x^2\hfill \end{array}$$

Acknowledgements

Authors thank the referees for useful suggestions in improving the presentation of the paper.

Additional information

Funding

The authors received no direct funding for this research.

Notes on contributors

Srinivasarao Thota

Srinivasarao Thota completed his MSc in Mathematics from Indian Institute of Technology Madras, India. Now he is pursuing his PhD in Mathematics from Motilal Nehru National institute of Technology Allahabad, India. Srinivasarao Thota’s area of research interest is Computer Algebra, precisely symbolic methods for ordinary differential equations. He attended and presented research papers in several national and international conferences.