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<title>Second derivative General Linear Method in Nordsieck form: Second derivative General Linear Method in Nordsieck form</title>
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<h1>Second derivative General Linear Method in Nordsieck form</h1>
<p class="authors">
<span class="author">Robert I. Okuonghae\(^\ast \) Monday Ndidi Oziegbe Ikhile\(^\ast \)</span>
</p>
<p class="date">November 24, 2017. Accepted: May 16, 2019. Published online: October 10, 2019.</p>
</div>
<p>\(^\ast \)Department of Mathematics, University of Benin, P.M.B 1154, Benin City, Nigeria, <br />e-mail: <span class="tt">okunoghae01@yahoo.co.uk</span>, <span class="tt">mnoikhilo@yahoo.com</span>. </p>

<div class="abstract"><p> This paper considers the construction of <i class="it">second derivative general linear methods</i> (SD-GLM) from hybrid LMM and their transformation to Nordsieck GLM. We show how the Runge-Kutta starters for the methods can be derived. The representation of the methods in Nordsieck form has the advantage of easy implementation in variable stepsize. </p>
<p><b class="bf">MSC.</b> 65L05, 65L06 </p>
<p><b class="bf">Keywords.</b> <i class="it">General linear methods</i> (GLM), Nordsieck vectors/methods, second derivative GLM, RK-starters, variable stepsize. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> The representation of second derivative hybrid linear multistep methods (SD-HLMM) as SD-GLM and transformation to its Nordsieck form for easy implementation will be the main focus herein. Nordsieck method was introduced by Arnold Nordsieck in 1962 as an easy means of changing step size in LMM <span class="cite">
	[
	<a href="#Ar1" >1</a>
	]
</span>. Variable step-size implementation is readily achieved for formulas of Nordsieck kinds and in fact, LMM can be represented by Nordsieck formulas, details can be found in <span class="cite">
	[
	<a href="#Ar1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#Ha1" >10</a>
	]
</span>, <span class="cite">
	[
	<a href="#Hu1" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#Ez1" >18</a>
	]
</span> and <span class="cite">
	[
	<a href="#Mo1" >19</a>
	]
</span>. Hybrid LMM which can be with one or more off-step points are aimed at bypassing the Dahlquist <span class="cite">
	[
	<a href="#Da" >7</a>
	]
</span> order barrier for the conventional LMM in computing the numerical solution of the autonomous initial value problems (IVPs), </p>
<div class="equation" id="equation1">
<p>
  <div class="equation_content">
    \begin{equation} \label{equation1} \left\{ \begin{array}{ll} y’= f(y(x)),\quad \textrm{$x\in [x_{0},X]$}& \\ y(x_{0})=y_{0},\quad y\in \Re ^{m},\quad f: \Re \times \Re ^{m}, m\geq 1,& \end{array}\right. \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p>Examples of hybrid LMM are in <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span> <span class="cite">
	[
	<a href="#Ok2" >13</a>
	]
</span> and <span class="cite">
	[
	<a href="#Ok7" >15</a>
	]
</span>. Indeed, it is somewhat difficult to implement the hybrid LMM in variable step-size because of the problems associated with attempt to put the input and output data in their standard representations at each step of approximations of the solution of <a href="#equation1" class="eqref">1</a>. The problems associated with the direct implementation of the hybrid LMM make it less attractive compared to the Nordsieck GLM <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>,<span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span>. Indeed, over the years there is now a greater sophistication in the understanding of the GLM, </p>
<div class="displaymath" id="a0000000003">
  \[ \begin{array}{ll} Y=hAF(Y)+Uy^{[n-1]},& \\ y^{[n]}=hBF(Y)+Vy^{[n-1]}, \end{array} \leqno {(2a)} \]
</div>
<p> where \(n=1, 2,\ldots \); \(h=\left( x_{n+1}-x_{n}\right)\) is the stepsize. The GLM (2a) is readily represented also as, </p>
<div class="displaymath" id="a0000000004">
  \[ \left( \begin{array}{c} Y \\ \hline y^{[n]} \\ \end{array} \right)= \left( \begin{array}{c|cc} A& U \\ \hline B& V \\ \end{array} \right)\left( \begin{array}{c} hF(Y) \\ \hline y^{[n-1]} \\ \end{array} \right)\leqno {(2b)} \]
</div>
<p> where, \(k\) is the number of input quantities, \(s\) is the number of stages, \(Y=(Y_{1}, Y_{2},. . .,Y_{s})^{T}\) denote the stage values computed in step \(n\), then \(y^{[n-1]}=(y_{1}^{[n-1]}, y_{2}^{[n-1]}\cdots ,y_{k}^{[n-1]})^{T}\) represent the vector of input quantities at the start of step \(n\), while \(y^{[n]}=(y_{1}^{[n]}, y_{2}^{[n]}\cdots ,y_{k}^{[n]})^{T}\) is the output solution and denotes the vector quantities exported to the next step, \(n+1\). The \(F(Y)=(f(Y_{1}), f(Y_{2}),. . .,f(Y_{s}))^{T}\) is the stage derivatives computed in step \(n\) and the matrices defining the structure of the method are \(A=\{ a_{ij}\} \in \Re ^{(s\times s)}\), \(B=\{ b_{ij}\} \in \Re ^{(k\times s)}\), \(U=\{ u_{ij}\} \in \Re ^{(s\times k)}\), and \(V=\{ v_{ij}\} \in \Re ^{(k\times k)}\). The \(c=\left[c_{1},\ldots ,c_{s}\right]^{T}\) is a vector of the stage abscissas. The Nordsieck form of (2b) is </p>
<div class="equation" id="f.3">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c} Y \\ \hline \overline{y}^{[n]} \\[1mm]\end{array} \right)= \left( \begin{array}{c|cc} A & UT \\ \hline T^{-1}B & T^{-1}VT \\[1mm]\end{array} \right)\left( \begin{array}{c} hF(Y) \\ \hline \overline{y}^{[n-1]} \\[1mm]\end{array} \right), \tag {3a}\label{f.3} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> where, </p>
<div class="displaymath" id="a0000000005">
  \[ y^{[n-1]}=T\overline{y}^{[n-1]}, \quad y^{[n]}=T\overline{y}^{[n]},\quad T=\left[t_{ij}\right]_{ij=0}^{p,p},\leqno {(3a)} \]
</div>
<p> with the Nordsieck vectors in the past and present being respectively, </p>
<div class="displaymath" id="a0000000006">
  \[  \overline{y}^{[n-1]}=\left[ \begin{smallmatrix} y(x_{n-1})\\[1mm] hy'(x_{n-1})\\[1mm] h^{2}y''(x_{n-1})\\[1mm] h^{3}y^{(3)}(x_{n-1})\\[1mm] \vdots \\ h^{p}y^{(p)}(x_{n-1})\\[1mm] \end{smallmatrix} \right], \quad \overline{y}^{[n]}=\left[ \begin{array}{cc|c} y(x_{n})\\ hy’(x_{n})\\ h^{2}y”(x_{n})\\ h^{3}y^{(3)}(x_{n})\\ \vdots \\ h^{p}y^{(p)}(x_{n})\\ \end{array} \right]. \leqno {(3b)} \]
</div>
<p>Hybrid LMM can in general be easily represented in the GLM formalism as in (2b) and when it is possible to define Nordsieck vector in (3b), then it is readily implementable as a variable stepsize Nordsieck GLM in <a href="#f.3" class="eqref">2</a>. The <span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span> has given a practical example of the transformation of hybrid LMM into the Nordsieck form <a href="#f.3" class="eqref">2</a> of a GLM in (2b).We are therefore motivated following [4] to consider the transformation of a certain class of hybrid second derivative LMM into its Nordsieck SD-GLM for the purpose of easy implementation in variable step-size. In this regard, the multi-derivative GLM is </p>
<div class="displaymath" id="a0000000007">
  \[  \left( \begin{smallmatrix} Y \\ \hline y^{[n]} \\ \end{smallmatrix} \right)= \left( \begin{smallmatrix} A^{(1)}& \cdots &  A^{(\ell -1)}&  A^{(\ell )} & U \\[1mm] \hline B^{(1)}& \cdots & B^{(\ell -1)}& B^{(\ell )}& V \\[1mm] \end{smallmatrix} \right)\left( \begin{smallmatrix} hF(Y) \\ h^{2}F'(Y) \\ \vdots \\ h^{(\ell )}F^{(\ell -1)}(Y)\\ \hline y^{[n-1]} \\ \end{smallmatrix} \right), \quad \ell \geq 1,\leqno {(4)} \]
</div>
<p> in [11], <span class="cite">
	[
	<a href="#Ok8" >16</a>
	]
</span>. This method engages directly the higher order derivatives of the solution of (<a href="#equation1">1</a>) in the implementation of (2b). In particular, the SD-GLM in the Butcher sense is </p>
<div class="displaymath" id="a0000000008">
  \[  \left( \begin{array}{c} Y \\ \hline y^{[n]} \\ \end{array} \right)= \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V \\ \end{array} \right)\left( \begin{array}{c} hF(Y) \\ h^{2}F’(Y) \\ \hline y^{[n-1]} \\ \end{array} \right). \leqno {(5a)} \]
</div>
<p> This class of methods introduced in [2] which incorporates second derivatives of the solution of <a href="#equation1" class="eqref">1</a>, enhances (2) in order and stability wise for a comparable number of stages with respect to (2b). The terms in (5a) which extend (2) are the matrices \(A^{(1)}=\{ a_{ij}^{(1)}\} \in \Re ^{(s\times s)}\), \(A^{(2)}=\{ a_{ij}^{(2)}\} \in \Re ^{(s\times s)}\), \( B^{(1)}=\{ b_{ij}^{(1)}\} \in \Re ^{(k\times s)}\), \(B^{(2)}=\{ b_{ij}^{(2)}\} \in \Re ^{(k\times s)}\), \(U=\{ u_{ij}^{(j)}\} \in \Re ^{(s\times k)}\), \(V=\{ v_{ij}\} \in \Re ^{(k\times k)}\) and the second derivatives stages \(F'(Y)=(f'(Y_{1}), f'(Y_{2}),. . .,f'(Y_{s}))^{T}\) respectively. The picture of (5a) in Nordsieck form is now </p>
<div class="displaymath" id="a0000000009">
  \[ \left( \begin{array}{c} Y \\ \hline \overline{y}^{[n]} \\ \end{array} \right)= \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & UT \\ \hline T^{-1}B^{(1)}& T^{-1}B^{(2)} & T^{-1}VT \\ \end{array} \right)\left( \begin{array}{c} hF(Y) \\ h^{2}F’(Y) \\ \hline \overline{y}^{[n-1]} \\ \end{array} \right).\leqno {(5b)} \]
</div>
<p> The transformation matrix \(T\) is as in equation (3a). The stability matrix of (5a) is </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  M(z)=V+z(B^{(1)}+zB^{(2)})(I-zA^{(1)}-z^{2}A^{(2)})^{-1}U,\quad z=\lambda h, \end{equation*}
</div>
<p> and the stability polynomial is given by </p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  \Pi (w,z)=\det (wI-M(z)). \end{equation*}
</div>
<p> It is readily obvious that (5a) and (5b) exhibit the same stability characteristics. The same for (2b) and <a href="#f.3" class="eqref">2</a>. An example of the explicit SD-GLM (5a) is from <span class="cite">
	[
	<a href="#Ok3" >14</a>
	]
</span>, </p>
<div class="displaymath" id="a0000000012">
  \[ \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V \\ \end{array} \right) =\left( \begin{smallmatrix} 0& 0& 0& 0& 0& 0& 1\\[1mm] \frac{1}{3}& 0& 0&  \frac{1}{27}& 0& 0& 1 \\[1mm] \frac{7}{16}&  \frac{9}{16}& 0&  \frac{1}{16}& \frac{1}{16}& 0& 1 \\[1mm] \hline \\[0.5mm] \frac{7}{16}&  \frac{9}{16}& 0&  \frac{1}{16}& \frac{1}{16}& 0& 1 \\[1mm] \end{smallmatrix}\right),\quad c=\left(\begin{smallmatrix} 0\\[1mm] \frac{2}{3}\\[1mm] 1 \end{smallmatrix}\right). \]
</div>
<p> The interval of absolute stability of this SD-GLM <span class="cite">
	[
	<a href="#Ok3" >14</a>
	]
</span> is \(\left(-12, 0 \right)\). Again, see <span class="cite">
	[
	<a href="#Bu2" >2</a>
	]
</span> and <span class="cite">
	[
	<a href="#Ok3" >14</a>
	]
</span> for some further examples of the SD-GLM in (5a). Sequential SD-GLM have been derived in [17] and inherent Runge-Kutta stable methods are proposed in [18]. <br />A direct generalisation of GLM of Butcher and O’Sullivan <span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span> incorporating multiple hybrid points is </p>
<div class="displaymath" id="a0000000013">
  \[ {\begin{split}  y(x_{n-1}+c_{j}h)& =\sum _{i=1}^{k}u_{ji}^{(j)}y_{n-i}+h\sum _{i=1}^{k}u_{ji}^{(j+1)}f_{n-i}+h\sum _{i=\ell }^{j+\ell -1}a_{ji}^{(1)}f_{n+c_{i}-1},\\ p& =2k+2(j+\ell -1),\quad 0{\lt}c_{j}\leq 1,\quad j=\ell (1)s-1;\quad \ell =0, 1, \end{split}}\leqno {(6a)} \]
</div>
<p> with the last stage and also serving as an output method is given as </p>
<div class="displaymath" id="a0000000014">
  \[ {\begin{split}  y(x_{n-1}+th)& =\sum _{i=1}^{k}v_{1i}(t)y_{n-i}+h\sum _{i=1}^{k}v_{2i}(t)f_{n-i}+h\sum _{i=\ell }^{s+\ell -1}b_{ji}^{(1)}(t)f_{n+c_{i}-1},\\ i& =2(1)k,\quad p=2k+2(s+\ell -1)-(k-1)-1,\quad v_{11}(t)=1,\\ \lbrace v_{1i}(t)& =0\rbrace _{i=2}^{k},\quad t=c_{s}=1, \end{split}} \leqno {(6b)} \]
</div>
<p> where \(f_{n+c_{i}-1}=f(y(x_{n-1}+c_{i}h))\) and \(x_{n-1}=x_{n}-h\). A new SD-GLM extension of this class of methods becomes </p>
<div class="displaymath" id="a0000000015">
  \[ {\begin{split}  y(x_{n-1}+c_{j}h)=& \sum _{i=1}^{k}u_{ji}^{(j)}y_{n-i}+h\sum _{i=1}^{k}u_{ji}^{(j+1)}f_{n-i}+h^{2}\sum _{i=1}^{k}u_{ji}^{(j+2)}f’_{n-i}\\ & +h\sum _{i=\ell }^{j+\ell -1}a_{ji}^{(1)}f_{n+c_{i}-1} +h^{2}\sum _{i=\ell }^{j+\ell -1}a_{ji}^{(2)}f’_{n+c_{i}-1},\\ p& =3k+2(j+\ell -1)-1,\quad 0{\lt}c_{j}\leq 1,\quad f’=f_{y}f, \end{split}}\leqno {(7a)} \]
</div>
<p> with last stage as the method, </p>
<div class="displaymath" id="a0000000016">
  \[ {\begin{split}  y(x_{n-1}+th)& =\sum _{i=1}^{k}v_{1i}(t)y_{n-i}+h\sum _{i=1}^{k}v_{2i}(t)f_{n-i}+h^{2}\sum _{i=1}^{k}v_{3i}(t)f’_{n-i}\\ & \quad +h\sum _{i=\ell }^{s+\ell -1}b_{ji}^{(1)}(t)f_{n+c_{i}-1}+h^{2}\sum _{i=\ell }^{s+\ell -1}b_{ji}^{(2)}(t)f’_{n+c_{i}-1},\quad i=2(1)k,\\ p& =3k+2(s+\ell -1)-(k-1)-1,\quad v_{11}(t)=1,\\ \lbrace v_{1i}(t)& =0\rbrace _{i=2}^{k},\quad t=c_{s}=1. \end{split}} \leqno {(7b)} \]
