# For the given function, find each fixed point and decide whether fixed point iteration is locally convergent to it

A theorem in my book states:

Let $g$ be a function and $r$ a number fixed by the function (i.e. $g(r) = r$). Assume $g$ is continuously differentiable, $g(r) = r$ and $|g'(r)| < 1$, then the fixed point iteration is locally convergent to $r$.

In other words, the sequence defined by $g(r_i) = r_{i+1}$ converges to $r$.

Let $g(x) = 0.5x^2 + 0.5x$

First I try to find the fixed points. I would like to find $x$ such that:

$x = 0.5x^2 + 0.5x$

Upon simplifying and solving I get that $x = 0, 1$.

By testing it on a calculator, I find that both values are the limit of the sequences of numbers I get by repeated iteration. However, when I try to apply the theorem, I find that they shouldn't be!

By the theorem:

$g'(x)$ is continuous, thus the function $g(x)$ is continuously differentiable. Next, $g(0) = 0$ and $g(1) = 1$. BUT, $|g'(0)| = 1$ and $|g'(1)| = 2$. Neither are less than $1$, like I expected.

What am I doing wrong?

• Isn't $g'(0)=1/2$? As for the other point, it might depend on your starting point. – Mark McClure Jan 29 '17 at 19:43

A small first remark : this theorem is a sufficient contition to have convergence, not a necessary one.

Now, as others have said, $|g'(0)| = \frac{1}{2} < 1$ so you have convergence.

For the second fixed point, it's easy to show that there isn't local convegence :

$\psi(x) = g(x)-x$ is strictly increasing on $[1,+\infty[$, so for every $\epsilon > 0$, $g(1+\epsilon) > 1+ \epsilon$, and by a quick induction, $\forall n > 0, g^n(1+\epsilon) > 1+ \epsilon$ (here it is the n-th composition) : it prove that starting from $1+\epsilon$, you cannot converge to $1$.

Hence no matter what neighborhood of $1$ you take, you'll find an $x=1+\epsilon$ such that the iteration will not converge to $1$

I guess you have made a mistake in your differentiation: $g'(x) = x + 0.5$, so $g'(0) = 0.5$, but $g'(1) = 1.5$ which is indeed not smaller than 1

Moreover the theorem says it converges only locale, so for some starting value in a neighbourhood of the fixed Point. It just says that given a starting point close to the fixed point the method is garanteed tot converge.

So in this case it Will only converge in a neighbourhood of 0 and not for a neighbourhood of 1