Showing method has same convergence as Newton's? I am using the following method to compute the solution to a nonlinear equation. Start by computing the value from Newton's method: $\hat x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$ and then use it to compute the next iteration $x_{k+1} = \frac{x_k + \hat x_{k+1}}{2}$. I want to show that this method converges under the same condition's as newton's method but I am not sure how to do that. Would showing how an $f(x) = 0$ that converges for Newton's method also work for this method?
 A: We can think about Newton's method mapping a rootfinding problem in $f$ to a fixed point problem in $g$. We do this by letting $g(x) := x - f(x)/f^{\prime}(x)$. 
To analyze the convergence behavior of Newton, it's convenient to consider $g^{\prime}$. We are led to think about this because $f(x) = x \implies g^{\prime}(x) = 0$. In this case, if $g^{\prime}(x) = 0$, then we'll have second order convergence (due to Taylor: consider $g(y) - x = 0 + \frac{1}{2}(y-x)^2 g^{\prime\prime}(\xi)$ for $\xi \in (x,y)$). Thus the error is like $e_{k+1} = \frac{1}{2}(e_k)^2 g^{\prime\prime}(\xi)$. Below we'll try and get something similar for your method.
Here's a start: consider the new fixed point function $\tilde{g}(x) = \frac{x + x - f(x)/f^{\prime}(x)}{2}$. This boils down to
$$
\tilde{g}(x) = x - \frac{1}{2} \frac{f(x)}{f^{\prime}(x)}.
$$
We can see this iteratively with
$$
x_{k+1} = \tilde{g}(x_{k}) = \frac{x_{k} + x_k - \frac{f(x_k}{f^{\prime}(x_k)}}{2} = x_k - \frac{1}{2}\frac{f(x_k)}{f^{\prime}(x_k)}.
$$
Now what can we say about the error $|\tilde{g}(x_{k}) - x|$ if $|g(x_k) - x|$ converges?
