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My question is motivated by this What is wrong with this funny proof that 2 = 4 using infinite exponentiation? discussion, namely an example of a map $x \mapsto f(x)$ is given, with $$f(x)= (\sqrt{2})^{x}.$$ In that thread, it is stated the map has 2 fixed points, at $x = 2$ and $x=4$. Now to examine their stability we look at the derivative $$f'(x) = (\sqrt{2})^{x} \log \sqrt{2}.$$

Now we have $f'(4) <1$, so $x=4$ should be stable. However, it is claimed, on the contrary, that $x=4$ is, in fact, $\textbf{unstable}$. Why is this?

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    $\begingroup$ Around $x=4$? The derivative is $2 \log 2$, or roughly $1.4$. $\endgroup$ – Patrick Stevens Nov 7 '18 at 21:40
  • $\begingroup$ My apologies! Was doing computations in terms of logs to base 10, rather than natural logarithms. May close the thread. @PatrickStevens $\endgroup$ – Alex Nov 7 '18 at 21:48
  • $\begingroup$ I'd like to have the question closed/deleted because it's on a false premise. For the case this is not wanted I added a formal answer which is "acceptable" for closing the case. $\endgroup$ – Gottfried Helms Dec 26 '18 at 10:26
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The question was based on a false premise and is superflous now (see comments below question) as based on a simple error as understood by the OP.

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