Table 2.

Absolute errors for Application 1, case II.

${\mathit{t}}_{\mathit{i}}$	${\mathit{E}}_8\left({\underset{\_}{\mathit{\omega }}}_{\mathit{\rho }}\right)=\left|{\underset{\_}{\mathit{\omega }}}_{\mathit{\rho }}\left(\mathit{t}\right)-{\underset{\_}{\mathit{\omega }}}_{\mathit{\rho }}^8\left(\mathit{t}\right)\right|$
	$\mathit{\rho }=0$	$\mathit{\rho }=0.5$	$\mathit{\rho }=1$
$0.16$	$0.0$	$0.0$	$0.0$
$0.32$	$4.00\times {10}^{-10}$	$0.0$	$0.0$
$0.48$	$1.10\times {10}^{-8}$	$1.00\times {10}^{-8}$	$8.00\times {10}^{-9}$
$0.64$	$1.53\times {10}^{-7}$	$1.29\times {10}^{-7}$	$1.06\times {10}^{-7}$
$0.80$	$1.15\times {10}^{-7}$	$9.75\times {10}^{-7}$	$8.04\times {10}^{-7}$
$0.96$	$5.96\times {10}^{-6}$	$5.09\times {10}^{-6}$	$4.22\times {10}^{-6}$
$t_i$	$E_8\left({\overline{\omega }}_{\rho }\right)=\left|{\overline{\omega }}_{\rho }\left(t\right)-{\overline{\omega }}_{\rho }^8\left(t\right)\right|$
	$\rho =0$	$\rho =0.5$	$\rho =1$
$0.16$	$0.0$	$0.0$	$0.0$
$0.32$	$0.0$	$0.0$	$0.0$
$0.48$	$4.00\times {10}^{-9}$	$6.00\times {10}^{-9}$	$8.00\times {10}^{-9}$
$0.64$	$5.90\times {10}^{-8}$	$8.30\times {10}^{-8}$	$1.06\times {10}^{-7}$
$0.80$	$4.61\times {10}^{-7}$	$6.32\times {10}^{-7}$	$8.04\times {10}^{-7}$
$0.96$	$2.48\times {10}^{-6}$	$3.35\times {10}^{-6}$	$4.22\times {10}^{-6}$