</div>
<p> This class of methods is of interest in the next section. This will be implemented as <i class="it">first same as last</i> (FSAL), see [3]. To construct an explicit or diagonally implicit SD-GLM from (6a), (6b), and (7a), (7b) set the flag \(\ell \) accordingly as, </p>
<div class="displaymath" id="a0000000017">
  \[ \ell =\left\{ \begin{array}{lccccr}0;\quad \textrm{gives an explicit SD-GLM}\\ 1;\quad \textrm{gives an implicit SD-GLM} \end{array}\right. \]
</div>
<p> By setting \(\ell =0\), \(k=2\), \(s=3\) and the coefficients \(\lbrace a_{ji}^{(1)}\rbrace _{j=0}\), \(\lbrace u_{ji}^{(j+2)}\rbrace _{j=0}^{k}\), \(\lbrace a_{ji}^{(2)}\rbrace _{j=0}^{k}\), \(\lbrace v_{3i}(t)\rbrace _{j=0}^{k}\) and \(\lbrace a_{ji}^{(2)}(t)\rbrace _{j=0}^{k}\) in (6a) and (6b) respectively to zero, [4] using the Cauchy integral formula we obtain the hybrid LMM </p>
<div class="displaymath" id="a0000000018">
  \[ {\begin{split} \begin{array}{ll} y_{n-\frac{7}{15}}= -\frac{529}{3375}y_{n-1}+\frac{3904}{3375}y_{n-2}+h\left(\frac{4232}{3375}f_{n-1}+\frac{1472}{3375}f_{n-2}\right),\quad p=3,& \\ \widehat{y}_{n}= \frac{152}{25}y_{n-1}-\frac{127}{25}y_{n-2}+h\left(\frac{189}{92}f_{n-\frac{7}{15}}-\frac{419}{100}f_{n-1}-\frac{1118}{575}f_{n-2}\right),\quad p=3,& \\ y_{n}= y_{n-1}+h\left(\frac{3375}{5152}f_{n-\frac{7}{15}}+\frac{25}{168}\widehat{f}_{n}+\frac{19}{96}f_{n-1}-\frac{1}{552}f_{n-2}\right),\quad p=5. \end{array}\end{split}} \leqno {(8)} \]
</div>
<p> By (3a) and (3b) the Nordsieck form of this becomes, </p>
<div class="equation" id="f.9">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c|cc} A^{(1)}& UT \\ \hline T^{-1}B^{(1)}& T^{-1}VT \\ \end{array} \right) =\left( \begin{smallmatrix} 0& 0& 0& 1& \frac{8}{15}& \frac{64}{225}& \frac{512}{3375}& \frac{-1984}{3375}& \frac{128}{125} \\[1mm] \frac{189}{92}& 0& 0& 1& \frac{-97}{92}& \frac{-137}{115}& \frac{-433}{575}& \frac{1551}{575}& \frac{-2669}{575} \\[1mm] \frac{3375}{5152}& \frac{25}{168}& 0& 1& \frac{433}{2208}& \frac{1}{276}& \frac{-1}{184}& \frac{1}{138}& \frac{-5}{552} \\[1mm] \hline \frac{3375}{5152}& \frac{25}{168}& 0& 1& \frac{433}{2208}& \frac{1}{276}& \frac{-1}{184}& \frac{1}{138}& \frac{-5}{552} \\[1mm] 0& 0& 1& 0& 0& 0& 0& 0& 0\\[1mm] \frac{-3375}{896}& \frac{-575}{672}& 3& 0& \frac{623}{384}& \frac{35}{48}& \frac{-7}{32}& \frac{-7}{24}& \frac{77}{96} \\[1mm] \frac{-111375}{20608}& \frac{-275}{224}& \frac{13}{4}& 0& \frac{9957}{2944}& \frac{633}{368}& \frac{-335}{736}& \frac{-149}{184}& \frac{1527}{736} \\[1mm] \frac{-57375}{20608}& \frac{-425}{672}& \frac{3}{2}& 0& \frac{16927}{8832}& \frac{1363}{1104}& \frac{-167}{736}& \frac{-431}{552}& \frac{3949}{2208} \\[1mm] \frac{-10125}{20608}& \frac{-25}{224}& \frac{1}{4}& 0& \frac{1039}{2944}& \frac{91}{368}& \frac{3}{736}& \frac{-47}{184}& \frac{373}{736} \\[1mm] \end{smallmatrix} \right), \tag {9}\label{f.9} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p> The abscissa vector for the GLM in <a href="#f.9" class="eqref">16</a> is \(c = \left(\frac{8}{15}, 1, 1\right)^{T}\). The \(T\) transformation in this representation is </p>
<div class="equation" id="f.10">
<p>
  <div class="equation_content">
    \begin{equation}  T=\left( \begin{smallmatrix} 1 &  0 &  0 &  0 &  0 &  0 \\[1mm] 1 &  -1 &  1 &  -1 &  1 &  -1 \\[1mm] 1 &  -2 &  4 &  -8 &  16 &  -32 \\[1mm] 0 &  1 &  0 &  0 &  0 &  0 \\[1mm] 0 &  1 &  -2 &  3 &  -4 &  5 \\[1mm] 0 &  1 &  -4 &  12 &  -32 &  80\\[1mm] \end{smallmatrix} \right).\tag {10}\label{f.10} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> The interval of absolute stability of <a href="#f.9" class="eqref">16</a> is \((-1.231, 0)\), see Fig. 1 for its stability plot. The region of absolute stability is the closed region in the <i class="it">left half plane</i> (LHP) of the complex plane in Fig. 1. </p>
<figure >
  <div class="centered"> <img src="img-0001.png" alt="\includegraphics[width=9cm]{Okuonghae/Bucher1.png}" style="width:9cm" />
<br /></div>
<p> Fig. 1. Region of absolute stability (closed region in the LHP) of the method in (9). </p>

</figure>
<p>An implicit counterpart of <a href="#f.9" class="eqref">16</a> is </p>
<div class="displaymath" id="a0000000019">
  \[ \left( \begin{smallmatrix} A^{(1)}& UT \\ \hline T^{-1}B^{(1)}& T^{-1}VT \\ \end{smallmatrix} \right) = \left( \begin{smallmatrix} \frac{92}{465} &  0 &  0 &  1 &  \frac{52}{155} &  \frac{512}{6975} &  \frac{-1792}{104625} &  \frac{-4096}{104625} &  \frac{3328}{34875} \\[1mm] \frac{-189 }{2300} &  \frac{22}{75} &  0 &  1 &  \frac{1303}{6900} &  \frac{112}{8625} &  \frac{-96}{14375} &  \frac{16}{43125} &  \frac{256}{43125} \\[1mm] \frac{3375 }{5152} &  0 &  \frac{25 }{168} &  1 &  \frac{433}{2208} &  \frac{1}{276} &  \frac{-1}{184} &  \frac{1}{138} &  \frac{-5}{552}\\[1mm] \hline \frac{3375}{5152} &  0 &  \frac{25}{168} &  1 &  \frac{433}{2208} &  \frac{1}{276} &  \frac{-1}{184} &  \frac{1}{138} &  \frac{-5}{552} \\[1mm] 0 &  0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] \frac{-3375}{896} &  0 &  \frac{1441}{672} &  0 &  \frac{623}{384} &  \frac{35}{48} &  \frac{-7}{32} &  \frac{-7}{24} &  \frac{77}{96} \\[1mm] \frac{-111375}{20608} &  0 &  \frac{453}{224} &  0 &  \frac{9957}{2944} &  \frac{633}{368} &  \frac{-335}{736} &  \frac{-149}{184} &  \frac{1527}{736} \\[1mm] \frac{-57375}{20608} &  0 &  \frac{583}{672} &  0 &  \frac{16927}{8832} &  \frac{1363}{1104} &  \frac{-167}{736} &  \frac{-431}{552} &  \frac{3949}{2208} \\[1mm] \frac{-10125}{20608} &  0 &  \frac{31}{224} &  0 &  \frac{1039}{2944} &  \frac{91}{368} &  \frac{3}{736} &  \frac{-47}{184} &  \frac{373}{736}\\ \end{smallmatrix} \right). \]
</div>
<p> The \(c = \big(\frac{8}{15}, \frac{2}{5}, 1\big)^{T}\) is the abscissa vector for the implicit counterpart of the method in <a href="#f.9" class="eqref">16</a> with the angle of absolute stability, \(\alpha = 77^{\circ }\) and having <a href="#f.10" class="eqref">17</a> as its \(T\). </p>
<p>This paper is organized as follows. Section 2 is on the transformation of SD-HLMM(7) into its Nordsieck form which is illustrated for step numbers \(k=1, 2\). In section 3 we derive the SD-HLMM (7) of step number \(k\leq 3\), while section 4 presents the variable step-size implementation. The section 5 presents how the RK-starters for the methods can be obtained and with examples. Section 6, presents results of numerical experiments. </p>
<h1 id="a0000000020">2 The SD-HLMM (7a,b)</h1>
<p>Consider an extension of the methods of [5] provided in (7a) and (7b) to SD-GLM. The explicit stage predictors are, </p>
<div class="equation" id="f.11">
<p>
  <div class="equation_content">
    \begin{equation}  {y(x_{n-1}+c_{1}h)=\sum _{j=1}^{k}u_{1,j}^{(1)}y_{n-j}+h\sum _{j=1}^{k}u_{1,j}^{(2)}f_{n-j}+h^{2}\sum _{j=1}^{k}u_{1,j}^{(3)}f’_{n-j},} \tag {11}\label{f.11} \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="a0000000021">
  \begin{align}  \widehat{y}(x_{n-1}+c_{2}h)=& \sum _{j=1}^{k}u_{2,j}^{(1)}y_{n-j}+h\sum _{j=1}^{k}u_{2,j}^{(2)}f_{n-j}+ha_{2,1}^{(1)}f_{n+c_{1}-1}\nonumber \\ & +h^{2}\sum _{j=1}^{k}u_{2,j}^{(3)}f’_{n-j}+h^{2}a_{2,1}^{(2)}f’_{n+c_{1}-1}, \tag {12}\label{f.12} \end{align}
</div>
<p> for computing the off-step points at \(x_{n+c_{1}-1}\) and \(x_{n+c_{2}-1}\) respectively. The last stage that will also serve as the output method is, </p>
<div class="displaymath" id="f.13">
  \begin{align}  y(x_{n-1}+th)= & \sum _{j=1}^{k}v_{1,j}(t)y_{n-j}+h\sum _{j=1}^{k}v_{2,j}(t)f_{n-j}+h b_{11}^{(1)}(t)f_{n+c_{1}-1}+ \tag {13}\label{f.13} \end{align}
</div>
<div class="displaymath" id="a0000000022">
  \begin{align*} & +h b_{12}^{(1)}(t)f_{n+c_{2}-1} +h^{2}\sum _{j=1}^{k}v_{3,j}(t)f’_{n-j}+h^{2}b_{11}^{(2)}(t)f’_{n+c_{1}-1}\\ \nonumber & +h^{2}b_{12}^{(2)}(t)f’_{n+c_{2}-1},\quad t=1.\nonumber \end{align*}
</div>
<p> The \(c=\left(c_{1}, c_{2}\right)\in \left[0, 1\right]\), is the stage abscissas vector, the continuous coefficients \(v_{1,1}(t)\) and \(v_{1,2}(t)\) are assumed to be one and zero respectively. The first and the second stage derivatives are \(F(Y)=\left[f(y_{n+c_{1}-1})\quad f(y_{n+c_{2}-1})\right]^{T}\) and \(F'(Y)=\left[f'(y_{n+c_{1}-1})\quad f'(y_{n+c_{2}-1})\right]^{T}\) respectively. Here \(f_{n+c_{i}-1}=f(y(x_{n-1}+c_{i}h)), i=1(1)s\), \(f'_{n+c_{i}-1}=f'(y(x_{n-1}+c_{i}h)), i=1(1)s\) and \(x_{n-1}=x_{n}-h\). The order of the algorithms in <a href="#f.11" class="eqref">18</a>, <a href="#f.12" class="eqref">19</a> and <a href="#f.13" class="eqref">20</a> are \(3k-1\), \(3k+1\) and \(3k+2\) respectively. <br /></p>
<p>Similarly, the equivalent methods of the implicit case are </p>
<div class="displaymath" id="f.14">
  \begin{align}  y(x_{n-1}+c_{1}h) = & \sum _{j=1}^{k}u_{1,j}^{(1)}y_{n-j}+h\sum _{j=1}^{k}u_{1,j}^{(2)}f_{n-j}+h^{2}\sum _{j=1}^{k}u_{1,j}^{(3)}f’_{n-j} \tag {14}\label{f.14} \\ &  +h a_{1,1}^{(1)}f_{n+c_{1}-1}+h^{2}\widehat{a}_{1,1}^{(2)}f’_{n+c_{1}-1},\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="f.15">
  \begin{align}  \widehat{y}(x_{n-1}+c_{2}h)= & \sum _{j=1}^{k}u_{2,j}^{(1)}y_{n-j}+\sum _{j=1}^{k}u_{2,j}^{(2)}f_{n-j}+h a_{2,1}^{(1)}f_{n+c_{1}-1}+h a_{2,2}^{(1)}f_{n+c_{2}-1} \tag {15}\label{f.15} \\ &  +h^{2}\sum _{j=1}^{k}u_{2,j}^{(3)}f’_{n-j}+h^{2}a_{2,1}^{(2)}f’_{n+c_{1}-1}+h^{2}a_{2,2}^{(2)}f’_{n+c_{2}-1},\nonumber \end{align}
</div>
<p> with the last stage as an output method, </p>
<div class="displaymath" id="a0000000023">
  \begin{align}  y(x_{n-1}+th)= & \sum _{j=1}^{k}v_{1,j}(t)y_{n-j}+h\sum _{j=1}^{k}v_{2,j}(t)f_{n-j}+h b_{11}^{(1)}(t)f_{n+c_{1}-1}\nonumber \\ & +h b_{12}^{(1)}(t)f_{n+c_{2}-1}+h b_{13}^{(1)}(t)f_{n+t-1}+h^{2}\sum _{j=1}^{k}v_{3,j}(t)f’_{n-j}\nonumber \\ & +h^{2}b_{11}^{(2)}(t)f’_{n+c_{1}-1}\! +\! h^{2}b_{12}^{(2)}(t)f’_{n+c_{2}-1}\! +\! h^{2}b_{13}^{(2)}(t)f’_{n+t-1},\  \  t=1.\tag {16}\label{f.16} \end{align}
</div>
<p> The prospect of these classes of SD-HLMM is the fact that the explicit <a href="#f.13" class="eqref">20</a> and implicit <a href="#f.16" class="eqref">21</a> methods have a common \(T\) transformation (3a) for a fixed \(k\) in their Nordsieck GLM. Moreover, the explicit method <a href="#f.11" class="eqref">18</a>-<a href="#f.13" class="eqref">20</a> can serve as a starter for the implicit one in <a href="#f.14" class="eqref">19</a>-<a href="#f.16" class="eqref">21</a>. The SD-HLMM can be transformed into Nordsieck methods. To get started, the SD-GLM representation of the SD-HLMM <a href="#f.11" class="eqref">18</a>, <a href="#f.12" class="eqref">19</a>, <a href="#f.13" class="eqref">20</a> when \(k=2\) is </p>
<div class="displaymath" id="a0000000024">
  \[  \left( \begin{smallmatrix} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V\\ \end{smallmatrix} \right) =\left( \begin{smallmatrix} 0 &  0 & 0&  0 &  0& 0& u_{11}^{(1)}& u_{12}^{(1)}& u_{11}^{(2)}& u_{12}^{(2)}& u_{11}^{(3)}& u_{12}^{(3)}\\[1mm] a_{21}^{(1)} & 0&  0 & a_{21}^{(2)} & 0& 0& u_{21}^{(1)}& u_{22}^{(1)}& u_{21}^{(2)}& u_{22}^{(2)}& u_{21}^{(3)}& u_{22}^{(3)} \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  0 &  b_{11}^{(2)} & b_{12}^{(2)}& 0& 1& 0& v_{11}^{(2)}& v_{12}^{(2)}& v_{11}^{(3)}& v_{12}^{(3)} \\[1mm] \hline b_{11}^{(1)} &  b_{12}^{(1)} &  0 & b_{11}^{(2)} & b_{12}^{(2)}& 0& 1& 0& v_{11}^{(2)}& v_{12}^{(2)}& v_{11}^{(3)}& v_{12}^{(3)} \\[1mm] 0 &  0 & 0& 0& 0& 0& 1& 0& 0& 0& 0& 0\\[1mm] 0& 0&  1 & 0 & 0& 0& 0& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 1& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 1& 0& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 0& 0& 1& 0 \\[1mm] \end{smallmatrix} \right),\leqno {(17a)} \]
</div>
<p> while the implicit SD-GLM partitioned matrix of the SD-HLMM <a href="#f.14" class="eqref">19</a>, <a href="#f.15" class="eqref">20</a>,<a href="#f.16" class="eqref">21</a> when \(k=2\) is </p>
<div class="displaymath" id="a0000000025">
  \[  \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V\\ \end{array} \right) =\left( \begin{smallmatrix}a_{11}^{(1)} &  0 & 0&  a_{11}^{(2)} &  0& 0&  & \\[1mm] a_{21}^{(1)} & a_{22}^{(1)}&  0 & a_{21}^{(2)} & a_{22}^{(2)} &  0&  U& \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  b_{13}^{(1)} & b_{11}^{(2)}& b_{12}^{(2)}& b_{13}^{(2)}&  &  \\[1mm] \hline &  B^{(1)} &  &  & B^{(2)}&  & V& \\[1mm] \end{smallmatrix} \right),\leqno {(17b)} \]
</div>
