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Suppose $u_{i+1}=f(u_i)$, for $i=0, 1, 2,...$, and $z_i=u_i-\alpha\cdot(u_{i+1}-u_i)$. Furthermore, let $\Delta u_i=f(u_i)-u_i$, and let $v_i=f(f(u_i))- 2f(u_i)+u_i.$ Can it be shown that $\Delta z_i= \Delta u_i - \alpha\cdot v_i$

The question was raised from (p. 26): http://biostats.bepress.com/cgi/viewcontent.cgi?article=1063&context=jhubiostat

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$\begin{split}\Delta z_i & = z_{i+1} - z_i = u_{i+1} - \alpha(u_{i+2} - u_{i+1}) - u_i + \alpha (u_{i+1} - u_i)\\ &= \Delta u_i - \alpha (u_{i+2} - 2u_{i+1} + u_i) = \Delta u_i - \alpha v_i\end{split}$

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    Thanks; I was wrongly assuming that $\alpha$ is changing (ie, not fixed).2012-10-30