Differentiation of exponential function?

Question

How to solve derivative $\lim_{n\to\infty}e^{{}^n(x)}$ with respective of $x$ ? Here, ${}^n(x)$ is a tetration function $$ {}^n(x)= \begin{cases} x^{[{}^{n-1}(x)]} & \mbox{ if } {\;n>1}\\ x & \mbox{ if } {\;n=1}\\ \end{cases} $$ Anyone give me any idea how can resolve this problem ?

Thanks

Answer 1

If you have meant $$\displaystyle e^{(x^{x^{\cdots\infty}})}$$

I will write $$\displaystyle x^{x^{\cdots\infty}}=y\implies x^y=y\ \ \ \ (1)$$

Now, the problem becomes $$\frac{d(e^y)}{dx}$$ $$\text{ which is }=e^y\cdot\frac{dy}{dx}$$

Can you find $\displaystyle\frac{dy}{dx}$ from $(1)$

Answer 2

The derivative $e^{{}^n(x)}$ will be given by some recursive formula. First all note that, for $n\geq 2$ we have that \begin{align} \log\Big({}^n(x)\Big)=& x\cdot \log\Big({}^{n-1}(x)\Big)\\ = & xx\cdot \log\Big({}^{n-2}(x)\Big)\\ = & xxx\cdot \log\Big({}^{n-3}(x)\Big)\\ \vdots\;& \hspace{2cm}\vdots \\ = & x^k\cdot \log\Big({}^{n-k}(x)\Big), \quad k=1,\ldots n-1\\ \vdots\;& \hspace{2cm}\vdots \\ = & x^{n-1}\cdot \log\Big({}^{1}(x)\Big), \\ = & x^{n-1}\cdot \log(x), \\ \end{align} implies \begin{align} D_x\Big({}^{n}(x)\Big)=&D_x\big(e^{\log\big({}^{n}(x)\big)}\big)\\ =& e^{\log\big({}^{n}(x)\big)}D_x\big( x^{n-1}\cdot \log(x) \big)\\ =& {}^{n}(x)\cdot \bigg( (n-1)\cdot x^{n-2}\cdot\log(x)+x^{n-1}\cdot \frac{1}{x} \bigg)\\ =& {}^{n}(x)\cdot \bigg( (n-1)\cdot x^{n-2}\cdot\log(x)+x^{n-2} \bigg)\\ =& {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)\\ \end{align} Finally, $$ D_x\Big[ e^{{}^n(x)}\Big]=e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg) $$ But what we want is derivative of $$ D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big] %=e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg) $$ So we need to ensure the convergence of series of functions. The ideal would be uniform rather than just pointwise convergence convergence. In uniform convergence we have $$ D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big] =\lim_{n\to \infty} \left[ e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg) \right] $$ It's easy to see that sequence $\{e^{{}^n(x)} \}_n$ is monotone in itervals $(0,1),(1,\infty)$. If $\{e^{{}^n(x)} \}_n$ converge monotonically in any this intervals we use the Dini's theorem

THEOREM(Dini) Let $f_n\to f$ pointwise and monotonically over $[a,b]$, with each $f_n$ continuous, and $f$ continuous. Then $f_n\to f$ uniformly.

to conclude $$ D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big] =\lim_{n\to \infty} \left[ e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg) \right] $$ Analyse the pontual convergence of $\lim_{n\to\infty}e^{{}^n(x)}$.

• For $x> 1$ the limit is infinite.
• For $0 <x <1$ the limit is zero $\lim_{n\to\infty}e^{{}^n(x)}=0$ and converge monotonically.

What can you conclude?