# How many iterations are required to reduce the convergence error by a factor of 10?

I've the following function

$$g(x) = x^2 + \frac{3}{16}$$

for which I found the two fixed points $x_1 = \frac{1}{4}$ and $x_2 = \frac{3}{4}$.

I noticed that the fixed-point iteration $$x_{k+1} = g(x_k)$$ actually converges for $g$ around $x_1$, but not around $x_2$, since I basically found that the absolute value of the derivative of $g$ is less than $1$ in the range $$A = \left(\frac{-1}{2}, \frac{1}{2} \right)$$, and I can find a sub-range of $A$ from where I can pick an initial guess for the fixed-point iteration method to find $x_1$. I can't, as I said above, do the same thing for $x_2$, because, first of all, it lies outside $A$.

Now, I need to find roughly how many iterations will be required to reduce the convergence error by a factor of $10$, but I'm not sure what it means and therefore how to find it.

I've looked around for explanations, but I'm still very confused.

2. By a factor of $10$ means that the convergence error is initial, say $e = x$, and we want to find the number of iterations required so that it is $e = \frac{x}{10}$? But still, this for me doesn't make much sense, since I don't know what's this convergence error, and how it is related to the fixed point iteration in general?
The convergence error is just the error between your current approximation $x_i$ and the root you are converging to, $\frac 14$, so it is $|x_i-\frac 14|$. When you are close to the root, the error will be multiplied by approximately $g'(\frac 14)=\frac 12$ each step. You can verify this experimentally with a spreadsheet. You can justify it theoretically by expanding $g(x)$ around the root in a Taylor series, so $x_{i+1}=g(x_i) \approx g'(\frac 14)(x_i-\frac 14)+\dots$ You are asked how many steps it takes to multiply the error by $\frac 1{10}$
• So, my answer depends on my choices of the range from which I can pick the initial value $x_0$, right? – nbro Apr 9 '16 at 23:42
• If you want to be exact about it, yes. The convergence will be slower at the ends of $A$. Often in error analysis we assume you are "close enough" to the root so we can neglect the higher order terms. – Ross Millikan Apr 9 '16 at 23:50