Consider the function, $\rm g(x) = e^{-x} (x-1)^m$ which has a single root $\rm m$ at $\rm x_*$ at $\rm 1$.
Now, if we consider the following modified iteration: $$\rm x_{k+1} = x_k + \frac{g(x_k)}{g`(x_k)\lambda} $$ how do we find the values for which the constant $\lambda$ provides the fastest convergence?
So I have the newton method for the above eqn. $$\rm g`(x) = (m-1) e^{-x} (x-1)^m, $$ and the Newton iterate simplifies to this: $\rm x_k - \frac{1}{m-1}$ for the unmodified version, while the modified version gives an amplification by scale factor of $\lambda$. Are we supposed to equate the latter term to the root $\rm x_*$?? Are there any intuitive explanations to this solution? I should also say that I am utterly new and confused to numerical analysis.
Thank you.