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Any help with the following:

Problem: Consider the fixed point problem: $x=f(x)$ and given: $x_{n+1}=\frac{n}{n+1}f\left ( x_{n} \right )$. If $x_{0}$ is a fixed point where $\left | f^{'}\left ( x_{0} \right ) \right |< 1$, prove convergence.

Now consider the case $\left | f^{'}\left ( x_{0} \right ) \right |=1$. In some cases this iteration converges then. Formulate such a theorem and prove it.

For the first part:here is what I did: $x_{n+1}=g(x_{n},n)=\frac{n}{n+1}g\left ( x_{n} \right )$ and then $\left | g^{'}(x_{0},n) \right |=\left | \frac{n}{n+1}g^{'}\left ( x_{0} \right ) \right |< 1 $ since $\left | f^{'}\left ( x_{0} \right ) \right |< 1$ and $\frac{n}{n+1}< 1$. So, the convergence follows from the contraction mapping theorem.

For the second part, I don't have any idea at all. Any help please with the second part? and also do let me know if what I did in the first part is correct or not.

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up vote 1 down vote accepted

The problem is wrong as you stated it. Consider the function $f(x) = 1 + (x-1)/2 - 10 (x-1)^2$ with $x_0 = 1$ and $f'(x_0) = 1/2$. But starting at $x_0 = 1$, we have $x_1 = 0$, $x_2 = \frac{1}{2} f(0) = -4.75$, and it is easy to prove by induction $x_n < -n$ for $n \ge 2$, so this doesn't converge. Perhaps there's some assumption you're leaving out?

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Well then, if this is homework you should contact your instructor. Probably he or she intended something other than what was written. – Robert Israel Jan 25 '12 at 22:33

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