## 2 Numerical Algorithms Applied to Continuous Newton's Method
The corresponding root-finding algorithm can be written as  
\[z_{n+1}=F_{2}(z_{n})=z_{n}-h\frac{p(\omega_{2}(z_{n}))}{p^{\prime}(\omega_{2} (z_{n}))},\quad n\geq 0,\quad\omega_{2}(z)=z-\frac{h}{2}\frac{p(z)}{p^{ \prime}(z)} \tag{12}\]  
If \(z^{*}\) is a simple root of \(p(z)=0\), we have \(\omega^{\prime}(z)=1-h(1-L_{p}(z))/2\), where  
\[L_{p}(z)=p(z)p^{\prime\prime}(z)/p^{\prime}(z)^{2} \tag{13}\]  
Therefore, \(\omega(z^{*})=z^{*}\), \(\omega^{\prime}(z^{*})=1-h/2\) and then  
\[F_{2}(z^{*})=z^{*},\quad F_{2}^{\prime}(z^{*})=\frac{1+(1-h)^{2}}{2}\]  
As a consequence, iterative method Equation (12) is consistent only for \(h\in(0,2)\). All the methods deduced in this case have linear convergence (the lower value for the asymptotic constant error is obtained for \(h=1\)). This fact together with a number of function evaluations equals to four,
Something similar happens with [PERSON]'s method given in Equation (7) and [PERSON] method of second order given in Equation (8). They have different iteration maps, respectively \(z_{n+1}=F_{3}(z_{n})\) and \(z_{n+1}=F_{4}(z_{n})\), where  
\[F_{3}(z)=z-\frac{h}{2}\left(\frac{p(\omega_{3}(z))}{p^{\prime}(\omega_{3}(z))}+ \frac{p(z)}{p^{\prime}(z)}\right)\quad n\geq 0,\quad\omega_{3}(z)=z-h\frac{p(z)}{p^{ \prime}(z)}\]  
\[F_{4}(z)=z-\frac{h}{4}\left(\frac{p(\omega_{4}(z))}{p^{\prime}(\omega_{4}(z))}+ \frac{p(z)}{p^{\prime}(z)}\right),\quad n\geq 0,\quad\omega_{4}(z)=z-\frac{2h}{3} \frac{p(z)}{p^{\prime}(z)}\]  
In both cases, we have  
\[F_{j}(z^{*})=z^{*},\quad F_{j}^{\prime}(z^{*})=\frac{1+(1-h)^{2}}{2},\quad j=3,4\]  
for a simple root \(z^{*}\) of \(p(z)=0\). So, as in the case of the refined Euler's method, only linear convergence is achieved for \(h\in(0,2)\) and then these two methods are inefficient from a numerical perspective. The study of the Runge-Kutta method of order fourth defined in Equation (9) applied to Equation (4) leads us to the iteration scheme \(z_{n+1}=F_{5}(z_{n})\), where  
\[F_{5}(z)=z-\frac{h}{6}\left(\frac{p(z)}{p^{\prime}(z)}+2\frac{p(z+k_{1}(z)/2) }{p^{\prime}(z+k_{1}(z)/2)}+2\frac{p(z+k_{2}(z)/2)}{p^{\prime}(z+k_{2}(z)/2)} +\frac{p(z+k_{3}(z))}{p^{\prime}(z+k_{3}(z))}\right)\]  
and  
\[z_{n+1}=F_{1}(z_{n})=z_{n}-h\frac{p(z_{n})}{p^{\prime}(z_{n})},\quad n\geq 0 \tag{11}\]0  
Note that, for a simple root \(z^{*}\), we have \(k_{1}(z^{*})=0\), \(k_{1}^{\prime}(z^{*})=-h\), \(k_{2}(z^{*})=0\), \(k_{2}^{\prime}(z^{*})=-h(1-h/2)\), \(k_{3}(z^{*})=0\), \(k_{3}^{\prime}(z^{*})=-h(1-h/2+h^{2}/4)\) and then  
\[z_{n+1}=F_{1}(z_{n})=z_{n}-h\frac{p(z_{n})}{p^{\prime}(z_{n})},\quad n\geq 0 \tag{11}\]1  
There are no real values of \(h\) such that \(F_{5}^{\prime}(z^{*})=0\), then only linear convergence can be attained. In addition, we obtain consistent methods, that is \(z^{*}\) is an attracting fixed point of \(F_{5}\) if \(|F_{5}^{\prime}(z^{*})|<1\). These inequalities happens for \(h\in(0,2.7853)\) where \(2.7853\) is, approximately, the only real root of the polynomial \(-24+12h-4h^{2}+h^{3}\). The optimum value for \(h\), for which the asymptotic error constant has the lowest value, is \(h^{*}=1.5961\). In this case \(F_{5}^{\prime}(1.5961)=0.2704\). The analysis of the Adams-Bahforth second order method given in Equation (10) leads us to the following two-step multipoint method:  
\[z_{n+1}=F_{1}(z_{n})=z_{n}-h\frac{p(z_{n})}{p^{\prime}(z_{n})},\quad n\geq 0 \tag{11}\]2  
To study the local order of convergence related to Equation (14), we consider a simple root \(z^{*}\) of \(p(z)=0\) and the error in the \(n\)-th step \(e_{n}=z_{n}-z^{*}\). Then,  
\[z_{n+1}=F_{1}(z_{n})=z_{n}-h\frac{p(z_{n})}{p^{\prime}(z_{n})},\quad n\geq 0 \tag{11}\]3  
where the terms of order higher than two in the error have been neglected.