6
$\begingroup$

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

  • 1
    Is there even an intuitive reason why $\int \frac{1}{x} dx = \ln x$?2010-11-05
  • 0
    Presumably an intuitive reason would have something to do with the various interpretations of the harmonic and logarithmic means.2010-11-05
  • 0
    The logarithmic mean is a piecewise function (e.g. $L(a,b) = a$ when $a=b$)?2010-11-05
  • 0
    @Trevor: Right.2010-11-05
  • 1
    @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)$2011-05-14

1 Answers 1