<p> where \(B^{(1)}\), \(B^{(2)}\), \(U\), \(V\) are of the same structure as in (17a). The stages, stage derivatives, the inputs and the output approximations for the implicit method in (17a, b) are: </p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  Y=& [y(x_{n-1}+c_{1}h), y(x_{n-1}+c_{2}h), y(x_{n-1}+th)]^{T}, \\ F(Y)=& [f(y(x_{n-1}+c_{1}h)), f(y(x_{n-1}+c_{2}h)), f(y(x_{n-1}+th))]^{T}, \\ F’(Y)=& [f’(y(x_{n-1}+c_{1}h)), f’(y(x_{n-1}+c_{2}h)), f’(y(x_{n-1}+th))]^{T}, \  {\rm and} \\ y^{[n]}=& [y_{n}, y_{n-1}, hf_{n}, hf_{n-1}, h^{2}f’_{n}, h^{2}f’_{n-1}]^{T} \end{align*}
</div>
<p> respectively. To achieve our aim we seek a transformation of (17a) into the form </p>
<div class="displaymath" id="a0000000027">
  \[ \left( \begin{smallmatrix} A^{(1)}&  A^{(2)} & UT \\ \hline T^{-1}B^{(1)}& T^{-1}B^{(2)} & T^{-1}VT \\ \end{smallmatrix} \right) =\left( \begin{smallmatrix} 0 &  0 & 0&  0 &  0& 0& 1& u_{11}& \cdots & u_{19}\\[1mm] a_{21}^{(1)} & 0&  0 & a_{21}^{(2)} & 0& 0& 1& u_{21}& \cdots & u_{29} \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  0 &  b_{11}^{(2)} & b_{12}^{(2)}& 0& 1& v_{11}& \cdots & v_{19} \\[1mm] \hline b_{11}^{(1)} &  b_{12}^{(1)} &  0 & b_{11}^{(2)} & b_{12}^{(2)}& 0& 1& v_{11}& \cdots & v_{19} \\[1mm] 0 &  0 & 1& 0& 0& 0& 0& 0& \cdots & 0\\[1mm] 0& 0&  0 & 0 & 0& 1& 0& 0& \cdots & 0\\[1mm] b_{41} &  b_{42}& b_{43}& b_{41} &  b_{42}& b_{43}& 0& v_{41}& \cdots & v_{49}\\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\[1mm] b_{91} &  b_{92}& b_{93}& b_{91} &  b_{92}& b_{93}& 0& v_{91}& \cdots & v_{99}\\[1mm] \end{smallmatrix} \right),\leqno {(18a)} \]
</div>
<p> using the Nordsieck vectors of (3b). Similarly, the implicit case (17b) is </p>
<div class="displaymath" id="a0000000028">
  \[ \left( \begin{smallmatrix} A^{(1)}&  A^{(2)} & UT \\ \hline T^{-1}B^{(1)}& T^{-1}B^{(2)} & T^{-1}VT \\ \end{smallmatrix} \right) =\left( \begin{smallmatrix} a_{11}^{(1)} &  0 & 0&  a_{11}^{(2)} &  0& 0&  & \\[1mm] a_{21}^{(1)} & a_{22}^{(1)}&  0 & a_{21}^{(2)} & a_{22}^{(2)} &  0&  UT& \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  b_{13}^{(1)} & b_{11}^{(2)}& b_{12}^{(2)}& b_{13}^{(2)}&  &  \\[1mm] \hline &  T^{-1}B^{(1)} &  &  & T^{-1}B^{(2)}&  & T^{-1}VT& \\[1mm] \end{smallmatrix} \right). \leqno {(18b)} \]
</div>
<p> The \(UT\), \(T^{-1}B^{(1)}\), \(T^{-1}B^{(2)}\) and \(T^{-1}VT\) are of similar form to that in (18a). To arrive at this the stage values and the stage derivatives at step number \(n\) are </p>
<div class="displaymath" id="a0000000029">
  \[ Y=\left(\begin{smallmatrix} y_{n+c_{1}-1}\\[1mm] y_{n+c_{2}-1}\\[1mm] y_{n+t-1}\\[1mm] \end{smallmatrix} \right),\quad F(Y)=\left(\begin{array}{c} f_{n+c_{1}-1}\\ f_{n+c_{2}-1}\\ f_{n+t-1}\\ \end{array} \right),\quad F'(Y)=\left(\begin{array}{c} f’_{n+c_{1}-1}\\ f’_{n+c_{2}-1}\\ f’_{n+t-1}\\ \end{array} \right). \]
</div>
<p> Thus, the \((3+9)\times (3+3+9)\) partitioned matrix of the SD-GLM (17a) is </p>
<div class="displaymath" id="a0000000030">
  \[ \left( \begin{smallmatrix} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V\\ \end{smallmatrix} \right)= \left( \begin{smallmatrix} 0 &  0 & 0&  0 &  0& 0& u_{11}^{(1)}& u_{12}^{(1)}& 0& u_{11}^{(2)}& u_{12}^{(2)}& 0& u_{11}^{(3)}& u_{12}^{(3)}& 0\\[1mm] a_{21}^{(1)} & 0&  0 & a_{21}^{(2)} & 0& 0& u_{21}^{(1)}& u_{22}^{(1)}& 0& u_{21}^{(2)}& u_{22}^{(2)}& 0& u_{21}^{(3)}& u_{22}^{(3)}& 0 \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  0 &  b_{11}^{(2)} &  b_{12}^{(2)} & 0& 1& 0& 0& v_{11}^{(2)}& v_{12}^{(2)}& 0& v_{11}^{(3)}& v_{12}^{(3)}& 0 \\[1mm] \hline b_{11}^{(1)} &  b_{12}^{(1)} &  0 &  b_{11}^{(2)} &  b_{12}^{(2)} & 0& 1& 0& 0& v_{11}^{(2)}& v_{12}^{(2)}& 0& v_{11}^{(3)}& v_{12}^{(3)}& 0 \\[1mm] 0 &  0 & 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\[1mm] 0& 0&  0 & 0 & 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  1 & 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0 \\[1mm] 0 &  0 &  0 & 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0 \\[1mm] \end{smallmatrix} \right). \leqno {(19a)}  \]
</div>
<p> In the same manner, </p>
<div class="displaymath" id="a0000000031">
  \[ \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}& B^{(2)} & V\\ \end{array} \right) =\left( \begin{smallmatrix} a_{11}^{(1)} &  0 & 0&  a_{11}^{(2)} &  0& 0&  & \\[1mm] a_{21}^{(1)} & a_{22}^{(1)}&  0 & a_{21}^{(2)} & a_{22}^{(2)} &  0&  U& \\[1mm] b_{11}^{(1)} &  b_{12}^{(1)} &  b_{13}^{(1)} & b_{11}^{(2)}& b_{12}^{(2)}& b_{13}^{(2)}&  &  \\[1mm] \hline &  B^{(1)} &  &  & B^{(2)}&  & V& \\[1mm] \end{smallmatrix} \right),\leqno {(19b)} \]
</div>
<p> for the implicit method in (18b), where \(y_{n-3}\), \(hf_{n-3}\) and \(h^{2}f'_{n-3}\) are the additional input data with coefficients zeros, which will be included to write the method (19a,b) in this \((3+9)\times (3+3+9)\) partitioned matrix, following the ideas of <span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span>. The incoming and the corresponding outgoing vectors are now, </p>
<div class="equation" id="f.20">
<p>
  <div class="equation_content">
    \begin{equation}  y^{[n-1]}=\left( \begin{smallmatrix} y_{1}^{[n-1]}\\[1mm] y_{2}^{[n-1]}\\[1mm] y_{3}^{[n-1]}\\[1mm] y_{4}^{[n-1]}\\[1mm] y_{5}^{[n-1]}\\[1mm] y_{6}^{[n-1]}\\[1mm] y_{7}^{[n-1]}\\[1mm] y_{8}^{[n-1]}\\[1mm] y_{9}^{[n-1]}\\[1mm] \end{smallmatrix} \right)=\left( \begin{smallmatrix} y_{n-1}\\[1mm] y_{n-2}\\[1mm] y_{n-3}\\[1mm] hf_{n-1}\\[1mm] hf_{n-2}\\[1mm] hf_{n-3}\\[1mm] h^{2}f'_{n-1}\\[1mm] h^{2}f'_{n-2}\\[1mm] h^{2}f'_{n-3}\\[1mm] \end{smallmatrix} \right), \quad y^{[n]}=\left( \begin{smallmatrix} y_{1}^{[n]}\\[1mm] y_{2}^{[n]}\\[1mm] y_{3}^{[n]}\\[1mm] y_{4}^{[n]}\\[1mm] y_{5}^{[n]}\\[1mm] y_{6}^{[n]}\\[1mm] y_{7}^{[n]}\\[1mm] y_{8}^{[n]}\\[1mm] y_{9}^{[n]}\\[1mm] \end{smallmatrix} \right)=\left( \begin{smallmatrix} y_{n}\\[1mm] y_{n-1}\\[1mm] y_{n-2}\\[1mm] hf_{n}\\[1mm] hf_{n-1}\\[1mm] hf_{n-2}\\[1mm] h^{2}f'_{n}\\[1mm] h^{2}f'_{n-1}\\[1mm] h^{2}f'_{n-2}\\ \end{smallmatrix} \right). \tag {20}\label{f.20} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p> The \(T\) obtained through Taylor’s expansion of \(y^{[n]}\) in <a href="#f.20" class="eqref">22</a> about \(x_{n}\) using (3a) is </p>
<div class="equation" id="f.21">
<p>
  <div class="equation_content">
    \begin{equation}  T=\left( \begin{smallmatrix} 1 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 1 &  -1 &  1 &  -1 &  1 &  -1 &  1 &  -1 &  1 \\[1mm] 1 &  -2 &  4 &  -8 &  16 &  -32 &  64 &  -128 &  256 \\[1mm] 0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  1 &  -2 &  3 &  -4 &  5 &  -6 &  7 &  -8 \\[1mm] 0 &  1 &  -4 &  12 &  -32 &  80 &  -192 &  448 &  -1024 \\[1mm] 0 &  0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  2 &  -6 &  12 &  -20 &  30 &  -42 &  56 \\[1mm] 0 &  0 &  2 &  -12 &  48 &  -160 &  480 &  -1344 &  3584\\[1mm] \end{smallmatrix} \right).\tag {21} \label{f.21} \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> Carrying out the operations \(UT\), \(T^{-1}B^{(1)}\), \(T^{-1}B^{(2)}\), and \(T^{-1}VT\) on (19a,b) yields the Nordsieck algorithms in (18a,b). The transformations above are illustrated by practical examples in the next section. </p>
<h1 id="a0000000032">3 The transformation of SD-HLMM to SD-Nordsieck methods</h1>
<p> First, consider the derivation of the SD-HLMM in <a href="#f.11" class="eqref">18</a> for \(k=1\) and its transformation to the explicit Nordsieck methods (18a). For the derivation let the polynomial solution of <a href="#equation1" class="eqref">1</a> be </p>
<div class="equation" id="f.22">
<p>
  <div class="equation_content">
    \begin{equation}  y(x)=\sum _{j=0}^{N}a_{j}x^{j},\quad N= 3k+1,\tag {22}\label{f.22} \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> to set up the following interpolatory system of linear equations, </p>
<div class="equation" id="f.23">
<p>
  <div class="equation_content">
    \begin{equation}  XA=B, \tag {23}\label{f.23} \end{equation}
  </div>
  <span class="equation_label">25</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000033">
  \[ X=\left(\begin{smallmatrix} 1 &  x_{n-1} &  x_{n-1}^{2} &  . &  . &  . &  x_{n-1}^{N} \\[1mm] 0 &  1 &  2x_{n-1} &  . &  . &  . &  Nx_{n-1}^{N-1}\\[1mm] 0 &  1 &  2x_{n+c_{1}-1} &  . &  . &  . &  Nx_{n+c_{1}-1}^{N-1}\\[1mm] 0 &  0 &  2 &  . &  . &  . &  N(N-1)x_{n-1}^{N-2}\\[1mm] 0 &  0 &  2 &  . &  . &  . &  N(N-1)x_{n+c_{1}-1}^{N-2}\\[1mm] \end{smallmatrix}\right),\quad A=\left(\begin{smallmatrix} a_{0} \\[1mm] a_{1} \\[1mm] a_{2} \\[1mm] a_{3} \\[1mm] a_{4} \\[1mm] \end{smallmatrix}\right),\quad B=\left(\begin{smallmatrix} y_{n-1} \\[1mm] f_{n-1} \\[1mm] f_{n+c_{1}-1} \\[1mm] f'_{n-1} \\[1mm] f'_{n+c_{1}-1} \\[1mm] \end{smallmatrix}\right),  \]
</div>
<p> for the algorithm in <a href="#f.13" class="eqref">20</a>. For \(k=1\) in <a href="#f.11" class="eqref">18</a> the <a href="#f.23" class="eqref">25</a> reduces to </p>
<div class="equation" id="f.24">
<p>
  <div class="equation_content">
    \begin{equation}  \left(\begin{smallmatrix} 1 &  x_{n-1} &  x_{n-1}^{2} \\[1mm] 0 &  1 &  2x_{n-1} \\[1mm] 0 &  0 &  2 \\[1mm] \end{smallmatrix}\right)\left(\begin{smallmatrix} a_{0} \\[1mm] a_{1} \\[1mm] a_{2} \\[1mm] \end{smallmatrix}\right)=\left(\begin{smallmatrix} y_{n-1} \\[1mm] f_{n-1} \\[1mm] f'_{n-1} \\[1mm] \end{smallmatrix}\right),\quad N=2.\tag {24}\label{f.24} \end{equation}
  </div>
  <span class="equation_label">26</span>
</p>
</div>
<p> Solving this resulting system of linear equations for \(a_{j}, j=0(1)2\) and substituting the resulting values into <a href="#f.11" class="eqref">18</a> with \(c_{1}=(x-x_{n-1})/h\) yields the following coefficients: </p>
<div class="equation" id="f.25">
<p>
  <div class="equation_content">
    \begin{equation}  u_{11}^{(1)}=1, u_{11}^{(2)}=c_{1},  u_{11}^{(3)} = \tfrac {c_1^2}{2}. \tag {25}\label{f.25} \end{equation}
  </div>
  <span class="equation_label">27</span>
</p>
</div>
<p> A predictor exists for any choice of \(c_{1}\). Consider for example, fixing \(c_{1}=\frac{1}{2}\) into <a href="#f.25" class="eqref">27</a> which gives thus </p>
<div class="equation" id="f.26">
<p>
  <div class="equation_content">
    \begin{equation}  y_{n-\frac{1}{2}}=y_{n-1}+\tfrac {h}{2} f_{n-1}+\tfrac {h^2}{8} f'_{n-1},\quad p=2. \tag {26}\label{f.26} \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p> Similarly, setting \(N=4\) in <a href="#f.22" class="eqref">24</a> yields <a href="#f.23" class="eqref">25</a>. Solving <a href="#f.23" class="eqref">25</a> for \(a_{j}, j=0(1)4\) and substituting the result into <a href="#f.22" class="eqref">24</a> lead to </p>
<div class="displaymath" id="a0000000034">
  \begin{align}  u_{11}^{(1)}& =1,&  u_{11}^{(2)}& =c_2-\tfrac {c_2^3}{c_1^2}+\tfrac {c_2^4}{2 c_1^3}, &  u_{11}^{(3)}& = \tfrac {c_2^2}{2}-\tfrac {2 c_2^3}{3 c_1}+\tfrac {c_2^4}{4 c_1^2},\nonumber \\ t& =c_{2},&  b_{11}^{(1)}& =\tfrac {c_2^3}{c_1^2}-\tfrac {c_2^4}{2 c_1^3},&  b_{11}^{(2)}& = -\tfrac {c_2^3}{3 c_1}+\tfrac {c_2^4}{4 c_1^2}. \tag {27}\label{f.27} \end{align}
</div>
<p> These coefficients in <a href="#f.27" class="eqref">29</a> show that an infinite number of output methods <a href="#f.13" class="eqref">20</a> exists for step number \(k=1\). Inserting \(c= \big[\frac{1}{2}, 1\big]\) into <a href="#f.27" class="eqref">29</a> yields </p>
<div class="equation" id="f.28">
<p>
  <div class="equation_content">
    \begin{equation}  y_n=y_{n-1}+h f_{n-1}+ h^2 \big(\tfrac {1}{6}f'_{n-1}+\tfrac {1}{3} f'_{n-\frac{1}{2}}\big), \quad p=4.\tag {28}\label{f.28} \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p> Following the approach in (17)–<a href="#f.20" class="eqref">22</a> the Nordsieck form of <a href="#f.26" class="eqref">28</a>,<a href="#f.28" class="eqref">30</a> is </p>
<div class="equation" id="f.29">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & UT \\ \hline T^{-1}B^{(1)}& T^{-1}B^{(2)} & T^{-1}VT\\ \end{array} \right) =\left( \begin{smallmatrix} 0 &  0 &  0 &  0 &  1 &  \frac{1}{2} &  \frac{1}{8} &  0 &  0 &  0 \\[1mm] 0 &  0 &  \frac{1}{3} &  0 &  1 &  1 &  \frac{1}{6} &  0 &  0 &  0\\[1mm] \hline 0 &  0 &  \frac{1}{3} &  0 &  1 &  1 &  \frac{1}{6} &  0 &  0 &  0 \\[1mm] 0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  -6 &  \frac{10}{3} &  3 &  0 &  6 &  \frac{7}{6} &  0 &  0 &  0 \\[1mm] 0 &  -8 &  5 &  3 &  0 &  8 &  \frac{3}{2} &  0 &  0 &  0 \\[1mm] 0 &  -3 &  2 &  1 &  0 &  3 &  \frac{1}{2} &  0 &  0 &  0\\[1mm] \end{smallmatrix} \right).\tag {29}\label{f.29} \end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p> The transformation matrix \(T\) in <a href="#f.29" class="eqref">31</a> is </p>
<div class="equation" id="f.30">
<p>
  <div class="equation_content">
