I've been thinking about different ways to average a set of numbers the two that I know are the geometric mean and the arithmetic mean I thought of a knew type of Mean. where instead of adding or multiplying to find it you do another operation which I don't know the name of but I will call $\rho$ $$a_1\,\rho \,a_2\,\rho \,a_3\,\rho \,\dots\,\rho \,a_n=\frac{1}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots+\frac{1}{a_n}}$$
When you do arithmetic mean you make it so if it's all the same number $a$ that it should output $a$ same with the geometric mean
Arithmetic Mean
$$a+a+...+a=na$$
$$\frac{a+a+...+a}{n}=a$$
$$\frac{a_1+a_2+...+a_n}{n}$$
Geometric Mean
$$a\times a\times ...\times a=a^n$$
$$\sqrt[n]{a\times a\times ...\times a}=a$$
$$\sqrt[n]{a_1\times a_2\times ...\times a_n}$$
Harmonic Mean
If I do this for
$\rho$ I would get
$\frac{1}{na}$
$$a\,\rho \,a\,\rho \,a\,\rho \,\dots\,\rho \,a=\frac{1}{\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\dots+\frac{1}{a}}=\frac{1}{\frac{n}{a}}=\frac{a}{n}$$
if I wanted the This Mean to act like the others I would need to multiply by
$n$
$$(a\,\rho \,a\,\rho \,a\,\rho \,\dots\,\rho \,a) \times n =a$$
So if I wanted to use
$\rho$ for a Mean it would have to be
$$(a_1\,\rho \,a_2\,\rho \,a_3\,\rho \,\dots\,\rho \,a_n)\times n$$
Now for Functions
If I wanted to know what is the Arithmetic Mean length of a spring that follows
$f(x)$ between
$a$ units of time and
$b$ units of time we would change it into this form
$(\int_a ^b f(x))/(a-b)$. When you do the Geometric Mean talked about in
here you change it into this form
$\exp\left(\frac{1}{n(b-a)} \int_a^b \log f(x) \,\mathrm{d}x \right)$
How would you extent The Harmonic Mean to work with functions, and how would you do it in general for $\phi$?
The discrete harmonic mean is defined as $$H = {n \over \sum_{k=1}^n \frac{1}{x_k}} \quad (x_k \neq 0)$$
Partitioning an interval $[a,b]$ into $n$ equal subintervals of length $\Delta x = \frac{b-a}{n}$ gives
$$H = \frac{b-a}{\Delta x \sum_{i=1}^n \frac{1}{f(x_i)}}$$
So the continuous harmonic mean is
$$\lim_\limits{n \to \infty} \frac{b-a}{\Delta x \sum_{i=1}^n \frac{1}{f(x_i)}} = {b-a \over \int_a^b \frac{dx}{f(x)}}$$
This even works with functions that are zero on $[a,b]$ as long as the integral exists.
The general idea for transforming a discrete "mean"-formula into a continuous one is to rewrite it as a sum and then take the limit to get the integral.