I am reading this paper. In some point of the analysis the non linear equation
$$\phi_1(\lambda) = \frac{1}{\sqrt{\psi(\lambda)}} - \frac{\sigma}{\lambda} \tag{A}$$
is studied, i.e. $(6.7)$ in the paper. For $(A)$ we have that $\phi'_1(\lambda)> 0$. The root of $\phi_1(\lambda)$ is computed using Newton's method, i.e., Algorithm 6.1 in the paper. Then, it is stated that the speed of convergence may be improved by only linearizing the term $\omega(\lambda) = \frac{1}{\sqrt{\psi(\lambda)}}$ and not the $\sigma / \lambda$ term as does Newton's method -- when computing a correction $\Delta \lambda^c$ to the estimate $\lambda$ of the required root of $\phi_1$. The resulting correction thus satisfies the equation
$$\omega(\lambda) + \omega' (\lambda) \Delta \lambda^c = \frac{1}{\psi^{1/2}(\lambda)} - \frac{1}{2} \frac{\psi' (\lambda)}{\psi^{3/2} (\lambda)} \Delta \lambda^c = \frac{\sigma}{\lambda + \Delta \lambda^c},\tag{B}$$
i.e., $(6.12)$ in the paper. I suppose that the authors means that $\lambda$ in $(B)$ is the root of $\phi_1 (\lambda) = 0$. Please correct me If I am wrong. Then, for clarity lets denote this root by $\lambda^M$. Then the Taylor expansion of $\omega$ near the root $\lambda^M $ is
$$\omega (\lambda ) = \omega (\lambda^M) + \omega' (\lambda^M) \Delta \lambda^c = \frac{1}{\psi^{1/2}(\lambda^M)} - \frac{1}{2} \frac{\psi' (\lambda^M)}{\psi^{3/2} (\lambda^M)} \Delta \lambda^c \tag{C}$$ where $\Delta \lambda^c = \lambda - \lambda^M$. Using $\phi_1(\lambda^M) = 0 \stackrel{(A)}{\Leftrightarrow} \lambda^M = \sigma \sqrt{\psi (\lambda^M)}$ we get
$$\begin{aligned}\omega (\lambda ) = & \omega (\lambda^M) + \omega' (\lambda^M) \Delta \lambda^c = \frac{1}{\psi^{1/2}(\lambda^M)} - \frac{1}{2} \frac{\psi' (\lambda^M)}{\psi^{3/2} (\lambda^M)} \Delta \lambda^c \\ =& \frac{\sigma}{\lambda^M} - \frac{\sigma}{(\lambda^M)^2} \Delta \lambda^c \\ = & \frac{\sigma}{\lambda^M + \Delta \lambda^c} \left( \frac{(\lambda^M)^2 - (\Delta\lambda^c)^2}{(\lambda^M)^2}\right) \end{aligned}$$
As it seems that we need to prove that $\frac{(\lambda^M)^2 - (\Delta\lambda^c)^2}{(\lambda^M)^2} = 1$ but I do not see it. Can we say that $\Delta\lambda^c$ is small and thus $\frac{(\lambda^M)^2 - (\Delta\lambda^c)^2}{(\lambda^M)^2} \approx 1$? Could you please someone cast some light?