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The harmonic mean of a finite set of positive real numbers $\{x_1, x_2, \ldots, x_n\}$ is defined to be $$H(\{x_1, x_2, \ldots, x_n\}) = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}.$$

The logarithmic mean of two distinct positive real numbers $a$ and $b$ is defined to be $$L(a,b) = \frac{b - a}{\ln b - \ln a}.$$

One of the first applications of integration that students often see is the extension of the arithmetic mean of a finite set of real numbers to the arithmetic mean of a function $f(x)$ on a continuous interval $[a,b]$ via $\frac{1}{b-a} \int_a^b f(x) dx.$

In the same way, you can extend the harmonic mean so that it applies to a positive function $f(x)$ over a continuous interval $[a,b]$. You get $$\frac{b-a}{\int_a^b \frac{dx}{f(x)}} .$$

Thus, if you take $f(x) = x$ you obtain the harmonic mean of the continuous interval $[a,b]$. This is $$H([a,b]) = \frac{b-a}{\int_a^b \frac{1}{x} dx} = \frac{b-a}{\ln b - \ln a} = L(a,b).$$

My question is this: Is there an intuitive reason why $H([a,b]) = L(a,b)$?

For comparison purposes, note that if $A$ denotes the arithmetic mean, then $A([a,b]) = A(a,b) = \frac{a+b}{2}$.

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Is there even an intuitive reason why $\int \frac{1}{x} dx = \ln x$? – Rahul Nov 5 '10 at 2:40
Presumably an intuitive reason would have something to do with the various interpretations of the harmonic and logarithmic means. – Mike Spivey Nov 5 '10 at 2:54
The logarithmic mean is a piecewise function (e.g. $L(a,b) = a$ when $a=b$)? – PEV Nov 5 '10 at 3:20
@Trevor: Right. – Mike Spivey Nov 5 '10 at 3:56
@Rahul : yes, I saw it in Amercian Mathemtical Monthly, $\displaystyle\lim_{n\to-1} \left( \int {x}^n dx = \frac {x^{n+1}}{n+1} \right)$ – Arjang May 14 '11 at 0:03

The harmonic mean can be generalized to an arbitrary invertible function $w$: define the mean of $x_1,\ldots,x_n$ to be

$w^{-1}\left(\frac{w(x_1) + \cdots + w(x_n)}{n}\right)$

The logarithmic mean can also be generalized re its mean value interpretation for an arbitrary function $f$ such that $f'$ is invertible: define the mean of $x,y$ as

$(f')^{-1}\left(\frac{f(x) - f(y)}{x - y}\right)$

We can generalize your observation by taking $w = f'$. The corresponding mean, before taking $w^{-1}$, is

$\frac{1}{y-x}\int_x^y f'(t) \, dt = \frac{f(y)-f(x)}{y-x}$

and so the (former) mean of the interval with weight $f'$ is the same as the (latter) mean with respect to $f$.

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Nice generalization. I would still like to see an explanation for my observation based on usual interpretations of the harmonic and logarithmic means, though. – Mike Spivey Nov 5 '10 at 4:09
What are their usual interpretations? I had never heard of the logarithmic mean before, and the interpretation above is the first one in Wikipedia. – Yuval Filmus Nov 5 '10 at 4:43
See, for instance, this article: More information can be found via a search on "logarithmic mean," too. – Mike Spivey Nov 5 '10 at 4:59
That paper gives an engineering problem from which the logarithmic mean arises through a calculation very similar to yours (equation 15 and forward). What is intuitive interpretation of the logarithmic mean do you have in mind? – Yuval Filmus Nov 5 '10 at 5:23
I don't have a specific one in mind; I was just hoping that there was some way of connecting the concepts. For instance, an interpretation of the geometric mean is that "the geometric mean of growth over periods yields the equivalent constant growth rate that would yield the same final amount." The harmonic mean "is appropriate for situations when the average of rates is desired" or when calculating equivalent resistance for resistors in parallel. Something like that. (Quotes from Wikipedia.) – Mike Spivey Nov 5 '10 at 6:10

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