Math Help - num. analysis #2

1. num. analysis #2

Hello everybody.

Suppose that g is continuously differentiable on some interval (c, d) that contains the fixed point p of g. Show that if $|g'(p)|<1$, then there exists $\delta>0$ such that if $|p_0-p|<\delta$, then the fixed-point iteration converges.

2. Clear Definitions

If $|g'(p)|<1$, then by definition of the derivative, $\left|\frac{\Delta y}{\Delta x}\right|<1$ for a sufficiently small $|\Delta x|<\delta$. Since $\left|\frac{\Delta y}{\Delta x}\right|=\frac{|\Delta y|}{|\Delta x|}$ , this means $|\Delta y|<|\Delta x|<\delta$. So define $\Delta x=p_0-p$ such that $|p_0-p|<\delta$, and we have just shown that $|g(p_0)-g(p)|=|g(p_0)-p|<|p_0-p|$ , which is exactly the definition of convergent. So when $p_0\in(p-\delta,p+\delta)$, the value of each iteration $p_i=g(p_{i-1})$ converges to $p$.

Notice that this proof allows you to predict the value of $\delta$ for a given function. Notably, for some fixed point $g(p)=p$ where $|g'(p)|<1$, if the approximation of $g'(p)\approx\left|\frac{g(p_0)-g(p)}{p_0-p}\right|<1$ at some point $p_0$, then $\delta=|p_0-p|$ and the iteration of any point in the interval $(p-\delta,p+\delta)$ will converge on $p$.