\begin{tabular}{|c|c|} \hline & Iterations at p0=1 \\ \hline a=1.1 & 5 \\ \hline a=1.2 & 6 \\ \hline a=1.3 & 6 \\ \hline a=1.4 & 6 \\ \hline a=1.5 & 7 \\ \hline a=1.6 & 7 \\ \hline a=1.7 & 8 \\ \hline a=1.8 & 8 \\ \hline a=1.9 & 10 \\ \hline Average & 7 \\ \hline \end{tabular} (e). Explain why the cases of larger values of a in part (c) need more iterations by discussing the term y used in the convergent analysis (see Appendix below for the details), in which we have obtained pn=g(pn1))=p0(1y+y2y3++y2n2y2n1)=a1ay2n for n=1,2,3,, where y=(a1) for p0=1( see part (c)). \begin{tabular}{|c|c|} \hline & Iterations at p0=1 \\ \hline a=1.1 & 5 \\ \hline a=1.2 & 6 \\ \hline a=1.3 & 6 \\ \hline a=1.4 & 6 \\ \hline a=1.5 & 7 \\ \hline a=1.6 & 7 \\ \hline a=1.7 & 8 \\ \hline a=1.8 & 8 \\ \hline a=1.9 & 10 \\ \hline Average & 7 \\ \hline \end{tabular} (e). Explain why the cases of larger values of a in part (c) need more iterations by discussing the term y used in the convergent analysis (see Appendix below for the details), in which we have obtained pn=g(pn1))=p0(1y+y2y3++y2n2y2n1)=a1ay2n for n=1,2,3,, where y=(a1) for p0=1( see part (c))