# Remarkable limit involving $m_p=\log_p(p^{x_1} + \cdots + p^{x_n})-\log_p(n)$

It is easy to prove that $$\lim_{p\rightarrow 1} m_p = (x_1 + \cdots + x_n)/n$$. The following fact about the derivative of $$m_p$$ with respect to $$p$$ is also elementary: $$m'_p =\frac{dm_p}{dp} =\frac{1}{p \log p}\cdot\Big[\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}}-m_p\Big].$$ My interest in this is to create an alternative to the power mean, called the exponential mean: see here and here. The limit I am interested in is $$\lim_{p\rightarrow 1} m'_p$$. Using WolframAlpha, I computed the limit for $$n=2,3,4,5$$ (see here) and the following remarkable pattern emerges: $$\lim_{p\rightarrow 1} m'_p=\frac{1}{2n^2}\sum_{1\leq i How do you go about formally proving this fact? It does not sound elementary to me. Also, it sounds like $$m_p$$ is a strictly increasing function of $$p$$ (its derivative beeing positive everywhere, with $$m'_0 =+\infty$$ and $$m'_\infty =0$$) unless all the $$x_i$$'s are identical.

Update

In short, $$m_1$$ is the arithmetic mean and $$m'_1$$ is half the empirical variance of $$x_1,\cdots,x_n$$. I tried to see if such simple formulas existed for the power mean $$M_p$$, but I could not find anything interesting other than the well known fact that $$M_1=m_1$$ is the arithmetic mean. It would be interesting to see how the second and third derivatives of $$m_p$$ at $$p=1$$ are linked to the higher empirical moments of $$x_1,\cdots,x_n$$.

• The 5th example of the quasi-arithmetic mean wikipedia page appears to be your expression $m_p$, which may be useful.
– Mark
Aug 25, 2020 at 17:45
• Thanks for asking a well written and very polite question. I think, though, that this is going to just yield to L'Hospital's rule. Aug 25, 2020 at 17:54
• Thanks Mark. I actually researched the mean in question in more details before posting, but I could not find anything leading to either a statement or proof of my result regarding $m'_1$. Aug 25, 2020 at 17:56
• I found it a bit confusing when reading this question that it started with a computation involving an apparently undefined quantity $m_p$. I eventually noticed that it was defined in the title, but, if you ever have occasion to edit, it may be appropriate to reproduce the definition in the body. Jan 16 at 15:57

$$\newcommand\bar\overline$$ Letting $$t:=\ln p$$, we see that the limit in question is the limit of $$d(t):=\frac1t\Big(\sum_1^n x_j e^{tx_j}\Big/\sum_1^n e^{tx_j}-m_{e^t}\Big)$$ as $$t\to0$$. Next, letting $$\bar x:=\frac1n\,\sum_1^n x_j$$, $$\bar{x^2}:=\frac1n\,\sum_1^n x_j^2$$, and $$s^2=\bar{x^2}-\bar x^2$$, we have $$\sum_1^n x_j e^{tx_j}=\sum_1^n x_j (1+tx_j+o(t)) =n(\bar x+t\bar{x^2})+o(t),$$ $$\sum_1^n e^{tx_j}=\sum_1^n (1+tx_j+o(t)) =n(1+t\bar x)+o(t),$$ $$m_{e^t}=\log_{e^t}\Big(\frac1n\,\sum_1^n e^{tx_j}\Big) \\ =\log_{e^t}(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\tfrac1t\,\ln(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\bar x+ts^2/2+o(t).$$ So, $$d(t)=\frac1t\Big(\frac{\bar x+t\bar{x^2}}{1+t\bar x}+o(t)-\bar x-ts^2/2\Big) \\ =\frac1t\Big((\bar x+t\bar{x^2})(1-t\bar x)+o(t)-\bar x-ts^2/2\Big) \\ =s^2/2+o(1).$$ So, the limit in question is $$s^2/2 =\frac1{4n^2}\sum_{1\le i,j\le n}(x_i-x_j)^2 \\ =\frac1{4n^2}\sum_{1\le i,j\le n,\ i\ne j}(x_i-x_j)^2 \\ =\frac1{2n^2}\sum_{1\le i as conjectured.
Details on the first equality in the last three-line display: The left-hand side of that equality is $$\frac12\,Var\,X$$, where $$X$$ is any random variable whose distribution is $$\frac1n\,\sum_1^n\delta_{x_j}$$, where $$\delta_a$$ is the Dirac probability measure at point $$a$$. The right-hand side of that equality is $$\frac14\,E(X-X')^2=\frac14\,Var(X-X')=\frac12\,Var\,X,$$ where $$X'$$ is an independent copy of $$X$$.
• Thanks Losif. Still trying to figure out why WA gives half your value, see dsc.news/2FXbHpX when $n=2$. Aug 25, 2020 at 18:55
• @VincentGranville : Sorry, I thought $m_p$ was defined as $\bar x$. The necessary correction has now been made, and your conjecture holds. Aug 25, 2020 at 19:52
• @Losif: thank you. One of my other problems is to prove (or disprove) that $m_p \leq m_q$ if $p<q$. This inequality holds for the power mean $M_p$, wondering if it is true too for the expo mean $m_p$. I'm wondering if I should open another question regarding this, or maybe you have a sense that it might be trivial. Aug 25, 2020 at 20:26
• @VincentGranville : Yes, $m_p$ is nondecreasing in $p>0$, because $L(t):=\ln Ee^{tX}$ is convex in $t$, with $L(0)=0$. (The first letter in my first name is, not L, but the upper case of i.) Aug 25, 2020 at 20:45