Tag Info

New answers tagged

2

No. If $|z|<1$ the limit is $0$, if $|z|>1$ the limit is $\infty$ or does not exist since $|z|^n\to\infty$. If $|z|=1$ then $z=e^{i\theta}$ for $\theta\in[0,2\pi)$ and $z^n=e^{in\theta}$. Unless $\theta=0$, i.e. $z=1$, the sequence will rotate the point on the unit circle by $\theta$ counterclockwise at each step, and there is no limit.


4

WLOG, let $y = f(x)$. You know that $$y =x^y$$ So $$\ln(y) =y\ln(x)$$ $$\frac{dy}{dx}*\frac{1}{y} = \frac{dy}{dx}*\ln(x) + \frac{y}{x}$$ $$\frac{dy}{dx} = \frac{y^2}{x-xy\ln(x)}$$


5

This function is known as infinite tetration. The sequence is undefined for $x\leq 0$ except at some rational points, so there is no point talking about the derivative there. Ditto for large $x$, where the sequence diverges to $\infty$. Rather surprisingly for $x>0$ the limit only exists for $x\in[e^{-e},e^{1/e}]$ as shown by Euler, see here and here. ...


3

It is not only true, but mentioned explicitly in the Wikipedia article on the subject. :-) In fact, it has been known since the time of Euler. What you forgot to add is that infinite tetration only converges for $x\in\Big[e^{-e}~,~\sqrt[e]e\Big]$.


1

I think I know what you're trying to say, but I'm also not sure if it is possible to say with any rigor. I think that you essentially want to essentially know whether "simplification" is in some sense "complete": so that if two recurrence relations which appear in different forms always give the same outputs given the same inputs, then they can both be ...


1

Any relation that produces this sequence can of course be 'ultimately' simplified to $a_{n+1}=2^{a_n}$ just because this is how consecutive elements in the sequence are related. You do not however specify which identities are 'allowed' in a simplification. There are analytic functions that vanish on all integers, $\sin(\pi x)$ for example, so ...



Top 50 recent answers are included