    \begin{equation}  T=\left( \begin{smallmatrix} 1 &  0 &  0 &  0 &  0 &  0 \\[1mm] 1 &  -1 &  1 &  -1 &  1 &  -1 \\[1mm] 0 &  1 &  0 &  0 &  0 &  0 \\[1mm] 0 &  1 &  -2 &  3 &  -4 &  5 \\[1mm] 0 &  0 &  1 &  0 &  0 &  0 \\[1mm] 0 &  0 &  2 &  -6 &  12 &  -20 \end{smallmatrix} \right). \tag {30}\label{f.30} \end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p> The eigenvalues of the stability matrix of (29) are </p>
<div class="displaymath" id="a0000000035">
  \[ \lbrace {1+z+\tfrac {z^2}{2}+\tfrac {z^3}{6}+\tfrac {z^4}{24}, 0, 0, 0, 0, 0, 0\rbrace } \]
</div>
<p> and the fact that only one of them is not identically zero means that the method will behave like a Runge-Kutta method with stability function \(R(z)= 1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}\), see <span class="cite">
	[
	<a href="#Bu9" >6</a>
	]
</span>. The interval of absolute stability of the algorithm in <a href="#f.29" class="eqref">31</a> is thus \([-2.785, 0]\). For the stability plot, see Fig. 2. </p>
<p>Similarly, the abscissa vector \(c=[\frac{2}{3},1]^{T}\) in <a href="#f.25" class="eqref">27</a> and <a href="#f.27" class="eqref">29</a> gives another fourth order SD-HLMM for step number \(k=1\) as </p>
<div class="equation" id="f.31">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{array}{ll} y_{n-\frac{1}{3}}=y_{n-1}+\frac{2h}{3} f_{n-1}+\frac{2h^2}{9}f_{n-1},\quad p=2,& \\ y_n=y_{n-1}+\frac{h}{16}\left(9f_{n-\frac{1}{3}}+ 7f_{n-1}\right)+\frac{h^2}{16} \left(f’_{n-\frac{1}{3}}+ f’_{n-1} \right),\quad p=4. \end{array} \tag {31}\label{f.31} \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> The Nordsieck form of this is </p>
<div class="equation" id="f.32">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & UT \\ \hline T^{-1}B^{(1)}& T^{-1}B^{(2)} & T^{-1}VT\\ \end{array} \right) =\left( \begin{smallmatrix} 0 &  0 &  0 &  0 &  1 &  \frac{2}{3} &  \frac{2}{9} &  0 &  0 &  0 \\[1mm] \frac{9}{16} &  0 &  \frac{1}{16} &  0 &  1 &  \frac{7}{16} &  \frac{1}{16} &  0 &  0 &  0\\[1mm] \hline \frac{9}{16} &  0 &  \frac{1}{16} &  0 &  1 &  \frac{7}{16} &  \frac{1}{16} &  0 &  0 &  0 \\[1mm] 0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  1 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] \frac{45}{8} &  -6 &  \frac{5}{8} &  3 &  0 &  \frac{3}{8} &  \frac{1}{8} &  0 &  0 &  0 \\[1mm] \frac{135}{16} &  -8 &  \frac{15}{16} &  3 &  0 &  -\frac{7}{16} &  -\frac{1}{16} &  0 &  0 &  0 \\[1mm] \frac{27}{8} &  -3 &  \frac{3}{8} &  1 &  0 &  -\frac{3}{8} &  -\frac{1}{8} &  0 &  0 &  0\\[1mm] \end{smallmatrix} \right).\tag {32}\label{f.32} \end{equation}
  </div>
  <span class="equation_label">34</span>
</p>
</div>
<p> The transformation matrix \(T\) in <a href="#f.32" class="eqref">34</a> is as in <a href="#f.30" class="eqref">32</a>. The eigenvalues of stability matrix of <a href="#f.32" class="eqref">34</a> are \(\lbrace {1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{72}, 0, 0, 0, 0, 0, 0\rbrace }\), see [19]. There is only one non-zero eigenvalue and is thus RK is nearly stable because of the term \(\frac{z^4}{72}\). The interval of absolute stability of the method in (32) is \([-3.117, 0]\), see <br />Fig. 2. </p>
<figure >
  <div class="centered"> <img src="img-0002.png" alt="\includegraphics[width=12cm]{Okuonghae/Nordsieck1b.png}" style="width:12cm" />
<br />Fig. 2. Region of absolute stability of the Nordsieck methods in (29) and (32). </div>

</figure>
<p> The coefficients of the first predictor of the SD-HLMM<a href="#f.11" class="eqref">18</a> for step number \(k=2\) are: </p>
<div class="displaymath" id="a0000000036">
  \begin{align}  u_{11}^{(1)}& =\left(1+c_1\right){}^3 \left(1-3 c_1+6 c_1^2\right),&  u_{12}^{(1)}& =-c_1^3 \left(10+3 c_1 \left(5+2 c_1\right)\right), \nonumber \\ u_{11}^{(2)}& = -c_1 \left(1+c_1\right){}^3 \left(-1+3 c_1\right),&  u_{12}^{(2)}& = -c_1^3 \left(1+c_1\right) \left(4+3 c_1\right), \tag {33}\label{f.33} \\ u_{11}^{(3)}& = \tfrac {1}{2} c_1^2 \left(1+c_1\right){}^3,&  u_{12}^{(3)}& = -\tfrac {1}{2} c_1^3 \left(1+c_1\right){}^2.\nonumber \end{align}
</div>
<p> If \(c_{1}\neq -1\), a family of predictors exist and \(0\leq c_{1} \leq 1\). Inserting \(c_{1}=\frac{2}{3}\) into the coefficients and substituting the resulting values into <a href="#f.11" class="eqref">18</a> gives the first predictor scheme </p>
<div class="displaymath" id="f.34">
  \begin{align}  y_{n-\frac{1}{3}}=&  \tfrac {625 }{81}y_{n-1}-\tfrac {544 }{81}y_{n-2}+h\left(-\tfrac {250}{81} f_{n-1}-\tfrac {80}{27} f_{n-2}\right)\tag {34}\label{f.34} \\ & +h^2\left(\tfrac {250}{243} f’_{n-1}-\tfrac {100}{243} f’_{n-2}\right),\quad p=5.\nonumber \end{align}
</div>
<p> The relevant equations for the second predictor <a href="#f.12" class="eqref">19</a> using <a href="#f.22" class="eqref">24</a> are thus obtained and following the above process leads to the coefficients in (A) of the appendix. Fixing \(c_{1}=\frac{2}{3}\) and \(c_{2}= 1\) into (A) gives <a href="#f.12" class="eqref">19</a> as </p>
<div class="displaymath" id="f.35">
  \begin{align}  \widehat{y}_{n}=&  2 y_{n-1}-y_{n-2}+h\left(\tfrac {243 }{1000}f_{n-\frac{1}{3}}+\tfrac {1}{8} f_{n-1}-\tfrac {46}{125}f_{n-2}\right)\tag {35}\label{f.35} \\ & +h^2\left(\tfrac {27}{200}f’_{n-\frac{1}{3}}+\tfrac {3}{8}f’_{n-1}-\tfrac {1}{25}f’_{n-2}\right),\quad p=7.\nonumber \end{align}
</div>
<p> The coefficients in terms of \((c, t)\) are very large expressions hence we fix the abscissa value of the output method as \(t=1\) as defined in section 2 to give the coefficients of the output method in <a href="#f.13" class="eqref">20</a> in terms of the other abscissa values \(c= \left[c_{1}, c_{2}\right]\). The expressions are in (B) of the appendix. Now inserting \(c=[\frac{2}{3}, 1]\) into (B) and substituting the resulting values into <a href="#f.13" class="eqref">20</a> gives </p>
<div class="displaymath" id="f.36">
  \begin{align}  y_{n}=& y_{n-1}+h\left(\tfrac {60993}{140000}f_{n-\frac{1}{3}}+\tfrac {883}{3360} \widehat{f}_{n}+\tfrac {1009 }{3360} f_{n-1}+\tfrac {521}{420000} f_{n-2}\right)\tag {36}\label{f.36} \\ & +h^2\left(\tfrac {-1863}{28000} f’_{n-\frac{1}{3}}-\tfrac {59 }{3360}\widehat{f'}_{n}+\tfrac {107 }{3360} f’_{n-1}+\tfrac {1}{4000}f’_{n-2}\right),\quad p=8.\nonumber \end{align}
</div>
<p> Composing the methods in <a href="#f.34" class="eqref">36</a>, <a href="#f.35" class="eqref">37</a> and <a href="#f.36" class="eqref">38</a> and reorganizing their data in SD-GLM format as in (19a) yields </p>
<div class="equation" id="f.37">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c|c|c} A^{(1)}&  A^{(2)} & U \\ \hline B^{(1)}&  B^{(2)} & V\\ \end{array} \right)= \left( \begin{smallmatrix} 0 &  0 & 0&  0 &  0& 0& \\[1mm] \frac{243}{1000} &  0 &  0 & \frac{27}{200} & 0& 0&  U\\[1mm] 1 &  0 &  \frac{1009}{3360} &  \frac{521 }{420000} &  \frac{107}{3360} &  \frac{1}{4000}\\[1mm] \hline \frac{60993}{140000} &  \frac{883 }{3360} &  0 & \frac{-1863}{28000}& \frac{-59}{3360}& 0&  \\[1mm] 0 &  0 & 0& 0& 0& 0& \\[1mm] 0& 0&  1 & 0 & 0& 0&  \\[1mm] 0 &  0 &  0 & 0& 0& 0&  V \\[1mm] 0 &  0 &  0 & 0& 0& 1&  \\[1mm] 0 &  0 &  0 & 0& 0& 0&  \\[1mm] \end{smallmatrix} \right) \tag {37}\label{f.37} \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000037">
  \[ U=\left( \begin{smallmatrix} & \frac{625 }{81} &  -\frac{544 }{81} &  -\frac{250}{81} &  -\frac{80}{27} &  \frac{250}{243} &  -\frac{100}{243}\\[1mm] & 2 &  -1 &  \frac{1}{8} &  -\frac{46}{125} &  \frac{3}{8} &  -\frac{1}{25} \\[1mm] & 1 &  0 &  \frac{1009}{3360} &  \frac{521 }{420000} &  \frac{107}{3360} &  \frac{1}{4000}\\[1mm] \end{smallmatrix} \right),\  \  V=\left(\begin{smallmatrix} 1 &  0 &  \frac{1009}{3360} &  \frac{521 }{420000} &  \frac{107}{3360} &  \frac{1}{4000}\\[1mm] 1& 0& 0& 0& 0& 0\\[1mm] 0& 0& 0& 0& 0& 0 \\[1mm] 0& 0& 1& 0& 0& 0 \\[1mm] 0& 0& 0& 0& 0& 0 \\[1mm] 0& 0& 0& 0& 1& 0 \\[1mm] \end{smallmatrix} \right). \]
</div>
<p> Putting all these details together as in (18) with \(T\) in <a href="#f.21" class="eqref">23</a> we find the coefficient matrices of the Nordsieck method to be </p>
<div class="equation" id="f.38">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{smallmatrix} 0 &  0 &  0 & 0 &  0 &  0 \\[1mm] \frac{243 }{1000} &  0 &  0& \frac{27}{200} &  0 &  0& UT \\[1mm] \frac{60993 }{140000} &  \frac{883 }{3360} &  0& \frac{-1863 }{28000} &  \frac{-59 }{3360} &  0 &  \\[1mm] \hline & T^{-1}B^{(1)}& &  & T^{-1}B^{(2)}&  & T^{-1}VT \\[1mm] \end{smallmatrix} \right), \tag {38}\label{f.38} \end{equation}
  </div>
  <span class="equation_label">40</span>
</p>
</div>
<p> with \(k=2\), where \(UT\), \(T^{-1}B^{(1)}, T^{-1}B^{(2)}, T^{-1}VT\) are in appendix C. By verification \(\operatorname {Det}(wI-\overline{V})=w^{k-1}(w-1)\) thus guaranteeing zero stability of the Nordsieck method in (19a). The stability polynomial of <a href="#f.38" class="eqref">40</a> is </p>
<div class="displaymath" id="a0000000038">
  \begin{align*} & w(w-1)+\tfrac {51 z}{16}-\tfrac {67 w z}{16} +\tfrac {757 z^2}{560}+\tfrac {187 w z^2}{140}+\tfrac {3881 z^3}{10080}-\tfrac {50 w z^3}{63}+\tfrac {1507 z^4}{20160} \\ & +\tfrac {119 w z^4}{960}+\tfrac {11 z^5}{1890}-\tfrac {4769 w z^5}{120960}-\tfrac {59 z^6}{60480}+\tfrac {59 w z^6}{24192}. \end{align*}
</div>
<p> The interval of absolute stability is \(\left(-0.2342, 0\right)\), see Fig. 4. The stability interval of explicit Nordsieck methods as with explicit LMM generally are usually very severely limited compared to their implicit counterpart; this is compensated by the flexibility in their automatic stepsize implementation allowed by such methods. </p>
<p>From <a href="#f.33" class="eqref">35</a>, <a href="#f.35" class="eqref">37</a> and (B) with \(c=[\frac{1}{2}, 1, 1]^{T}\), the SD-HLMM for \(k=2\) is </p>
<div class="displaymath" id="f.39">
  \begin{align}  y_{n-\tfrac {1}{2}}\! =\!  &  \tfrac {27 }{8}y_{n-1}\! -\! \tfrac {19 }{8}y_{n-2}\! +\! h\Big(\! -\! \tfrac {27}{32} f_{n-1}\! -\! \tfrac {33}{32}f_{n-2}\Big) \tag {39}\label{f.39} \! +\! h^2\Big(\tfrac {27}{64}f’_{n-1}\! -\! \tfrac {9}{64}f’_{n-2}\Big),\quad p\! =\! 5,\nonumber \\ y_n=& \tfrac {128 }{29}y_{n-1}-\tfrac {99 }{29}y_{n-2}+h\Big(-\tfrac {1792}{783}f_{n-\tfrac {1}{2}}+\tfrac {32}{29}f_{n-1}-\tfrac {962}{783}f_{n-2}\Big)\nonumber \\ & +h^2\left.\Big(\tfrac {64}{87}f’_{n-\tfrac {1}{2}}+\tfrac {44}{29}f’_{n-1}-\tfrac {34}{261}f’_{n-2}\Big)\right.,\quad p=7,\nonumber \\ y_n =& y_{n-1}+h\Big(\tfrac {32}{63} f_{n-\tfrac {1}{2}}+\tfrac {23 }{112}f_n+\tfrac {2}{7} f_{n-1}+\tfrac {1}{1008}f_{n-2}\Big)\nonumber \\ & +h^2 \Big(\tfrac {8}{315} f’_{n-\tfrac {1}{2}}-\tfrac {1}{80} f’_n+\tfrac {1}{35} f’_{n-1}+\tfrac {1}{5040}f’_{n-2}\Big),\quad p=8.\nonumber \end{align}
</div>
<p> Its Nordsieck GLM form is </p>
<div class="equation" id="f.40">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{smallmatrix} 0 &  0 &  0 &  0 &  0 &  0 \\[1mm] -\frac{1792}{783} &  0 &  0 &  \frac{64}{87} &  0 &  0& UT \\[1mm] \frac{32}{63} &  \frac{23 }{112} &  0 &  \frac{8}{315} &  -\frac{1}{80} &  0 \\[1mm] \hline & T^{-1}B^{(1)}& &  & T^{-1}B^{(2)}&  & T^{-1}VT \\[1mm] \end{smallmatrix} \right), \tag {40}\label{f.40} \end{equation}
  </div>
  <span class="equation_label">41</span>
</p>
</div>
<p> where \(UT\), \(T^{-1}B^{(1)}, T^{-1}B^{(2)}, T^{-1}VT\) are in appendix D. To distinguish between the wide range of special subcases within <a href="#f.38" class="eqref">40</a> and <a href="#f.40" class="eqref">41</a>, with a view to making a choice for practical use, we consider the stability region for the given abscissae vector \(c=[\frac{1}{2}, 1, 1]^T\). The stability polynomial \(\det (wI-M(z))\) of <a href="#f.40" class="eqref">41</a> is </p>
<div class="displaymath" id="a0000000039">
  \begin{align*} & w(w-1)+\tfrac {79 z}{56}-\tfrac {135 w z}{56}-\tfrac {475 z^2}{1624}+\tfrac {2759 w z^2}{1624}+\tfrac {311 z^3}{6090}-\tfrac {1516 w z^3}{1015}+\tfrac {2419 z^4}{24360} \\ & +\tfrac {79 w z^4}{203}+\tfrac {769 z^5}{48720}-\tfrac {1357 w z^5}{16240}-\tfrac {3 z^6}{2320}+\tfrac {9 w z^6}{2320}. \end{align*}
</div>
<p> The interval of absolute stability is \(\left(-0.4, 0\right)\), see Fig. 3. </p>
<figure >
  <div class="centered"> <img src="img-0003.png" alt="\includegraphics[width=12cm]{Okuonghae/Nordsieck2.png}" style="width:12cm" />
<br />Fig. 3. Region of absolute stability of the Nordsieck methods in (38) <br />and (40). </div>

</figure>
<h1 id="a0000000040">4  Variable stepsize implementation of the Nordsieck GLM in (5b)</h1>
