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Math Help - num. analysis #2

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