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 $A([a,b]) = A(a,b) = \frac{a+b}{2}$.
Best Answer
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 $f$.