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suppose $\exists S \subset \mathbb{R}$ and a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x_0 \in S $ the sequence $x_{n+1} = f(x_n)$ converge to $x \in S$

now, let $\alpha \in (0,1)$ and $q \in \mathbb{R}$

under which condition (of $f$ and $S$), does the following sequence converge in $S$ ?

$y_{n+1} = (1- \alpha^n) f(y_n) + \alpha^n q $

does it converge to $x$ ?

$f$ is a $C^1$ function is a sufficient condition?

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    $\begingroup$ It's really confusing to use the same notation $x_n$ for two completely different sequences. $\endgroup$ Commented May 28, 2019 at 16:57

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A sufficient condition is that $f(x)=x$ and there exist $R > \max(|x_0 - x|, |q - x|)$ and $c$ with $0 < c < 1$ such that $|f'| < c$ on $[x-R, x+R]$.

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  • $\begingroup$ thhx, you mean when f is a contraction? is there a more general condition? $\endgroup$
    – SC_thesard
    Commented May 29, 2019 at 8:07

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