A generalized (MacLaurin's) average for functions

As requested in the comments, here is the proof : \begin{gather*} \begin{aligned} \binom nk^{\frac 1k} & = \left( \frac{n!}{k!(n-k)!} \right)^{\frac 1k} \sim \left( \frac{ \sqrt{2 \pi n} (n/e)^n }{ (\sqrt{ 2 \pi k } (k/e)^k ) ( \sqrt{2 \pi(n-k)} ((n-k)/e)^{(n-k)}}\right)^{\frac 1k} \\ & = \left( \frac 1{\sqrt{2\pi}} \right)^{\frac 1k} \left( \frac{n}{k(n-k)} \right)^{\frac 1{2k}} \left( \frac{n^n}{k^k(n-k)^{n-k}}\right)^{\frac 1k} \\ & = \left( \frac 1{\sqrt{2 \pi}} \right)^{\frac 1k} \left( \frac{n}{k(n-k)} \right)^{\frac 1{2k}} \left( \frac{ n^{\frac nk}}{k \, n^{((n/k) -1)} (1 - k/n)^{(n/k) - 1} } \right) \\ & = \left( \frac 1{\sqrt{2 \pi}} \right)^{\frac 1k} \left( \frac{n}{k(n-k)} \right)^{\frac 1{2k}} \left( \frac{ 1 }{ \frac kn (1 - k/n)^{(n/k) - 1} } \right) \\ & \longrightarrow \frac 1{\alpha (1-\alpha)^{1/\alpha -1}} = \frac{(1-\alpha)^{1-1/\alpha}}{\alpha} = \left( \frac 1{\alpha^{\alpha} (1-\alpha)^{1-\alpha} } \right)^{\frac 1{\alpha}} . \end{aligned} \end{gather*}

Note that even if the asymptotic is supposed to work when we don't take the $\frac 1k$ power, the asymptotic holds independently of the $\frac 1k$ power that we take (i.e. the bounds for the asymptotic can be chosen independently of the power that is taken), so it still holds.

Hope that helps,

Wow, so much inspiration.

Here's a generalization of your question : let us begin with a function $f : ]0,\infty[ \to ]\inf_{x > 0} f(x),\sup_{x > 0} f(x)[$ which is monotone. Let $g : [0,1] \to ]0,\infty[$. Define the $f$-average of $g$ by $$ A_f(g) = f^{-1} \left( \int_0^1 f(g(t)) dt \right). $$ Note that when $f(x) = x$ you get the arithmetic average and when $f(x) = \log(x)$ you get the geometric average. These functions are increasing. However, you can also take $f(x) = 1/x$ and get the harmonic mean. This function is decreasing ; perhaps that is why it behaves differently than the other averages.

This $f$-average is not completely randomly defined ; it has some properties. In a similar way that arithmetic and geometric averages are defined for finite sets instead of functions, we can define the $f$-mean of $n$ numbers by $$ M_f(x_1, \dots, x_n) = f^{-1} \left( \frac 1n \sum_{i=1}^n f(x_i) \right). $$ Again, taking $f$ as stated before gives us arithmetic, geometric and harmonic means. Now we can see that the $f$-average of the $f$-mean of $g_1(t), \dots, g_n(t)$ is the $f$-mean of the $f$-averages : $$ A_f( M_f(g_1,\dots,g_n)) = f^{-1} \left( \int_0^1 f(M_f(g_1,\dots,g_n)(t)) \, dt \right) \\ = f^{-1} \left( \int_0^1 \frac 1n \sum_{i=1}^n f(g_i(t)) \, dt \right) \\ = f^{-1} \left( \frac 1n \sum_{i=1}^n \int_0^1 f(g_i(t)) \, dt \right) \\ = f^{-1} \left( \frac 1n \sum_{i=1}^n f(A_f(g_i)) \right) \\ = M_f(A_f(g_1), \dots, A_f(g_n)). $$

We can also obtain a generalized QM-AM-GM-HM inequality :

Theorem. Suppose $f,g$ are twice differentiable, $f' \neq 0 \neq g'$, $f'' \le g''$ and $h(x) = g'(x) / f'(x)$ is injective. Then for all $x_1, \dots, x_n \in ]0,\infty[$, we have $M_f(x_1,\dots,x_n) \le M_g(x_1,\dots,x_n)$.

The proof is similar to the one with QM-AM-GM-HM except that the facts used for $f$ and $g$ are quite clear now. Let us minimize $M_g(x_1,\dots,x_n)$ subject to $M_f(x_1,\dots,x_n) = C$. It follows from the Lagrange Multipliers that there exists a $\lambda$ such that $$ \frac{g'(x_i)}{g'(M_g(x_1,\dots,x_n))} = \frac{\lambda f'(x_i)}{f'(M_f(x_1,\dots,x_n))} \quad \Longrightarrow \quad \frac{g'(x_i)}{f'(x_i)} = \frac{\lambda g'(M_g)}{f'(M_f)} = \frac{g'(x_j)}{f'(x_j)}, \quad i \neq j. $$ Since we assumed $h(x) = g'(x)/f'(x)$ to be injective, $x_1 = x_2 = \dots = x_n \overset{def}= x$, so it follows that $(x,\dots,x)$ is a minimum and $$ M_g(x_1,\dots,x_n) \ge M_g(x,\dots,x) = x = M_f(x,\dots,x) = C = M_f(x_1,\dots,x_n). $$ Note that letting $f \in \{x^2, x, \log x, 1/x\}$, this proves QM-AM-GM-HM.

What would really help for computations is if we could find the functions $f_{\alpha}$ corresponding to the $A_{\alpha}$-averages, if such functions exist. This would definitely be the way to go to compute these averages. Would those functions happen to be nicely differentiable, we could also show using this theorem that $A_{\alpha}(f)$ is a decreasing function of $\alpha$.