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  1. #1
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    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 $\displaystyle |g'(p)|<1$, then there exists $\displaystyle \delta>0 $ such that if $\displaystyle |p_0-p|<\delta$, then the fixed-point iteration converges.
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  2. #2
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    Clear Definitions

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

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