18
$\begingroup$

The average value of a function $y=f(x)$, on an interval $[a,b]$, is ${1\over {b-a}}\int_a^b f(t)dt$. This of course relates to the arithmetic average. It is easy to see that a corresponding formula for the geometric average is $\exp\left({1\over {b-a}}\int_a^b \ln(f(t))dt\right)$.

There are many other types of averages. In particular the ones motivated by the elementary symmetric polynomials are interesting as they "mix" the function values. My question is: How can we evaluate those averages?

To be specific, consider a real positive continuous function $y=f(x)$ on $[a,b]$. Create a partition of $n$ sub-intervals of width $\Delta x$. Let $Y=(y_1,y_2,\cdots, y_n)$ be the values of the function $f$ at some point in those intervals. Define the elementary symmetric polynomials $e_k=e_k(Y)$, for $1\le k \le n$, through
$$ \prod_{i=1}^n (t+y_i)= t^n+e_1t^{n-1}+\cdots+e_n. $$ Define the average $$ a_k(Y)={\root k \of {{e_k} \over {\left (n \atop k \right )}}}. $$ Define $A_\alpha(f)$, the $\alpha$-average of $f$ over $[a,b]$, as the limit of $a_k(Y)$ as $n \to \infty$, $\Delta x \to 0$, and $k/n \to \alpha$. Note $\alpha=0$ corresponds to the arithmetic average and $\alpha=1$ is the geometric average.

What do we know about $A_\alpha$ for $0<\alpha <1$? How can we compute it? For example if $f(x)=x$, $[a,b]=[1,2]$, and $\alpha=1/2$ what is $A_\alpha$?

Edit 1:

Some related inequalities are Maclaurin's and Newton's.

Edit 2:

I guess the requirement of continuity can be relaxed to piecewise continuity and still have a unique limit. Finding $A_\alpha$ for the following function, for a given $m>0$, will also be of interest: $$f(x)= \cases { 1 & if $ \quad 0 \le x \le 1/2$ \cr m & if $ \quad 1/2 < x \le 1$ }.$$

  • 0
    Is it obvious that $A_\alpha$ depends only on $\alpha$? Choosing $k=0$ for every $n$, or $k=1$ for every $n$, seems already to yield two different limits.2012-10-08
  • 0
    $k=0$ is there only for uniform definition of $e_k$. I will make the necessary correction. $k=1$ or $k=2$ or any finite number lead to the same $A_0$, i.e. the arithmetic average.2012-10-08
  • 0
    @Maesumi : I wish I could +1 this a thousand times, this is a reaaaaaaaaaaaally cool question.2012-10-10
  • 0
    @Maesumi : It is not clear at all to me how you work with $k=0$, I don't see any natural definition that is supposed to be implicit.2012-10-10
  • 0
    Do you assume you're working with positive functions? It would simply stuff like taking logs...2012-10-10
  • 1
    @PatrickDaSilva There is no case of $k=0$ in the final edited formulation. It won't make sense in the definition of $a_k(Y)$. As it is $1 \le k \le n$. The limit of $k/n$ can be zero in which case we have the arithmetic average.2012-10-10
  • 0
    @PatrickDaSilva Yes $f$ is positive. It can be relaxed to non-negative provided we deal with occasional improper integral as it is evident for the geometric average and presence of logarithm.2012-10-10
  • 0
    @Maesumi : I asked about $k=0$ because there could be some kind of $L^p$ norm analog where you could've taken the $\max$ for $k=0$ in some weird way but I couldn't imagine a right way to state it.2012-10-10
  • 1
    Just a suggestion ; I believe a first step to understanding this problem would be to be able to compute $$ \underset{k/n \to \alpha}{\lim_{n \to \infty}} \left( \binom nk \right)^{\frac 1k} $$ which will most probably involve the $\Gamma$ function to a certain extent... this coefficient will appear in your formulas independently of the function $f$ you work with so I think it will be important to understand.2012-10-10
  • 0
    It's also an interesting: Is your observation related to [Harmonic average](http://en.wikipedia.org/wiki/Harmonic_mean)?2012-10-10
  • 0
    @KvanTTT For harmonic average you get the reciprocal of the average of the reciprocal of the original function. It does not "mix" the function values the way symmetric average does.2012-10-10
  • 0
    @Maesumi : Using Stirling's formula, one can show that the denominator of $a_K(Y)$ goes to $(1-\alpha)^{(1-\alpha)}/\alpha$. Would you like me to detail the computations in an answer? It's not that long but too long for a comment.2012-10-10
  • 0
    @PatrickDaSilva Yes, every step helps.2012-10-10
  • 0
    Does exist the limit of the $a_k(Y)$ at all? Why do you take the $k$-th root?2012-10-16
  • 1
    @vesszabo I can prove the limit exists only for a few special cases, when #k# is finite, or $k=n-s$ and $s$ is finite. Numerically the limit seems to exists. We have to take the $k$-th root to produce a finite number.2012-10-16
  • 0
    @Maesumi : I like the Edit 1. This seems of interest. I'm actually amazed by those results and I would love to see more. It's sad that I don't have time to invest on those questions.2012-11-23
  • 0
    @Maesumi : Thinking of $\log$ as $x^0$ (in the sense of the integral of $x^{-1}$), I think the right conjecture would be that the $\alpha$-average of $f$ is really just $(\int f^{\alpha})^{1/\alpha}$, because the limit when $\alpha \searrow 0$ of this is the geometric average. I have no idea yet how to show this though.2012-12-01
  • 0
    @PatrickDaSilva A numerical experiment will help to decide this. But I doubt it.2012-12-02
  • 0
    @Maesumi : I don't. I'm actually thinking about doing that numerical experiment myself, I got really interested in this problem (and in the $f$-averages as well).2012-12-03
  • 0
    @PatrickDaSilva In my definition $\alpha=0$ corresponds to the arithmetic average.2012-12-03
  • 0
    @Maesumi : Yes... I'll have to give it some more thought, but I still think there's something there.2012-12-03

3 Answers 3