## 2 Fractional-Order Scheme Development, Construction, and Analysis
: Let \(\xi\) be a simple root of \(\hbar\), and define the current iterate as \(x^{[j]}=\xi+\varepsilon^{[j]}\). By expanding \(\hbar(x^{[j]})\) and \(\left[D_{\gamma_{1}}^{\gamma}\right]\hbar(x^{[j]})\) in Taylor series around \(x=\xi\) and noting that \(\hbar(\xi)=0\), we obtain  
\[\hbar(x^{[j]})=\frac{\left[D_{\xi}^{\gamma}\right]\hbar(\xi)}{\Gamma(\gamma+1 )}\bigg{[}\Big{[}\varepsilon^{[j]}\Big{]}^{\gamma}+c_{2}\Big{[}\varepsilon^{[ j]}\Big{]}^{2\gamma}+c_{3}\Big{[}\varepsilon^{[j]}\Big{]}^{3\gamma} \bigg{]}+O\bigg{(}\Big{[}\varepsilon^{[j]}\Big{]}^{4\gamma}\bigg{)}. \tag{14}\]  
Taking the fractional derivative of (14) yields  
\[\big{[}D_{\gamma_{1}}^{\gamma}\big{]}\hbar(x^{[j]})=\frac{\left[D_{\xi}^{ \gamma}\right]\hbar(\xi)}{\Gamma(\gamma+1)}\left[ \begin{array}{c}\Gamma( \gamma+1)+\frac{\Gamma(2\gamma+1)}{\Gamma(\gamma+1)}c_{2}\Big{[}\varepsilon^ {[j]}\Big{]}^{\gamma}\\ +\frac{\Gamma(3\gamma+1)}{\Gamma(2\gamma+1)}c_{3}\Big{[}\varepsilon^{[j]} \Big{]}^{2\gamma}\end{array} \right]+O\bigg{(}\Big{[}\varepsilon^{[j]}\Big{]} ^{3\gamma}\bigg{)}. \tag{15}\]  
By dividing (14) by (15), we obtain  
\[\frac{\hbar(x^{[j]})}{\left[D_{\gamma_{1}}^{\gamma}\right]\hbar(x^{[j]})}= \left[ \begin{array}{c}\frac{[\varepsilon^{[j]}]^{\gamma}}{\Gamma(\gamma) \gamma}+\Lambda_{1}\Big{[}\varepsilon^{[j]}\Big{]}^{2\gamma}+\Lambda_{2}\Big{[} \varepsilon^{[j]}\Big{]}^{3\gamma}+O\bigg{(}\Big{[}\varepsilon^{[j]}\Big{]} ^{4\gamma}\bigg{)}\end{array} \right], \tag{16}\]  
where  
\[\Lambda_{1} = \left(\frac{c_{2}}{\Gamma(\gamma)\gamma}-\frac{(2^{\kappa})^{2} \Gamma(\gamma+1/2)c_{2}}{\gamma^{2}\sqrt{\pi}(\Gamma(\gamma))^{2}}\right),\] \[\Lambda_{2} = \left( \begin{array}{c}\frac{c_{3}}{\Gamma(\gamma)\gamma}-\frac{( 2^{\kappa})^{2}\Gamma(\gamma+1/2)\,c_{2}^{2}}{\gamma^{2}\sqrt{\pi}(\Gamma( \gamma))^{2}}-\\ \frac{1}{2}\frac{c_{3}(3^{\gamma})^{2}\sqrt{3}(\gamma+1/3)\Gamma(\gamma+2/3)} {\gamma^{2}\sqrt{\pi}(\Gamma(\gamma))^{2}(2^{\gamma})^{2}\Gamma(\gamma+1/2)}+ \frac{(2^{\gamma})^{4}(\Gamma(\gamma+1/2))^{2}c_{2}}{(\Gamma(\gamma))^{3} \gamma\gamma^{3}}\end{array} \right).\]  
By substituting (16) into the first step of (10), we obtain  
\[y^{[j]}-\xi=x^{[j]}-\xi-\left(\Gamma(\gamma+1)\frac{\hbar(x^{[j]})}{\left[D_{ \gamma_{1}}^{\gamma}\right]\hbar(x^{[j]})}\right)^{1/\gamma}, \tag{17}\]  
\[y^{[j]}=\left(-c_{2}+\frac{(2\gamma)^{2}\Gamma(\gamma+1/2)c_{2}}{\Gamma( \gamma)\gamma\sqrt{\pi}}\right)\Big{[}\varepsilon^{[j]}\Big{]}^{\gamma+1}+ \Lambda_{3}\Big{[}\varepsilon^{[j]}\Big{]}^{2\gamma+1}+O\bigg{(}\Big{[} \varepsilon^{[j]}\Big{]}^{3\gamma+1}\bigg{)}, \tag{18}\]where  
\[\Lambda_{3}=\left( \begin{array}{c}\frac{c_{3}}{\Gamma(7)\gamma}-\frac{(2\gamma)^{ 2}\Gamma(\gamma+1/2)\ c_{2}^{2}}{\gamma^{2}\pi(\Gamma(\gamma))^{2}}-\\ \frac{1}{2}\,\frac{c_{3}(3\gamma)^{3}\sqrt{3}\Gamma(\gamma+1/3)\Gamma(\gamma+2 /3)}{\gamma^{2}\sqrt{\pi}(\Gamma(\gamma))^{2}(\sigma^{2})\Gamma(\gamma+1/2)}+ \frac{(2\gamma)^{4}(\Gamma(\gamma+1/2))^{2}c_{2}^{2}}{(\Gamma(\gamma))^{3\gamma 3}\pi}\end{array}  \right).\]  
At this stage, expanding \(\hbar\Big{(}y^{[j]}\Big{)}\) in a Taylor series around \(y=\xi\) yields  
\[\hbar\Big{(}y^{[j]}\Big{)}=\left[ \begin{array}{c}\Lambda_{4}\Big{[}\epsilon _{y}^{[j]}\Big{]}^{4\gamma+1}+\Lambda_{5}\Big{[}\epsilon_{y}^{[j]}\Big{]}^{3 \gamma+1}+-1/2\Lambda_{6}\Big{[}\epsilon_{y}^{[j]}\Big{]}^{2\gamma+1}\\ +\Lambda_{7}\Big{[}\epsilon_{y}^{[j]}\Big{]}^{\gamma+1}+O\bigg{(}\Big{[} \epsilon_{y}^{[j]}\Big{]}^{5\gamma+1}\bigg{)}\end{array} \right], \tag{19}\]  
where  
\[\Lambda_{4} = \left( \begin{array}{c}-c_{3}+\frac{(2\gamma)^{2}\Gamma(\gamma+ 1/2)c_{2}^{2}}{\Gamma(\gamma)\gamma\sqrt{\pi}}+\frac{1}{2}\,\frac{\Lambda_{3} (3\gamma)^{3}\sqrt{3}\Gamma(\gamma+1/3)\Gamma(\gamma+2/3)}{\Gamma(\gamma) \gamma\sqrt{\pi(2)^{2}\Gamma}(\gamma+1/2)}\\ -\frac{(2\gamma)^{4}(\Gamma(\gamma+1/2))^{2}c_{2}^{2}}{(\Gamma(\gamma))^{2} \gamma^{2}\pi}\end{array} \right),\] \[\Lambda_{5} = \left(