<p> Take note that Section 3 has shown how <a href="#f.11" class="eqref">18</a>-<a href="#f.13" class="eqref">20</a> can practically be represented as the Nordsieck method in (5b) and found in (18a,b). An effective implementation of (18a,b) requires step-size changing and rescaling of the vector of the external approximations as it becomes necessary, see <span class="cite">
	[
	<a href="#Wr2" >17</a>
	]
</span>. The variable stepsize form of (5b) is </p>
<div class="displaymath" id="a0000000041">
  \begin{align}  Y=& h_{n}A^{(1)}F(Y)+h_{n}^{2}A^{(2)}F’(Y)+UTy^{[n-1]},\nonumber \\ y^{[n]}=& h_{n}D(r)T^{-1}B^{(1)}F(Y)+h_{n}^{2}D(r)T^{-1}B^{(2)}F’(Y)+D(r)T^{-1}VTy^{[n-1]} \tag {41}\label{f.41} \end{align}
</div>
<p> where the rescaling matrix is given by </p>
<div class="equation" id="f.42">
<p>
  <div class="equation_content">
    \begin{equation}  D(r)=\operatorname {diag}\left(1, r, r^2,\cdots ,r^p \right),\quad r=h_{n}/h_{n-1}, \tag {42}\label{f.42} \end{equation}
  </div>
  <span class="equation_label">43</span>
</p>
</div>
<p> and </p>
<div class="equation" id="f.43">
<p>
  <div class="equation_content">
    \begin{equation}  h_{n}=rh_{n-1};\quad r=\min \Big\{  \max \Big\{  0.5, 0.9*\big(\tfrac {TOL}{\| \Delta _{n}\| }\big)^{1/(p+1)}\Big\}  ,2.0\Big\} ,\tag {43}\label{f.43} \end{equation}
  </div>
  <span class="equation_label">44</span>
</p>
</div>
<p> where \(n= 1, 2,\ldots ,\quad TOL=10^{-m};\quad m\gg 1\). <a href="#f.43" class="eqref">44</a> is the step-size required to advance from step \(n-1\) to \(n\), if the last step was successful, while \(h_{n-1}\) represents the step-size in the last attempt either a successful or a failed step, \(p\) denotes the order of the corrector formula and \(TOL\) is the allowable error estimate specified by the user. The \(\Delta _{n}\) in <a href="#f.43" class="eqref">44</a> is the local error estimate with the condition \(\| \Delta _{n}\| {\lt} TOL\). If it passes this condition, the step is successful and the next step-size will be obtained using the step-size increment formula in <a href="#f.43" class="eqref">44</a>. We compute the local error estimate using a linear combination of the stage derivatives of the current step only. That is, </p>
<div class="equation" id="f.44">
<p>
  <div class="equation_content">
    \begin{equation}  \Omega _{p+1}h_{n}^{p+1}y_{(x_{n-1})}^{(p+1)}=h_{n}\sum _{i=1}^{s}\Theta _{i}F(Y)+h_{n}^{2}\sum _{i=1}^{s}\psi _{i} F'(Y) \tag {44}\label{f.44} \end{equation}
  </div>
  <span class="equation_label">45</span>
</p>
</div>
<p> where \(\Omega _{p+1}\) is the error constant of the output point of the Nordsieck method. Expanding <a href="#f.44" class="eqref">45</a> using the Taylor series, it follows that the values of \(\Theta _{i}\) and \(\psi _{i}\) satisfy the resulting system of equations </p>
<div class="displaymath" id="a0000000042">
  \begin{align}  \sum _{i=1}^{p}\Theta _{i}=0,\quad \sum _{i=1}^{p}\psi _{i}+\sum _{i=1}^{p}\Theta _{i}c_{i}=0,\quad \sum _{i=1}^{p}\psi _{i}c_{i}+\sum _{i=1}^{p}\tfrac {\Theta _{i}c_{i}^{2}}{2!}=0,\nonumber \\ \sum _{i=1}^{p}\tfrac {\psi _{i}c_{i}^{2}}{2!}+\sum _{i=1}^{p}\tfrac {\Theta _{i}c_{i}^{3}}{3!}=0,\cdots , \sum _{i=1}^{p}\tfrac {\psi _{i}c_{i}^{q-2}}{(q-2)!}+\sum _{i=1}^{p}\tfrac {\Theta _{i}c_{i}^{q-1}}{(q-1)!}=0,\tag {45}\label{f.45}\\ \sum _{i=1}^{p}\tfrac {\psi _{i}c_{i}^{q-1}}{(q-1)!}+\sum _{i=1}^{p}\tfrac {\Theta _{i}c_{i}^{q}}{(q)!}=1, \quad q=5, 6, \ldots \nonumber \end{align}
</div>
<p> see [12], page 86-87. Consider for example, the method <a href="#f.29" class="eqref">31</a> of order \(p=4\), \(\Omega _{5}=\frac{1}{720}\) with abscissa vector \(c=[\frac{1}{2}, 1]^{T}\) in <a href="#f.43" class="eqref">44</a>. Solving the resulting system of linear equations in <a href="#f.45" class="eqref">46</a> yields no value for the variables \(\Theta _{1}\), \(\Theta _{2}\), \(\psi _{1}\) and \(\psi _{2}\). No local error estimate exist for the methods in <a href="#f.32" class="eqref">34</a>, <a href="#f.38" class="eqref">40</a> and <a href="#f.40" class="eqref">41</a> from <a href="#f.44" class="eqref">45</a>, therefore, an alternative approach to estimate the error is adopted. To control the error in <a href="#f.29" class="eqref">31</a> and <a href="#f.32" class="eqref">34</a>, \(N=3\) in <a href="#f.22" class="eqref">24</a> to obtain a method of order 3, one order less than the corrector. The resulting error estimators for <a href="#f.29" class="eqref">31</a>\((j=1)\) and <a href="#f.32" class="eqref">34</a> \((j=2)\) will therefore be based on \(\| y_{n}^{e_{j}}-y_{1}^{[n]}\| \), \(j=1, 2\) with </p>
<div class="displaymath" id="a0000000043">
  \begin{align*}  y_n^{e_{1}}= &  y_{n-1}+\tfrac {h}{3} \left(4 f_{n-\frac{1}{2}}-f_{n-1}\right)-\tfrac {h^2}{6}f’_{n-1},\quad p=3,& \\ y_n^{e_{2}}= &  y_{n-1}+ \tfrac {h}{4} \left(f_{n-1}+3 f_{n-\frac{1}{3}}\right),\quad f_n=f(x_{n},y_{1}^{[n]}),\quad p=2. \end{align*}
</div>
<p>Similarly, for the Nordsieck methods in <a href="#f.38" class="eqref">40</a> \((j=3)\) and <a href="#f.40" class="eqref">41</a> \((j=4)\) the methods adopted to control the error can be: </p>
<div class="displaymath" id="a0000000044">
  \begin{align*}  y_n^{e_{3}}=& y_{n-1}+h \left(\tfrac {1}{5250}f_{n-2}+\tfrac {473}{1680}f_{n-1}+\tfrac {5589}{14000}f_{n-\frac{1}{3}}+\tfrac {67}{210}f_n\right) \\ & +h^2 \left(\tfrac {43}{1680}f’_{n-1}-\tfrac {243}{2800}f’_{n-\tfrac {1}{3}}-\tfrac {1}{42}f’_n\right),\quad f_n=f(x_{n},y_{1}^{[n]}),\quad p=7, \\ y_n^{e_{4}}=& y_{n-1}+h \left(\tfrac {1}{7560}f_{n-2}+\tfrac {9}{35}f_{n-1}+\tfrac {61}{280}f_n+\tfrac {496}{945}f_{n-\frac{1}{2}}\right) \\ & +h^2 \left(\tfrac {3}{140}f’_{n-1}+\tfrac {4}{315}f’_{n-\frac{1}{2}}-\tfrac {1}{70}f_n\right), \quad p=7. \end{align*}
</div>
<p>The \(\Delta _{n}\) in <a href="#f.43" class="eqref">44</a> is now \(\| y_{n}^{e_{j}}-y_{1}^{[n]}\| \), \(j=3, 4\) and is the error estimate for the proposed Nordsieck methods in <a href="#f.38" class="eqref">40</a>, and <a href="#f.40" class="eqref">41</a>. This is defined as the difference between the output methods of proposed Nordsieck methods in <a href="#f.38" class="eqref">40</a>, and <a href="#f.40" class="eqref">41</a>, and the alternative scheme used to control the error. As a result of the asymptotic behaviour of \(\| y_{n}^{e_{j}}-y_{1}^{[n]}\| , j= 3, 4\) we use \(r\) in <a href="#f.43" class="eqref">44</a> to increase or decrease the error predicted in the next step by multiplying it by \(h_{n-1}\), (see <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>, pp. 130 - 131, Algorithm 271\(\alpha \)). The availability of initial data \(y_{i}^{[n-1]}\) facilitates easy adjustment of stepsize. The Nordsieck vectors (3b) in <a href="#f.41" class="eqref">42</a> are the outgoing and incoming approximation at step number \(n\) respectively. The application of <a href="#f.41" class="eqref">42</a> to the scalar test problem \(y'(x)=0\), yields </p>
<div class="equation" id="f.46">
<p>
  <div class="equation_content">
    \begin{equation}  y^{[n]}=D(r)T^{-1}VTy^{[n-1]}. \tag {46}\label{f.46} \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> The zero stability property of the method <a href="#f.41" class="eqref">42</a> is determined by the eigenvalues of the matrix \(D(r)T^{-1}VT\), see <span class="cite">
	[
	<a href="#Hu1" >12</a>
	]
</span> and <span class="cite">
	[
	<a href="#Wr2" >17</a>
	]
</span>. Consider for example the matrix \(T^{-1}VT\) in (29a) given by </p>
<div class="displaymath" id="a0000000045">
  \[ T^{-1}VT=\left( \begin{smallmatrix} 1 &  1 &  \frac{1}{6} &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  6 &  \frac{7}{6} &  0 &  0 &  0 \\[1mm] 0 &  8 &  \frac{3}{2} &  0 &  0 &  0 \\[1mm] 0 &  3 &  \frac{1}{2} &  0 &  0 &  0\\[1mm] \end{smallmatrix} \right). \]
</div>
<p> The eigenvalues of \(D(r)T^{-1}VT\) from (29a) are found to be \(\{ 1,0,0,0,0,0\} \). We obtain similarly the eigenvalues of the methods in <a href="#f.38" class="eqref">40</a> and <a href="#f.40" class="eqref">41</a> as<br />\(\lbrace 1,\frac{1}{3}, 0,0 ,0,0,0,0,0 \rbrace \) and \(\lbrace 1,\frac{1}{2}, 0,0 ,0,0,0,0,0 \rbrace \) respectively. This means that the Nordsieck methods are zero-stable. </p>
<h1 id="a0000000046">5 The RK and SD-RK starters</h1>
<p> To start up the methods (5b), we find an approximation to the Nordsieck vector </p>
<div class="equation" id="f.47">
<p>
  <div class="equation_content">
    \begin{equation}  y^{[0]}=\left(\begin{array}{cccccccccccccccc} y(x_{0})& h_{0}y’(x_{0})&  h_{0}^{2}y”(x_{0})& \ldots &  h_{0}^{p}y^{(p)}(x_{0}) \end{array}\right)^{T},\tag {47}\label{f.47} \end{equation}
  </div>
  <span class="equation_label">48</span>
</p>
</div>
<p> by use of a RK starter </p>
<div class="displaymath" id="a0000000047">
  \[ {\begin{split} \begin{array}{ll} Y= h\overline{A}F(Y)+\overline{U}y_{0},& \\ y^{[0]}= h\overline{B}F(Y)+\overline{V}y_{0}. \end{array}\end{split}} \leqno {(48a)}  \]
</div>
<p> In the search of a high order starter for a given stage number, the SD-RK starter </p>
<div class="displaymath" id="a0000000048">
  \[ {\begin{split} \begin{array}{ll} Y=h^{2}AF’(Y)+ h\overline{A}F(Y)+\overline{U}y_{0}, & \\ y^{[0]}= h^{2}BF’(Y)+h\overline{B}F(Y)+\overline{V}y_{0}, \end{array}\end{split}} \leqno {(48b)}  \]
</div>
<p> comes in handy to advance the solution at \(x_{0}\) through one single step to \(x_{0}+h\). The vectors of \(\overline{U}\) and \(\overline{V}\) in (48a) and (48b) are \(\left(1,1,\ldots ,1\right)^{T},\) and \(\left(1,0,\ldots ,0\right)^{T}\) respectively. In the Nordsieck vector, the first and second components of <a href="#f.47" class="eqref">48</a> are known from the initial values defining the ODE <a href="#equation1" class="eqref">1</a>, but we can determine the other components using a Runge-Kutta method with multiple outputs. To obtain (48a) we use the internal stages of the method given by </p>
<div class="displaymath" id="a0000000049">
  \[ Y_{i}=h\sum _{j=1}^{i-1}\overline{a}_{ij}F(Y_{j})+ y_{0}, \quad \widehat{c}_{i}\in [0, 1],\leqno {(49a)} \]
</div>
<p> and for (48b) we use </p>
<div class="displaymath" id="a0000000050">
  \[  Y_{i}=h^{2}\sum _{j=1}^{i-1}a_{ij}F'(Y_{j})+h\sum _{j=1}^{i-1}\overline{a}_{ij}F(Y_{j})+ y_{0}, \quad \widehat{c}_{i}\in [0, 1],\leqno {(49b)}  \]
</div>
<p> where \(Y_{i}=y(x_{0}+\widehat{c}_{i}h)\). Expanding (49a) by Taylor series and equating the powers of \(h\) yields the first stage order condition as </p>
<div class="equation" id="f.50">
<p>
  <div class="equation_content">
    \begin{equation}  \overline{u}_{11}=1, \overline{a}_{11}=0, \tag {50}\label{f.50} \end{equation}
  </div>
  <span class="equation_label">51</span>
</p>
</div>
<p> for the second stage we have </p>
<div class="equation" id="f.51">
<p>
  <div class="equation_content">
    \begin{equation}  \overline{u}_{21}=1, \sum _{j=1}^{1}\overline{a}_{2j}=\widehat{c}_{2}, \tag {51}\label{f.51} \end{equation}
  </div>
  <span class="equation_label">52</span>
</p>
</div>
<p> while the third stage gives </p>
<div class="equation" id="f.52">
<p>
  <div class="equation_content">
    \begin{equation}  \overline{u}_{31}=1, \sum _{j=1}^{2}\overline{a}_{3j}=\widehat{c}_{3}, \sum _{j=1}^{2}\overline{a}_{3j}\widehat{c}_{j}=\tfrac {\widehat{c}_{3}^{2}}{2!}, \tag {52}\label{f.52} \end{equation}
  </div>
  <span class="equation_label">53</span>
</p>
</div>
<p> and the fourth stage order conditions are </p>
<div class="equation" id="f.53">
<p>
  <div class="equation_content">
    \begin{equation}  \overline{u}_{41}=1, \sum _{j=1}^{3}\overline{a}_{4j}=\widehat{c}_{4}, \sum _{j=1}^{3}\overline{a}_{4j}\widehat{c}_{j}=\tfrac {\widehat{c}_{4}^{2}}{2!}, \tfrac {1}{2!}\sum _{j=1}^{3}\overline{a}_{4j}\widehat{c}_{j}^{2}=\tfrac {\widehat{c}_{4}^{3}}{3!}. \tag {53}\label{f.53} \end{equation}
  </div>
  <span class="equation_label">54</span>
</p>
</div>
<p> Also, the fifth stage order conditions are </p>
<div class="displaymath" id="f.54">
  \begin{align}  \overline{u}_{51}& =1, \sum _{j=1}^{4}\overline{a}_{5j}=\widehat{c}_{5}, \tag {54}\label{f.54} \\ \sum _{j=1}^{4}\overline{a}_{5j}\widehat{c}_{j}& =\tfrac {\widehat{c}_{5}^{2}}{2!}, \tfrac {1}{2!}\sum _{j=1}^{4}\overline{a}_{5j}\widehat{c}_{j}^{2}=\tfrac {\widehat{c}_{5}^{3}}{3!}, \tfrac {1}{3!}\sum _{j=1}^{4}\overline{a}_{5j}\widehat{c}_{j}^{3}=\tfrac {\widehat{c}_{5}^{4}}{4!}. \nonumber \end{align}
</div>
<p> Let \(\widehat{c}=[0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1]^{T}\) in <a href="#f.50" class="eqref">51</a>-<a href="#f.54" class="eqref">55</a> and solving the resulting system of linear equations we obtain the value \(\overline{a}_{ij}'s\), of matrix \(\overline{A}\). </p>
<p>Similarly, the components of the output method in (48a) are obtained using </p>
