Show that $g'(p) \approx ({p_n-p_{n-1}})/({p_{n-1}-p_{n-2}})$

Asked Sep 17 '13 at 18:54

Active Sep 17 '13 at 19:07

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Suppose the sequence {${p_n}$} is generated by the fixed point iteration scheme $p_n = g(p_n-1).$ Further, suppose that the sequence converges linearly to the fixed point $p.$

Show that $g'(p) \approx ({p_n-p_{n-1}})/({p_{n-1}-p_{n-2}})$

Show that $|e_n| \approx |(g'(p)/(g'(p)-1)| |p_n-p_{n-1}|$

edited Sep 17 '13 at 19:07

asked Sep 17 '13 at 18:54

PooperScooper

The first one is just the definition of derivative. $f'= (f(b)-f(a))/(b-a)$. Here $b=p_{n-1}, a=p_{n-2}$. So long as you have convergence (but points staying distinct) you get the indicated approximation. – Maesumi Sep 17 '13 at 19:27
For the second one use the fact that $e_n=p_n-p=g(p_{n-1})-g(p)\approx g'(p)(p_{n-1}-p)$. Now find $p$ from here, and see the rest. – Maesumi Sep 17 '13 at 19:43

Show that $g'(p) \approx ({p_n-p_{n-1}})/({p_{n-1}-p_{n-2}})$

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