## 2 Fractional-Order Scheme Development, Construction, and Analysis
\begin{array}{c}2\,c_{3}(\Gamma(\gamma))^{2\gamma 2}\pi^{3/2}(2 \gamma^{2})\Gamma(\gamma+1/2)-2\,(2\gamma^{2})^{4}\\ (\Gamma(\gamma+1/2))^{2}c_{2}^{2}\Gamma(\gamma)\gamma\,\pi\\ -c_{3}(3\gamma)^{3}\sqrt{3}\Gamma(\gamma+1/3)\Gamma(\gamma+2/3)\\ \Gamma(\gamma)\gamma\,\pi+2\,(2\gamma^{2})^{6}(\Gamma(\gamma+1/2))^{3}c_{2} ^{2}\sqrt{\pi}\end{array} \right),\] \[\Lambda_{6} = \left(\frac{c_{2}^{2}\Big{(}\Gamma(\gamma)\gamma\,\sqrt{\pi}-(2 \gamma)^{2}\Gamma(\gamma+1/2)\Big{)}}{(\Gamma(\gamma))^{3}\,\gamma^{3}\pi^{2} (2\gamma^{2})^{2}\Gamma(\gamma+1/2)}\right),\] \[\Lambda_{7} = \left(\frac{c_{2}^{3}\Big{(}\Gamma(\gamma)\gamma\,\sqrt{\pi}-(2 \gamma)^{2}\Gamma(\gamma+1/2)\Big{)}^{2}}{(\Gamma(\gamma))^{2}\gamma^{2}\pi }\right).\]  
By multiplying (14) by the inverse of \(\hbar\Big{(}x^{[j]}\Big{)}-2\hbar\Big{(}y^{[j]}\Big{)}\), we obtain  
\[\frac{\hbar\Big{(}x^{[j]}\Big{)}}{\hbar\Big{(}x^{[j]}\Big{)}-2\hbar\big{(}y^ {[j]}\Big{)}}=1+\Lambda_{8}\epsilon^{[j]}+\Lambda_{9}\Big{[}\epsilon^{[j]} \Big{]}^{\gamma+1}+\Lambda_{10}\Big{[}\epsilon^{[j]}\Big{]}^{2\gamma+1}+... \tag{20}\]  
where  
\[\Lambda_{8} = \left(2\,\frac{c_{2}(2\gamma)^{2}\Gamma(\gamma+1/2)}{\Gamma( \gamma)\gamma\sqrt{\pi}}-2\,c_{2}\right),\] \[\Lambda_{9} = \left( \begin{array}{c}\frac{c_{3}(3\gamma)^{3}\sqrt{3}}{\Gamma( \gamma+1/3)\Gamma(\gamma+2/3)}\\ \frac{c_{3}(3\gamma)^{3}\sqrt{3}\Gamma(\gamma+1/3)\Gamma(\gamma+2/3)}{\Gamma( \gamma+1)}\frac{\Gamma(\gamma+\frac{1}{2})}{\Gamma(\gamma+1)}\\ -8\,\frac{(2\gamma)^{2}\Gamma(\gamma+1/2)c_{2}^{2}}{\Gamma(\gamma)\gamma\sqrt {\pi}}+2\,\frac{(2\gamma)^{4}(\Gamma(\gamma+1/2))^{2}c_{2}^{2}}{(\Gamma(\gamma ))^{2}\gamma^{2}\pi}\end{array} \right),\] \[\Lambda_{10} = \left( \begin{array}{c}-5\,\frac{c_{2}c_{3}(3\gamma)^{3}\sqrt{3} \Gamma(\gamma+1/3)\Gamma(\gamma+2/3)}{\Gamma(\gamma)\gamma\sqrt{\pi}(2)^{2} \Gamma(\gamma+1/2)}+\\ 4\,\frac{c_{2}c_{3}(3\gamma)^{3}\sqrt{3}\Gamma(\gamma+1/3)\Gamma(\gamma+2/3)}{ \Gamma(\gamma)^{2}\gamma^{2}\pi}-12\,\frac{c_{3}^{2}(2\gamma)^{4}(\Gamma( \gamma+1/2))^{2}}{(\Gamma(\gamma))^{2}\gamma^{2}\pi}\\ +28\,\frac{c_{3}^{2}(2\gamma)^{2}\Gamma(\gamma+1/2)}{\Gamma(\gamma)\gamma\sqrt {\pi}}-16\,c_{2}{}^{3}+12\,c_{2}c_{3}-10\,\frac{c_{2}(2\gamma)^{2}\Gamma( \gamma+1/2)c_{2}}{\Gamma(\gamma)\gamma\,\sqrt{\pi}}\end{array} \right).\]  
Using the result above, we express  
\[\frac{\hbar\Big{(}y^{[j]}\Big{)}}{\big{[}D_{\gamma}^{\gamma}\big{]}\hbar\big{(} x^{[j]}\big{)}}=\xi=\Lambda_{11}\Big{[}\epsilon^{[j]}\Big{]}^{\gamma+1}+ \Lambda_{12}\Big{[}\epsilon^{[j]}\Big{]}^{2\gamma+1}+\Lambda_{13}\Big{[}\epsilon^ {[j]}\Big{]}^{3\gamma+1}+\cdots \tag{21}\]with coefficients  
\[\Lambda_{11} = \left(-\frac{c_{2}c_{3}(3^{7})^{3}\sqrt{3}}{(\Gamma(\gamma))^{7} \gamma^{3}\pi^{2}}\Gamma\Big{(}\ \gamma+\frac{1}{3}\Big{)}\ \Gamma\left(\ \gamma+\frac{2}{3}\right)\right),\] \[\Lambda_{12} = \left( \begin{array}{c}\frac{c_{3}(3^{7})^{3}\sqrt{3}}{2\left( \Gamma(\gamma)\right)^{7}2^{3}\sqrt{\pi}(2\gamma)^{2}}\frac{\Gamma\left(\ \gamma+\frac{1}{3}\right)\Gamma\left(\ \gamma+\frac{2}{3}\right)}{\Gamma\left(\ \gamma+\frac{1}{3}\right)}\\ -\frac{c_{3}}{\Gamma(\gamma)\gamma}-2\ \frac{(2\gamma^{2})^{4}(\Gamma(\gamma+1/2))^{2}c_{2} ^{2}}{(\Gamma(\gamma))^{3}\gamma^{3}\pi}+2\ \frac{(2\gamma^{2})^{2}\Gamma(\gamma+1/2)c_{2} ^{2}}{(\Gamma(\gamma))^{7}2^{7}\sqrt{\pi}}\end{array} \right),\] \[\Lambda_{13} = \left(