<div class="equation" id="f.55">
<p>
  <div class="equation_content">
    \begin{equation}  y^{[0]}=y(x_{0}+th)=\sum _{j=1}^{s}\overline{b}_{ij}F(Y_{j})+y_{0}.\tag {55}\label{f.55} \end{equation}
  </div>
  <span class="equation_label">56</span>
</p>
</div>
<p> The \(t=0\), gives the output value at \(x=x_{0}\). If we set \(t=1\) the output values will be for the first point \(x_{1}=x_{0}+h\). Expanding <a href="#f.55" class="eqref">56</a> by Taylor series and equating both sides in powers of \(h\), the first component \(y_{1}^{[0]}=y(x_{0}+th)\), yields </p>
<div class="displaymath" id="f.56">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{1j}& =t,&  \sum _{j=1}^{5}\overline{b}_{1j}\widehat{c}_{j}& =\tfrac {1}{2!}t^2,& \tag {56}\label{f.56} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{1j}\widehat{c}_{j}^{2}& =\tfrac {1}{3!}t^3,&  \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{1j}\widehat{c}_{j}^{3}& =\tfrac {1}{4!}t^4,&  \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{1j}\widehat{c}_{j}^{4}& =\tfrac {1}{5!}t^5.\nonumber \end{align}
</div>
<p> The system of linear equations for the second component \(y_{2}^{[0]}=hy'(x_{0}+th)\) is </p>
<div class="displaymath" id="f.57">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{2j}& =1,&  \sum _{j=1}^{5}\overline{b}_{2j}\widehat{c}_{j}& =t,& \tag {57}\label{f.57} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{2j}\widehat{c}_{j}^{2}& =\tfrac {t^2}{2!},&  \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{2j}\widehat{c}_{j}^{3}& =\tfrac {t^3}{3!},&  \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{2j}\widehat{c}_{j}^{4}& =\tfrac {t^4}{4!},\nonumber \end{align}
</div>
<p> The third component \(y_{3}^{[0]}=h^{2}y''(x_{0}+th)\) gives </p>
<div class="displaymath" id="f.58">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{3j}& =0, \sum _{j=1}^{5}\overline{b}_{3j}\widehat{c}_{j}=1,\tag {58}\label{f.58} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{3j}\widehat{c}_{j}^{2}& =t, \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{3j}\widehat{c}_{j}^{3}=\tfrac {t^2}{2!}, \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{3j} \widehat{c}_{j}^{4}=\tfrac {t^3}{3!}, \nonumber \end{align}
</div>
<p> In the same manner, the fourth component \(y_{4}^{[0]}=h^{3}y'''(x_{0}+th)\) of the output method (48a) gives </p>
<div class="displaymath" id="f.59">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{4j}& =0,&  \sum _{j=1}^{5}\overline{b}_{4j}\widehat{c}_{j}& =0,& \tag {59}\label{f.59} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{4j}\widehat{c}_{j}^{2}& =1,&  \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{4j}\widehat{c}_{j}^{3}& =t,&  \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{4j}\widehat{c}_{j}^{4}& =\tfrac {t^2}{2!}.\nonumber \end{align}
</div>
<p> Similarly, the fifth component \(y_{5}^{[0]}=h^{4}y^{(4)}(x_{0}+th)\) yields </p>
<div class="displaymath" id="f.60">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{5j}& =0,&  \sum _{j=1}^{5}\overline{b}_{5j}\widehat{c}_{j}& =0,& \tag {60}\label{f.60} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{5j}\widehat{c}_{j}^{2}& =0,&  \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{5j}\widehat{c}_{j}^{3}& =1,&  \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{5j}\widehat{c}_{j}^{4}& =t,\nonumber \end{align}
</div>
<p> while the sixth component \(y_{6}^{[0]}=h^{5}y^{(5)}(x_{0}+th)\) gives </p>
<div class="displaymath" id="f.61">
  \begin{align}  \sum _{j=1}^{5}\overline{b}_{6j}& =0,&  \sum _{j=1}^{5}\overline{b}_{6j}\widehat{c}_{j}& =0,&  \tag {61}\label{f.61} \\ \tfrac {1}{2!}\sum _{j=1}^{5}\overline{b}_{6j}\widehat{c}_{j}^{2}& =0,&  \tfrac {1}{3!}\sum _{j=1}^{5}\overline{b}_{6j}\widehat{c}_{j}^{3}& =0,&  \tfrac {1}{4!}\sum _{j=1}^{5}\overline{b}_{6j}\widehat{c}_{j}^{4}& =1.\nonumber \end{align}
</div>
<p> The GLM representation of (48b) is defined by </p>
<div class="equation" id="f.62">
<p>
  <div class="equation_content">
    \begin{equation}  \left( \begin{array}{c|c} \overline{A}&  \overline{U} \\  \overline{B_{t}}& \overline{V} \\ \end{array} \right);\quad \left( \begin{array}{c|c|c} \overline{A}& A&  \overline{U} \\  \overline{B_{t}}& B_{t}& \overline{V} \\ \end{array} \right).\tag {62}\label{f.62} \end{equation}
  </div>
  <span class="equation_label">63</span>
</p>
</div>
<p> respectively. Again, let \(\widehat{c}=[0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1]^{T}\) and \(t=0\) in <a href="#f.56" class="eqref">57</a>-<a href="#f.61" class="eqref">62</a> and solving the resulting system of linear equations respectively gives \(\overline{b}_{ij}'s\) and their values are in matrix \(\overline{B}\) in (62), given here as, </p>
<div class="displaymath" id="a0000000051">
  \[ \left( \begin{array}{c|cccccc} \overline{A}& \overline{U}\\  \overline{B}_{t=0}& \overline{V}\\ \end{array}\right)=\left(\begin{smallmatrix} 0& 0& 0& 0& 0& 1\\[1mm] \frac{1}{4}& 0& 0& 0& 0& 1\\[1mm] 0& \frac{1}{2}& 0& 0& 0& 1\\[1mm] \frac{3}{16}& 0& \frac{9}{16}& 0& 0& 1\\[1mm] 0& \frac{2}{3}& -\frac{1}{3}& \frac{2}{3}& 0& 1\\[1mm] \hline \\[0mm] & & \overline{B}_{t=0}& & & \overline{V}\\[1mm] \end{smallmatrix}\right), \]
</div>
<div class="displaymath" id="a0000000052">
  \[  \left(\begin{array}{c|ccc}\overline{B}_{t=0}& \overline{V}\end{array}\right)= \left(\begin{smallmatrix} 0& 0& 0& 0& 0& 1\\ 1& 0& 0& 0& 0& 0\\ -\frac{25}{3}&  16& -12&  \frac{16}{3}& -1& 0\\[1mm] \frac{140}{3}& -\frac{416}{3}& 152& -\frac{224}{3}& \frac{44}{3}& 0\\[1mm] -160 &  576 &  -768& 448& -96& 0\\[1mm] 256&  -1024& 1536&  -1024&  256& 0\\ \end{smallmatrix}\right). \leqno {(63a)}  \]
</div>
<p> At \(t=1\) we have </p>
<div class="displaymath" id="a0000000053">
  \[ \left( \begin{array}{c|cccccc} \overline{A}& \overline{U}\\  \overline{B}_{t=1}& \overline{V} \end{array}\right)=\left(\begin{smallmatrix} 0& 0& 0& 0& 0& 1\\[1mm] \frac{1}{4}& 0& 0& 0& 0& 1\\[1mm] 0& \frac{1}{2}& 0& 0& 0& 1\\[1mm] \frac{3}{16}& 0& \frac{9}{16}& 0& 0& 1\\[1mm] 0& \frac{2}{3}& -\frac{1}{3}& \frac{2}{3}& 0& 1\\[1mm] \hline \\[0mm] & & \overline{B}_{t=1}& & & \overline{V}\\[1mm] \end{smallmatrix}\right), \]
</div>
<div class="displaymath" id="a0000000054">
  \[ \left(\begin{array}{c|ccc}\overline{B}_{t=1}& \overline{V}\end{array}\right)= \left(\begin{smallmatrix} \frac{7}{90}& \frac{16}{45}& \frac{2}{15}& \frac{16}{45}& \frac{7}{90}& 1\\[1mm] 0& 0& 0& 0& 1& 0\\[1mm] 1& -\frac{16}{3}& 12&  -16& \frac{25}{3}& 0\\[1mm] \frac{44}{3}& -\frac{224}{3}& 152& -\frac{416}{3}& \frac{140}{3}& 0\\[1mm] 96 & -448 &  768& -576& 160& 0\\[1mm] 256&  -1024& 1536&  -1024&  256& 0\\[1mm] \end{smallmatrix}\right). \leqno {(63b)}  \]
</div>
<p>Following this, a starting method for the Nordsieck method in <a href="#f.38" class="eqref">40</a> and <a href="#f.40" class="eqref">41</a> is derived. The outputs of the scheme <a href="#f.62" class="eqref">63</a> are the starter for the explicit Nordsieck method in <a href="#f.29" class="eqref">31</a> and <a href="#f.32" class="eqref">34</a> respectively. As was noted in <span class="cite">
	[
	<a href="#Hu1" >12</a>
	]
</span>, this idea of using a RK method with multiple outputs as starter for a Nordsieck method was introduced by Gear<span class="cite">
	[
	<a href="#Ge3" >8</a>
	]
</span>. This can readily and analogously extended to SD-RK starters through (48b). </p>
<h1 id="a0000000055">6 Numerical Experiments and Conclusion</h1>
<p>To illustrate the application of the schemes, and for the purpose of comparison consider implementation of the methods </p>
<p>(i). order four Adams-Bashforth methods (ABF4) in <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>, page 116, Table 244(I), </p>
<p>(ii). order four Runge-Kutta method (RK4) in <span class="cite">
	[
	<a href="#Bu5" >3</a>
	]
</span>, page 102, (235i), </p>
<p>(iii). order four second derivative Nordsieck method (SDNM4) in <a href="#f.32" class="eqref">34</a>, </p>
<p>(iv). order four second derivative GLM with nearly Almost Runge-Kutta methods stability (SDNARK4) in <span class="cite">
	[
	<a href="#Ok9" >20</a>
	]
</span> on the following DETEST problems: </p>
<p><em>Problem</em> 1: </p>
<div class="displaymath" id="a0000000056">
  \begin{eqnarray*}  y’(x) = -\tfrac {y^3}{2},\quad y(0) = 1,\quad y(x) = 1/\sqrt{1+x},\quad x \in [0, 5]. \end{eqnarray*}
</div>
<p><em>Problem</em> 2: </p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray*}  \left\{ \begin{array}{lll} y_{1}’(x) = y_{2}^2-2y_{1},&  y_{1}(0) = 0,&  y_{1}(x) = xe^{-2x},\\ y_{2}’(x) = y_{1}-y_{2}-xy_{2}^2,&  y_{2}(0) = 1,&  y_{2}(x) = e^{-x},\\ x \in [0, 1]. \end{array}\right. \end{eqnarray*}
</div>
<p><em>Problem</em> 3: A nonlinear chemical reaction. </p>
<div class="displaymath" id="a0000000058">
  \begin{eqnarray*}  \left\{ \begin{array}{lll} y_{1}’(x) = -y_{1},&  y_{1}(0) = 1,& \\ y_{2}’(x) = y_{1}-y_{2}^{2}, &  y_{2}(0) = 0,& \\ y_{3}’(x) = y_{2}^{2},&  y_{3}(0) = 0,& \\ x \in [0, 5]. \end{array}\right. \end{eqnarray*}
</div>
<p>In Tables 1–3, \(TOL\) is the tolerance, \(ns\) represent the number of total step, \(nfs\) is the number of failed step, \(nfe\) denotes the number of function evaluations, \(Error\) is the \(|y_{exact}-y_{n}|\), while \(\| \Delta _{n}\| _{\infty }\) represents the maximum error of the error estimate. The \(y_{exact}\) and \(y_{n}\) are the exact and the numerical solutions respectively. The results obtained by each of the fourth order methods for Problems 1–3 with \(Tol=10^{-j}, j=2,4,6\) are given in Tables 1–3. </p>
<p><small class="footnotesize"><div class="table"  id="tab:my_label">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(ns\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nrs\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nfe\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(11\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(40\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(9\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(32\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(8\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(21\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(64\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(252\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(39\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(160\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(25\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(100\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(16\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(45\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6351\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(9\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(25436\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(316\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(4\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1276\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(99\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(400\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(38\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(114\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\( 635065\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(11 \)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(2540300 \)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Results for solving Problem 1. Output \(x=5\), starting \(h=0.1\).</span> 
</figcaption>  </div>
</div></small> </p>
