Geometric mean of reals between 0 and 1
What is the geometric mean of all reals between $0$ and $1$?
I was thinking over this, but could not come up with anything useful. Please help me out.
It's not entirely clear what is meant by the geometric mean of an uncountable collection of numbers. But here is a possible interpretation.
The geometric mean of two numbers $x,y$ is given by $$g=\sqrt{xy}$$ which can be rewritten $$\ln g=\frac{\ln x+\ln y}{2}\ .$$ That is, the log of the geometric mean is the arithmetic mean of the logs. The arithmetic mean of the logs of all numbers from $a$ to $b$ is $$\frac1{b-a}\int_a^b \ln x\,dx=\frac{(b\ln b-a\ln a)-(b-a)}{b-a}$$ and the geometric mean is therefore $$\exp\Bigl(\frac{(b\ln b-a\ln a)-(b-a)}{b-a}\Bigr) =\frac1e\Bigl(\frac{b^b}{a^a}\Bigr)^{\frac1{b-a}}\ .$$ Note that if $a=0$ then $a^a$ is not defined; but you can use the limit as $a\to0^+$, which is $1$.
Specifically, for the interval $[0,1]$ this gives $e^{-1}$, and for the interval $[1,2]$ it gives $4e^{-1}$.
It would be
$$\exp\left(\int_{0}^{1}\ln x \, dx\right)=e^{-1}$$
This is a more general result. First consider the Generalised mean: $$\operatorname{GM}(n, m)=\sqrt[n]{\frac1m\sum_{k=1}^mx^n_k}$$ We can extend this definition to a function rather than a sequence. Let $I=[m_0,m_1]$ be a bound such that it is the domain $f(x)$. Now we can write the Generalised mean of the values of $f(x)$ as $$\operatorname{GM}(n, f)=\sqrt[n]{\frac1{m_1-m_0}\int_{m_0}^{m_1}f^n(x)\ \mathrm dx}$$ A property of the discrete Generalised mean was that $n=1$ was the arithmetic mean, $n=2$ was the quadratic mean, $n=0$ was the geometric mean, $n=-1$ was the harmonic mean, as $n\to \infty$ it approached the maximum of the values and as $n\to-\infty$ it approched the minimum of the values.
If we take $f(x)=x$, $[m_0, m_1]=[0, 1]$ and limit $n\to 0$ (just like the geometric mean for the discrete case) we get $$\lim_{n\to 0}\sqrt[n]{\int_0^1 x^n \mathrm dx}=\lim_{n\to0}\frac1{\sqrt[n]{n+1}}=\lim_{n\to\infty}\left(1+\frac1n\right)^{-n}=e^{-1}$$ as expected. This approach also allows you to find the other means. If you limit $n\to \infty$ or $\to -\infty$ you will get the upper and lower bounds of the domain.