\[\epsilon_k\]
of the Newton Raphson method, which estimates a root \[\alpha\]
of the equation \[f(x)=0\]
using the iteration formula \[x_{k+1}=x_k- \frac{f(x_k)}{f'(x_k)}\]
.We can write this as
\[x_{k+1}-x_k=- \frac{f(x_k)}{f'(x_k)}\]
(1)Write
\[x_{k+1}-\alpha=x_k-\alpha+(x_{k+1}-x_k)\]
\[\epsilon_{k+1}=\epsilon_k+(x_{k+1}-x_k)\]
Into this equation substitute (1) to give
\[\epsilon_{k+1}=\epsilon_k- \frac{f(x_k)}{f'(x_k)}\]
Now expand numerator and denominator of the second term in Taylor series about
\[\alpha\]
.\[\begin{equation} \begin{aligned} \epsilon_{k+1} &= \epsilon_k- \frac{f(\alpha)+f'(\alpha)(x_k-\alpha)+\frac{1}{2}f''(\alpha)(x_k-\alpha)^2+ ...}{f'(\alpha)+f''(\alpha)(x_k-\alpha)+\frac{1}{2}f''''(\alpha)(x_k-\alpha)^2+ ...} \\ & \simeq \epsilon_k- \frac{f'(\alpha)\epsilon_k+\frac{1}{2}f''(\alpha) \epsilon_k^2}{f'(\alpha)+f''(\alpha)\epsilon_k+\frac{1}{2}f''''(\alpha)\epsilon_k^2} \\ &= \epsilon_k- \epsilon_k \frac{f'(\alpha)+\frac{1}{2}f''(\alpha) \epsilon_k}{f'(\alpha)+f''(\alpha)\epsilon_k+\frac{1}{2}f''''(\alpha)\epsilon_k^2} \\ &\simeq \epsilon_k- \epsilon_k \frac{f'(\alpha)+\frac{1}{2}f''(\alpha) \epsilon_k}{f'(\alpha)+f''(\alpha)\epsilon_k+\frac{1}{2}f''''(\alpha)\epsilon_k^2} \\ &= \epsilon_k - \frac{f''(\alpha)}{2f'(\alpha)} \epsilon_k^2 \end{aligned} \end{equation}\]
Convergence is quadratic.