# Why is $x=4$ as a fixed point of a map $\sqrt{2}^{x}$ unstable?

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?

• Around $x=4$? The derivative is $2 \log 2$, or roughly $1.4$. – Patrick Stevens Nov 7 '18 at 21:40
• My apologies! Was doing computations in terms of logs to base 10, rather than natural logarithms. May close the thread. @PatrickStevens – Alex Nov 7 '18 at 21:48
• 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. – Gottfried Helms Dec 26 '18 at 10:26