<p><small class="footnotesize"><div class="table"  id="tab:my_label">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Error\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(||\Delta _{n}||_{\infty }\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Time\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(7.4843\times 10^{-2}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.2649\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.010317\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.3662\times 10^{-6}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.8535\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.006815\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.8506\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.4734\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.005878\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.7540\times 10^{-6}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.006422\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(7.5249\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.4075\times 10^{-5}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.010725\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.1057\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.4688\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.009433\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.2355\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.4720\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.006053\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.2965\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.204725\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.7454\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.3987\times 10^{-7}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.019703\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.1923\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.9825\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.010058\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.3229\times 10^{-6}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.2121\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.006593\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(3.7533 \times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(2525.875703\)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Continuation of Table 1.</span> 
</figcaption>  </div>
</div> </small> </p>
<p>Table 1 shows that the SDNM4 <a href="#f.32" class="eqref">34</a> (<i class="it">i.e.</i>, the order 4 second derivative Nordsieck method) performs better than ABF4, RK4 and SDNARK4 in terms of accuracy and computational cost. The main advantage over other RK methods is that its stage order is two. In addition, the transformation of the SDHLMM <a href="#f.11" class="eqref">18</a>–<a href="#f.13" class="eqref">20</a> to Nordsieck method (18a) also enhanced the performance of the SDNM4 (32) in variable step-size implementation. </p>
<p><small class="footnotesize"><div class="table"  id="Table 2:">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(ns\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nrs\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nfe\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(28 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(108\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(9\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(32\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(15\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(27\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(104\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(214 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(852\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(23\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(88\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(12\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(33\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1796\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(7188\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2078\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(3\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(8320\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(89\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(352\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(30\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(87\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(179623\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(9 \)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(718524\)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Results for solving problem 2. Output \(x=1\), starting \(h=0.001\).</span> 
</figcaption>  </div>
</div> </small> </p>
<p><small class="footnotesize"><div class="table"  id="tab:my_label">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Max Error</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Min Error</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.1963\times 10^{-2}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.0052\times 10^{-3}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.5545\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.0623\times 10^{-4}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\( 7.3033\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.8690\times 10^{-4}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.4695 \times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.7568 \times 10^{-3}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2.2598\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.4308\times 10^{-4}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.6210 \times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.4258 \times 10^{-7}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(9.7776 \times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.5894 \times 10^{-7}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.5733\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.6534 \times 10^{-5}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.2749\times 10^{-4}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(9.2037\times 10^{-5}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.5644\times 10^{-9}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.2320 \times 10^{-9}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(3.9004 \times 10^{-9}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.1259 \times 10^{-9}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(2.5277\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(6.6029\times 10^{-5}\)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Continuation of Table 2.</span> 
</figcaption>  </div>
</div> </small> </p>
<p><small class="footnotesize"><div class="table"  id="tab:my_label">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(\| \Delta _{n}\| _{\infty }\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Time\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.4844\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0210\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(9.3170\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0124\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.5921\times 10^{-4}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0134\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.54\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0149\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.4298\times 10^{-5}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0343\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(3.7849\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0172\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.3744\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0147\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0683\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.5476\times 10^{-7}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.1079\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.5693\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0187\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.6670 \times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.0174 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(287.43\)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">Continuation of Table 2.</span> 
</figcaption>  </div>
</div> </small> </p>
<p>The numerical results given in Table 2 are the results of evidences for the very good performance of the proposed method compared with the methods considered for comparison. In terms of computational cost, the SDNM4 in <a href="#f.32" class="eqref">34</a> outperform the ABF4, RK4, and SDNARK4, for details, see Table 2. </p>
<p><small class="footnotesize"><div class="table"  id="Table 3:">
   <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> Method</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Tol\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(ns\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nrs\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(nfe\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(\| \Delta _{n}||_{\infty }\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(Time\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(45 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(184\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.0653\times 10^{-3}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.026\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(13\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(48\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.4439\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.021\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(9\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(24\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(2.2116\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.015\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-2}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(110\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(3\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(448\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.5609\times 10^{-3}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.021\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(406\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(4\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1636\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.3484\times 10^{-5}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.040\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(44\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(176\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.9075\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.018\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(21\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(0\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(60\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(4.0231\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.020\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-4}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(10999\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(9\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(44028\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-5}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(1.254\)</p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>ABF4</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(4067\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(5\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(16284\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.5387\times 10^{-7}\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.342\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>RK4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(187\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(2\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(752\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(6.1335\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.028\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>SDNM4</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(57\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(1\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> \(171\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(5.4834 \times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(0.017 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>SDNARK4</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(10^{-6}\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(541521\) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(12 \)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(2166128\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(6.5610\times 10^{-7}\)</p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(19329.266\)</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">6</span> 
  <span class="caption_text">Results for solving problem 3. Output \(x=5\), starting \(h=0.1\).</span> 
</figcaption>  </div>
</div> </small> </p>
<p>The numerical results given in Table 3 confirm the improved performance of the SDNM4 over ABF4, RK4, and SDNARK4 in terms of accuracy and computational cost. In fact, in all the problems solved the SDNM4 has the smallest number of function evaluations when compared with other methods considered herein. However, the numerical results presented in Tables 1, 2 and 3 show the accuracy of the new scheme. Observe from Tables 1–3, as the tolerance becomes smaller, the better is the accuracy of the SDNM4 in <a href="#f.32" class="eqref">34</a>; however, SDNARK4 at high tolerance gives poor response time. </p>
