Let $x_0$ be the stationary point of the $\alpha$th derivative of the function $f(x)=x^\alpha-\alpha^x$, and let $$\lambda_\alpha=\frac{f(-x_0)}{-x_0}.$$

Does the limit $$\lim_{\alpha\to\infty}\bigg|\frac{\lambda_{\alpha+1}}{\alpha\lambda_\alpha}\bigg|$$ exist, and if so, what is its value?

Note that $\lambda_{\alpha+1}$ is obtained by simply replacing $\alpha$ by $\alpha+1$ in the expression for $\lambda_{\alpha}$, and assume that $\alpha\in\mathbb{N}$.

We write $$\begin{align}f(x)=x^{\alpha}-\alpha^x&\implies f'(x)=\alpha x^{\alpha-1}-\alpha^x\ln\alpha\\&\implies\quad\quad\qquad\cdots\\&\implies f^{(\alpha)}(x)=\left[\prod_{i=0}^{\alpha-1}(\alpha-i)\right]-\alpha^x\ln^\alpha\alpha=0\end{align}$$ for stationary points. Then $$-x_0=\frac1{\ln\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]\tag{1}$$ so $$\begin{align}&\lambda_{\alpha}=\frac{\frac1{\ln^\alpha\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]^\alpha-\frac{\ln^\alpha\alpha}{\prod_{i=0}^{\alpha-1}(\alpha-i)}}{\frac1{\ln\alpha}\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]}\\\implies&\lambda_{\alpha}=\frac{\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]^\alpha\prod_{i=0}^{\alpha-1}(\alpha-i)-\ln^{2\alpha}\alpha}{\ln^{\alpha-i}\alpha\left[\ln\ln^\alpha\alpha-\sum_{i=0}^{\alpha-1}\ln(\alpha-i)\right]\prod_{i=0}^{\alpha-1}(\alpha-i)}\tag{2}\end{align}$$ The expression for $|\lambda_{\alpha+1}/\alpha\lambda_{\alpha}|$ is extremely complicated so I will not give it here.

How should I continue? It seems that the limit does exist, but I would like a rigorous proof of it.

  • $\begingroup$ What is $\alpha$? A positive natural number, a positive real number, or something else? $\endgroup$ – Servaes Apr 25 '18 at 15:27
  • $\begingroup$ There exist derivatives of complex order, so being an $\alpha$-th derivative is not sufficient to assume that $\alpha$ is a natural number. See arxiv.org/ftp/arxiv/papers/1209/1209.0400.pdf. $\endgroup$ – AlexanderJ93 Apr 30 '18 at 9:22
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    $\begingroup$ I think the limit is -e. does that seem correct? $\endgroup$ – mathstackuser12 May 4 '18 at 9:50

Consider the following $$f\left( x,z \right)={{x}^{z}}-{{z}^{x}}$$ Then it is not difficult to show $$\frac{{{\partial }^{n}}}{\partial {{x}^{n}}}f\left( x,z \right)=\frac{\Gamma \left( z+1 \right)}{\Gamma \left( z-n+1 \right)}{{x}^{z-n}}-{{z}^{x}}\log {{\left( z \right)}^{n}}$$ Hence define $$\frac{\Gamma \left( z+1 \right)}{\Gamma \left( z-n+1 \right)}x_{0}^{z-n}-{{z}^{{{x}_{0}}}}\log {{\left( z \right)}^{n}}=0$$ Now consider the limit for $n\to z$
$$\Gamma \left( z+1 \right)-{{z}^{{{x}_{0}}}}\log {{\left( z \right)}^{z}}=0$$ This is the equation defining the stationary point ${{x}_{0}}$ . Solving for this and substituting into your definition for ${{\lambda }_{z}}$ we have after considerable algebra $$\frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}=\frac{\log \left( z+1 \right)\log \left( z{{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)\left\{ {{\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{\log \left( z+1 \right)} \right)}^{z+1}}-\frac{{{\log }^{z+1}}\left( z+1 \right)}{\left( z+1 \right)\Gamma \left( z+1 \right)} \right\}}{z\log \left( z \right)\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)\left\{ {{\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)} \right)}^{z}}-\frac{{{\log }^{z}}\left( z \right)}{z\Gamma \left( z \right)} \right\}}$$ Now observe that $\lim \frac{{{\log }^{z}}\left( z \right)}{z\Gamma \left( z \right)}=0$ , and many terms cancel in numerator and denominator (like for example the leading terms outside the { } brackets as they are of the same order in the limit). We find therefore $$\small\begin{align}\lim \frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}&=\lim {{\left( \frac{\log \left( z \right)\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{\log \left( z+1 \right)\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)} \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\\&=\lim {{\left( \frac{\log \left( z \right)+\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z \right) \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)} \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\\&=\lim {{\left( \frac{\log \left( z \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}+1 \right)}^{z}}\left( -\frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)} \right)\end{align}$$ What remains are two limits: note $$\lim \frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)}=\lim \frac{1}{z}-\frac{\log \left( \log \left( z \right) \right)}{\log \left( z \right)}+\frac{\log \left( \Gamma \left( z \right) \right)}{z\log \left( z \right)}$$ Now I think it is reasonable to assume that it is well known that $\log \left( \Gamma \left( z+1 \right) \right)=\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z+O\left( 1 \right)$ for large z (use Stirling). So we have $$\lim \frac{\log \left( \left( z+1 \right){{\log }^{-z-1}}\left( z+1 \right)\Gamma \left( z+1 \right) \right)}{z\log \left( z+1 \right)}=\lim \frac{\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z}{z\log \left( z \right)}=1$$ Where we have used the fact that $\log \left( \log \left( z \right) \right)$ grows more slowly than $\log \left( z \right)$. That’s one out of the way. Next: $$\lim \frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)}=1-\lim \frac{z\log \left( \log \left( z \right) \right)}{\log \left( z \right)}+\lim \frac{\log \left( \Gamma \left( z \right) \right)}{\log \left( z \right)}$$ Hence we have the asymptotic $$\lim \frac{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}{\log \left( z \right)}=1+\lim \frac{\left( z-\tfrac{1}{2} \right)\log \left( z \right)-z-z\log \left( \log \left( z \right) \right)}{\log \left( z \right)}\sim z$$

Hence $$\lim {{\left( \frac{\log \left( z \right)}{\log \left( \left( z+1 \right){{\log }^{-z}}\left( z \right)\Gamma \left( z \right) \right)}+1 \right)}^{z}}\simeq \lim {{\left( \frac{1}{z}+1 \right)}^{z}}=e$$ Putting it all together we have shown therefore $$\lim \bigg|\frac{{{\lambda }_{z+1}}}{z{{\lambda }_{z}}}\bigg|=e$$

  • $\begingroup$ Thanks for your answer! I'll have a good look at it and I'll award the bounty if there's nothing wrong. $\endgroup$ – TheSimpliFire May 4 '18 at 16:20

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