## 3 Integro Quintic Spline Interpolation with Five Artificial End Conditions
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline \(x\) & \(x_{i-2}\) & \(x_{i-1}\) & \(x_{i}\) & \(x_{i+1}\) & \(x_{i+2}\) \\ \hline \(B_{i}\) & \(\frac{1}{120}\) & \(\frac{26}{120}\) & \(\frac{66}{120}\) & \(\frac{26}{120}\) & \(\frac{1}{120}\) \\ \(B_{i}^{{}^{\prime}}\) & \(\frac{1}{24h}\) & \(\frac{10}{24h}\) & \(0\) & \(-\frac{10}{24h}\) & \(-\frac{1}{24h}\) \\ \(B_{i}^{{}^{\prime\prime}}\) & \(\frac{1}{6h^{2}}\) & \(\frac{2}{6h^{2}}\) & \(-\frac{6}{6h^{2}}\) & \(\frac{2}{6h^{2}}\) & \(\frac{1}{6h^{2}}\) \\ \(B_{i}^{(3)}\) & \(\frac{1}{2h^{2}}\) & \(-\frac{2}{2h^{2}}\) & \(0\) & \(\frac{2}{2h^{2}}\) & \(-\frac{1}{2h^{2}}\) \\ \(B_{i}^{(4)}\) & \(\frac{1}{h^{4}}\) & \(-\frac{4}{h^{4}}\) & \(\frac{6}{h^{4}}\) & \(-\frac{4}{h^{4}}\) & \(\frac{1}{h^{4}}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The values of \(B_{i}^{(k)},k=0,1,2,3,4\), at the knots lying in the interior of the support of \(B_{i}\). that is:  
\[\frac{1}{40}(c_{n-2}+10c_{n-1}+18c_{n}+10c_{n+1}+c_{n+2})=\theta_{n}. \tag{25}\]  
Finally, from \(s^{{}^{\prime}}(b)=\vec{y}_{n}^{{}^{\prime}}\), it follows that:  
\[-c_{n-2}-10c_{n-1}+10c_{n+1}+c_{n+2}=24h\vec{y}_{n}^{{}^{\prime}}.\]  
Therefore, we get the linear system:  
\[AC=Y, \tag{26}\]  
where  
\[A=\left( \begin{array}{ccccccccc}1&26&66&26&1&&&&\\ -1&-10&0&10&1&&&&\\ 1&2&-6&2&1&&&&\\ 1&57&302&302&57&1&&&\\ &1&57&302&302&57&1&&&\\ &&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&&\\ &&&1&57&302&302&57&1\\ &&&&1&57&302&302&57&1\\ &&&&1&10&18&10&1\\ &&&&-1&-10&0&10&1\end{array} \right)_{(n+5)\times(n+5)}, \tag{27}\]  
and:  
\[C=(c_{-2},c_{-1},c_{0},c_{1},\cdots,c_{n+1},c_{n+2})^{\text{T}},\]  
\[Y=(120\vec{y}_{0},24h\vec{y}_{0}^{{}^{\prime}},6h2\vec{y}_{0}^{{}^{\prime}}, \frac{720}{h}I_{0},\cdots,\frac{720}{h}I_{n-1},40\theta_{n},24h\vec{y}_{n}^{{} ^{\prime}})^{\text{T}}.\]  
**Theorem 1**. : _The coefficient matrix \(A\) (27) is invertible._  
Proof. : We will prove that the determinant of matrix \(A\) is nonzero. We will perform some proper elementary transformations to \(A\) in order to verify \(|A|\
eq 0\). Let \(C(i)\) denote the \(i\)-th column and \(R(i)\) denote the \(i\)-th row of a matrix obtained by an elementary row or column transformation. We first perform \(n+4\) elementary column transformations to \(A\). **Step 1:**: For \(i=n+4,n+3,\ldots,1\), \(C(i):=C(i)-C(i+1)\). Then, we get:  
\[A_{1}=\left( \begin{array}{ccccccccc}16&-15&41&25&1&&&&\\ 0&-1&-9&9&1&&&&\\ -8&9&-7&1&1&&&&\\ 0&1&56&246&56&1&&&\\ &0&1&56&246&56&1&&&\\ &&&\ddots&\ddots&\ddots&\ddots&\ddots&&\\ &&&1&56&246&56&1&\\ &&&&0&1&9&9&1\\ &&&&0&-1&-9&9&1\end{array} \right)_{(n+5)\times(n+5)}.\]  
We continue to perform the following elementary row transformations to \(A_{1}\). **Step 2:**: \(R(3):=\frac{2}{3}R(3)+\frac{1}{3}R(1)\);  
**Step 3:**: \(R(3):=R(3)+R(2)\), and \(R(n+4):=R(n+4)-R(n+5)\);
**Step 4:**: \(R(4):=R(4)+R(2)\), and \(R(n+3):=R(n+3)-R(n+5)\);
**Step 5:**: \(R(5):=R(5)-\frac{1}{47}R(4)\), and \(R(n+2):=R(n+2)-\frac{1}{47}R(n+3)\);
**Step 6:**: \(R(3)\rightleftharpoons R(4)\), and \(R(n+4)\rightleftharpoons R(n+3)\). Thus, we get:  
\[A_{2}=\left( \begin{array}{ccccccccc}16&-15&41&25&1&0&0&&&&\\ 0&-1&-9&9&1&0&0&&&&\\ 0&0&47&255&57&1&0&&&&\\ 0&0&0&18&2&0&0&&&&\\ &&\frac{2077}{4}&\frac{11505}{56}&\frac{2431}{246}&1&&&&\\ &&1&56&246&56&1&&&&\\ &&&&\ddots&\ddots&\ddots&\ddots&\ddots&&\\ &&&&1&56&246&56&1&&&&\\ &&&&1&56&246&56&1&&&&\\ &&&&1&1&231&\frac{11505}{2}&\frac{2077}{18}&0&0\\ &&&&0&1&57&255&47&0\\ &&&&0&0&-1&-9&9&1\end{array} \right).\]  
By the basic knowledge of linear algebra, we have:  
\[|A_{2}|=16\times(-1)\times 47\times|C_{nn}|\times 47\times 1,\]  
where \(C_{nn}\) is the central block matrix of \(A_{2}\). \(C_{nn}\) is strictly diagonally dominant, and so, \(|C_{nn}|\
eq 0\). It implies that \(|A_{2}|\
eq 0\) and, hence, \(|A|\
eq 0\). In other words, \(A\) (27) is invertible, and the theorem is proven. Theorem 1 guarantees the existence and uniqueness of the integro quintic spline \(s=\sum\limits_{i=-2}^{n+2}c_{i}B_{i}(x)\) determined by (17)-(19).