<p>Conclusively, we have described in this paper the construction of second derivative Nordsieck methods (5b) for nonstiff differential systems <a href="#equation1" class="eqref">1</a>. This approach can be extended to a higher step numbers and off-step points in <a href="#f.11" class="eqref">18</a>–<a href="#f.13" class="eqref">20</a>. We have shown thus that it is possible to extend the methods of <span class="cite">
	[
	<a href="#Bu7" >4</a>
	]
</span> to multiple hybrid points with Nordsieck implementation. <div class="acknowledgement_thmwrapper " id="a0000000059">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgements
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors wish to express their most sincere thanks to Prof. J.C. Butcher (Univ. of Auckland, Auckland, New Zealand) for the help to implement these methods. </p>

  </div>
</div> </p>
<h1 id="a0000000060">Appendix A</h1>
<div class="displaymath" id="a0000000061">
  \begin{align*}  u_{21}^{(1)} =&  \tfrac {2+7 c_1^2 \left(1+c_2\right){}^3 \left(1-3 c_2+6 c_2^2\right)+2 c_2^5 \left(21+5 c_2 \left(7+3 c_2\right)\right)}{2+7 c_1 \left(1+c_1\right)}\nonumber \tfrac {-7 c_1 \left(-1+c_2^4 \left(15+2 c_2 \left(12+5 c_2\right)\right)\right)}{2+7 c_1 \left(1+c_1\right)},\nonumber \\ u_{22}^{(1)}=&  \tfrac {c_2^3 \left(-7 c_1^2 \left(10+3 c_2 \left(5+2 c_2\right)\right)-2 c_2^2 \left(21+5 c_2 \left(7+3\right. c_2\right)\right)}{2+7 c_1 \left(1+c_1\right)}\nonumber +\tfrac {\left.7 c_1 c_2 \left(15+2 c_2 \left(12+5 c_2\right)\right)\right)}{2+7 c_1 \left(1+c_1\right)}, \nonumber \\ u_{21}^{(2)}=&  \tfrac {-\left(c_2 \left(1+c_2\right){}^3 \left(-c_1^3 \left(2+7 c_1 \left(1+c_1\right)\right)+3 c_1^3 \left(2+7 c_1 \left(1+c_1\right)\right) c_2\right.\right.}{\left(c_1^3 \left(2+7 c_1 \left(1+c_1\right)\right) \right)},\nonumber \\ & +\tfrac {\left.\left.c_1 \left(2+c_1 \left(2-c_1 \left(17+35 c_1\right)\right)\right) c_2^2+\left(-1+c_1 \left(-2+c_1 \left(4+15 c_1\right)\right)\right) c_2^3\right)\right)}{\left(c_1^3 \left(2+7 c_1 \left(1+c_1\right)\right) \right)},\nonumber \\ u_{22}^{(2)} =& \tfrac {(c_2^3 (c_1^2(20+c_1 (68+7 c_1 (11+4 c_1))) }{((1+c_1){}^3 (2+7 c_1 (1+c_1)))}+\tfrac {c_1 (-30+c_1 (-69+c_1 (-1+7 c_1 (13+7 c_1)))) c_2+12 c_2^2)}{((1+c_1){}^3 (2+7 c_1 (1+c_1)))}\nonumber \\ & -\tfrac {c_2^5 (3 c_1 (-4+c_1 (-41+7 c_1 (-7+ (-1+c_1 ) c_1 ) ) )}{((1+c_1){}^3 (2+7 c_1 (1+c_1)))}+\tfrac {22 c_2+c_1 (53-7 c_1 (-1+c_1 (9+5 c_1))) c_2}{((1+c_1){}^3 (2+7 c_1 (1+c_1)))}\nonumber \\ & +\tfrac {(10+c_1t(35+c_1 (41+15 c_1))) c_2^2))}{((1+c_1){}^3 (2+7 c_1 (1+c_1)) )},\nonumber \\ u_{21}^{(3)} =&  \tfrac {c_2^2 \left(1+c_2\right){}^3 \left(3 c_1^2 \left(2+7 c_1 \left(1+c_1\right)\right)\right.}{6 c_1^2 \left(2+7 c_1 \left(1+c_1\right)\right)}\- \tfrac {\left.c_1 \left(8+c_1 \left(31+35 c_1\right)\right) c_2+3 \left(1+c_1 \left(4+5 c_1\right)\right) c_2^2\right)}{6 c_1^2 \left(2+7 c_1 \left(1+c_1\right)\right)},\nonumber \\ u_{22}^{(3)}=&  -\tfrac {c_2^3 (1+c_2){}^2(c_1^2(10+7 c_1(4+3 c_1))}{c_1^3 (1+c_1){}^3 (2+7 c_1 (1+c_1))}-\tfrac {c_1 (15+c_1 (44+35 c_1)) c_2+3(2+c_1 (6+5 c_1)) c_2^2)}{6 (1+c_1){}^2 (2+7 c_1 (1+c_1))},\nonumber \\ a_{21}^{(1)}=& -\tfrac {c_2^3 \left(1+c_2\right){}^3 \left(c_2+c_1 \left(-2+5 c_2+c_1 \left(-8-7 c_1+5 c_2\right)\right)\right)}{c_1^3 \left(1+c_1\right){}^3 \left(2+7 c_1 \left(1+c_1\right)\right)},\nonumber \\ a_{21}^{(2)}=&  \tfrac {c_2^3 \left(1+c_2\right){}^3 \left(3 c_2+c_1 \left(-4-7 c_1+6 c_2\right)\right)}{6 c_1^2 \left(1+c_1\right){}^2 \left(2+7 c_1 \left(1+c_1\right)\right)}. \end{align*}
</div>
<h1 id="a0000000062">Appendix B</h1>
<div class="displaymath" id="a0000000063">
  \begin{align*}  b_{11}^{(1)}=&  \tfrac {(14 c_1^3 (142+7 c_2 (-49+31 c_2 ) )}{(420 (c_1-c_2){}^3 c_2^3 (1+c_2){}^3)}\tfrac {-7 c_1^2 (5+2 c_2 (103+c_2 (-253+155 c_2))))}{(420 c_1^3 (1+c_1){}^3 (c_1-c_2){}^3)}, \\ b_{12}^{(1)}=&  \tfrac {(c_2 (970+7 (5-284 c_2) c_2)+14 c_1^3 (-49+5 c_2 (-1+31 c_2))}{(420 (c_1-c_2){}^3 c_2^3 (1+c_2){}^3)}\! +\! \tfrac {2 c_1^2 (568+c_2 (793-7 c_2 (253+217 c_2)))}{(420 (c_1-c_2){}^3 c_2^3 (1+c_2){}^3)}\\ & +\tfrac {c_1 (-485+2 c_2 (-1195+7 c_2 (103+343 c_2 ))))}{(420 (c_1-c_2){}^3 c_2^3 (1+c_2){}^3)}, \\ b_{11}^{(2)}=&  \tfrac {(485+2 c_2 (-568+343 c_2)-4 c_1 (142+7 c_2 (-49+31 c_2 )))}{(840 c_1^2 (1+c_1){}^2 (c_1-c_2){}^2)}, \\ v_{11}^{(1)}=&  1, \qquad v_{12}^{(1)}= 0, \\ b_{12}^{(2)}=&  \tfrac {(485-568 c_2+2 c_1 (-568+7 c_1 (49-62 c_2)+686 c_2))}{(840 (c_1-c_2){}^2 c_2^2(1+c_2){}^2)}, \\ v_{11}^{(2)}=&  \tfrac {(8 c_1^2 (-1+c_2 ) (142+7 c_2 (-49+31 c_2))}{(420 c_1^3 c_2^3)}+\tfrac {c_2 (485+2 c_2 (-568+343 c_2))}{(420 c_1^3 c_2^3)}\\ & +\tfrac {c_1 (485+c_2 (-2473+4 (970-497 c_2) c_2))}{(420 c_1^3 c_2^3)}+\tfrac {14 c_1^3 (49+c_2 (-142+(124-15 c_2) c_2)))}{(420 c_1^3 c_2^3)}, \\ v_{12}^{(2)}=&  \tfrac {(735+2 c_2 (-595-86 c_2+350 c_2^2)}{(420 (1+c_1){}^3 (1+c_2){}^3)}+\tfrac {c_1 (-1190+c_2 (1957+2 (181-665 c_2) c_2))}{(420 (1+c_1){}^3 (1+c_2){}^3)}+ \end{align*}
</div>
<div class="displaymath" id="a0000000064">
  \begin{align*} & +\tfrac {2 c_1^2 (-86+c_2 (181+7 c_2 (-19+11 c_2)))}{(420 (1+c_1){}^3 (1+c_2){}^3)}+\tfrac {14 c_1^3 (50+c_2 (-95+c_2 (11+45 c_2))))}{(420 (1+c_1){}^3 (1+c_2){}^3)}, \\ v_{11}^{(3)}=&  \tfrac {(485+2 c_2 (-568+343 c_2)-8 c_1 (142+7 c_2 (-49+31 c_2))}{(840 c_1^2 c_2^2)}+\tfrac {14 c_1^2 (49+c_2 (-124+85 c_2)))}{(840 c_1^2 c_2^2)}, \\ v_{12}^{(3)}=&  \tfrac {(225+4 c_2 (-130+77 c_2)-4 c_1 (130+7 c_2 (-44+27 c_2))}{(840 (1+c_1){}^2 (1+c_2){}^2)}+\tfrac {14 c_1^2 (22+c_2 (-54+35 c_2)))}{(840 (1+c_1){}^2 (1+c_2){}^2)}. \end{align*}
</div>
<h1 id="a0000000065">Appendix C</h1>
<div class="displaymath" id="a0000000066">
  \[ T^{-1}B^{(1)}=\left( \begin{smallmatrix} \frac{60993}{140000} &  \frac{883}{3360} &  0 \\[1mm] 0 &  0 &  1 \\[1mm] 0 &  0 &  0 \\[1mm] \frac{6038307}{560000} &  \frac{29139}{4480} &  -12 \\[1mm] \frac{66055419}{2240000} &  \frac{318763}{17920} &  -\frac{117}{4} \\[1mm] \frac{75204369}{2240000} &  \frac{362913}{17920} &  -\frac{501}{16} \\[1mm] \frac{43488009}{2240000} &  \frac{629579}{53760} &  -\frac{279}{16} \\[1mm] \frac{12625551}{2240000} &  \frac{60927}{17920} &  -\frac{79}{16} \\[1mm] \frac{182979}{280000} &  \frac{883}{2240} &  -\frac{9}{16} \end{smallmatrix} \right),\qquad T^{-1}B^{(2)}= \left( \begin{smallmatrix} -\frac{1863}{28000} &  -\frac{59}{3360} &  0 \\[1mm] 0 &  0 &  0 \\[1mm] 0 &  0 &  1 \\[1mm] -\frac{184437}{112000} &  -\frac{1947}{4480} &  \frac{9}{2} \\[1mm] -\frac{2017629}{448000} &  -\frac{21299}{17920} &  \frac{33}{4} \\[1mm] -\frac{2297079}{448000} &  -\frac{24249}{17920} &  \frac{63}{8} \\[1mm] -\frac{1328319}{448000} &  -\frac{42067}{53760} &  \frac{33}{8} \\[1mm] -\frac{385641}{448000} &  -\frac{4071}{17920} &  \frac{9}{8} \\[1mm] -\frac{5589}{56000} &  -\frac{59}{2240} &  \frac{1}{8} \end{smallmatrix} \right),  \]
</div>
<div class="displaymath" id="a0000000067">
  \[ \quad UT= \left( \begin{smallmatrix} 1 &  \frac{2}{3} &  -\frac{142}{243} &  \frac{8}{27} &  \frac{16}{81} &  \frac{32}{243} &  -\frac{104}{81} &  \frac{88}{27} &  -\frac{1472}{243} \\[1mm] 1 &  \frac{757}{1000} &  \frac{31}{1000} &  \frac{17}{125} &  -\frac{1}{125} &  -\frac{1}{25} &  \frac{1}{125} &  \frac{13}{125} &  -\frac{37}{125} \\[1mm] 1 &  \frac{63323}{210000} &  \frac{4181}{140000} &  \frac{311}{140000} &  -\frac{103}{52500} &  \frac{101}{84000} &  \frac{1}{17500} &  -\frac{109}{60000} &  \frac{107}{26250} \end{smallmatrix} \right), \]
</div>
<div class="displaymath" id="a0000000068">
  \[ T^{-1}VT= \left( \begin{smallmatrix} 1 &  \frac{63323}{210000} &  \frac{4181}{140000} &  \frac{311}{140000} &  -\frac{103}{52500} &  \frac{101}{84000} &  \frac{1}{17500} &  -\frac{109}{60000} &  \frac{107}{26250} \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  -\frac{1480341}{280000} &  -\frac{286081}{560000} &  -\frac{669211}{560000} &  \frac{14101}{70000} &  \frac{31333}{112000} &  -\frac{17401}{70000} &  -\frac{23597}{80000} &  \frac{47281}{35000} \\[1mm] 0 &  -\frac{20190397}{1120000} &  -\frac{3171977}{2240000} &  -\frac{10163187}{2240000} &  \frac{190317}{280000} &  \frac{512461}{448000} &  -\frac{261417}{280000} &  -\frac{419349}{320000} &  \frac{782377}{140000} \\[1mm] 0 &  -\frac{25214247}{1120000} &  -\frac{2264827}{2240000} &  -\frac{14736537}{2240000} &  \frac{220167}{280000} &  \frac{825511}{448000} &  -\frac{366267}{280000} &  -\frac{764799}{320000} &  \frac{1295227}{140000} \\[1mm] 0 &  -\frac{45990701}{3360000} &  \frac{41053}{2240000} &  -\frac{10138257}{2240000} &  \frac{294061}{840000} &  \frac{1920013}{1344000} &  -\frac{226787}{280000} &  -\frac{2117717}{960000} &  \frac{3200041}{420000} \\[1mm] 0 &  -\frac{4590713}{1120000} &  \frac{445467}{2240000} &  -\frac{3295623}{2240000} &  \frac{10393}{280000} &  \frac{230969}{448000} &  -\frac{52293}{280000} &  -\frac{327521}{320000} &  \frac{436133}{140000} \\[1mm] 0 &  -\frac{67927}{140000} &  \frac{12543}{280000} &  -\frac{51567}{280000} &  -\frac{103}{35000} &  \frac{3601}{56000} &  \frac{3}{35000} &  -\frac{7609}{40000} &  \frac{8857}{17500} \end{smallmatrix} \right). \]
</div>
<div class="displaymath" id="a0000000069">
  \[ T^{-1}B^{(1)}= \left( \begin{smallmatrix} \frac{32}{63} &  \frac{23}{112} &  0 \\[1mm] 0 &  0 &  1 \\[1mm] 0 &  0 &  0 \\[1mm] \frac{88}{7} &  \frac{2277}{448} &  -12 \\[1mm] \frac{722}{21} &  \frac{24909}{1792} &  -\frac{117}{4} \\[1mm] \frac{274}{7} &  \frac{28359}{1792} &  -\frac{501}{16} \\[1mm] \frac{1426}{63} &  \frac{16399}{1792} &  -\frac{279}{16} \\[1mm] \frac{46}{7} &  \frac{4761}{1792} &  -\frac{79}{16} \\[1mm] \frac{16}{21} &  \frac{69}{224} &  -\frac{9}{16} \end{smallmatrix} \right) ,\quad T^{-1}B^{(2)}= \left( \begin{smallmatrix} \frac{8}{315} &  -\frac{1}{80} &  0 \\[1mm] 0 &  0 &  0 \\[1mm] 0 &  0 &  1 \\[1mm] \frac{22}{35} &  -\frac{99}{320} &  \frac{9}{2} \\[1mm] \frac{361}{210} &  -\frac{1083}{1280} &  \frac{33}{4} \\[1mm] \frac{137}{70} &  -\frac{1233}{1280} &  \frac{63}{8} \\[1mm] \frac{713}{630} &  -\frac{713}{1280} &  \frac{33}{8} \\[1mm] \frac{23}{70} &  -\frac{207}{1280} &  \frac{9}{8} \\[1mm] \frac{4}{105} &  -\frac{3}{160} &  \frac{1}{8} \end{smallmatrix} \right),  \]
</div>
<h1 id="a0000000070">7 Appendix D</h1>
<div class="displaymath" id="a0000000071">
  \[ \quad UT= \left( \begin{smallmatrix} 1 &  \frac{1}{2} &  -\frac{11}{64} &  \frac{1}{8} &  \frac{1}{16} &  \frac{1}{32} &  -\frac{13}{32} &  \frac{17}{16} &  -2 \\[1mm] 1 &  \frac{685}{783} &  \frac{2125}{783} &  -\frac{497}{261} &  \frac{1841}{783} &  -\frac{1987}{783} &  \frac{643}{261} &  -\frac{1667}{783} &  \frac{1201}{783} \\[1mm] 1 &  \frac{289}{1008} &  \frac{17}{630} &  \frac{1}{560} &  -\frac{1}{630} &  \frac{1}{1008} &  0 &  -\frac{1}{720} &  \frac{1}{315} \end{smallmatrix} \right), \]
</div>
<div class="displaymath" id="a0000000072">
  \[ T^{-1}VT=\left( \begin{smallmatrix} 1 &  \frac{289}{1008} &  \frac{17}{630} &  \frac{1}{560} &  -\frac{1}{630} &  \frac{1}{1008} &  0 &  -\frac{1}{720} &  \frac{1}{315} \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 \\[1mm] 0 &  -\frac{2533}{448} &  -\frac{163}{280} &  -\frac{2701}{2240} &  \frac{59}{280} &  \frac{123}{448} &  -\frac{1}{4} &  -\frac{91}{320} &  \frac{93}{70} \\[1mm] 0 &  -\frac{102311}{5376} &  -\frac{5413}{3360} &  -\frac{40917}{8960} &  \frac{2369}{3360} &  \frac{6073}{5376} &  -\frac{15}{16} &  -\frac{4921}{3840} &  \frac{4643}{840} \\[1mm] 0 &  -\frac{42391}{1792} &  -\frac{1381}{1120} &  -\frac{59247}{8960} &  \frac{913}{1120} &  \frac{3273}{1792} &  -\frac{21}{16} &  -\frac{3017}{1280} &  \frac{2571}{280} \\[1mm] 0 &  -\frac{231415}{16128} &  -\frac{1109}{10080} &  -\frac{40727}{8960} &  \frac{3697}{10080} &  \frac{22889}{16128} &  -\frac{13}{16} &  -\frac{25193}{11520} &  \frac{19099}{2520} \\[1mm] 0 &  -\frac{7689}{1792} &  \frac{181}{1120} &  -\frac{13233}{8960} &  \frac{47}{1120} &  \frac{919}{1792} &  -\frac{3}{16} &  -\frac{1303}{1280} &  \frac{869}{280} \\[1mm] 0 &  -\frac{341}{672} &  \frac{17}{420} &  -\frac{207}{1120} &  -\frac{1}{420} &  \frac{43}{672} &  0 &  -\frac{91}{480} &  \frac{53}{105} \end{smallmatrix} \right). \]
</div>
<p><small class="footnotesize">  </small></p>
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</a> </p>
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</dl>


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