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.

  • $\begingroup$ @MichaelHardy So is the problem not valid for $0$ to $1$? $\endgroup$ Oct 30, 2014 at 3:56
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    $\begingroup$ What do you mean when you say geometric mean of an uncountable set of numbers? $\endgroup$
    – Macavity
    Oct 30, 2014 at 4:02
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    $\begingroup$ $$\sqrt[n]{\frac1n\frac2n\frac3n\cdots\frac{n}{n}}=\frac{\sqrt[n]{n!}}{n}\to \frac1e.$$ $\endgroup$ Oct 30, 2014 at 4:16
  • $\begingroup$ @wei2912: I don't really think it's as simple or elegant as $\exp\left(\frac{1}{b-a}\int_a^b\log(x)\,dx\right)$, and generalizing it is messier. $\endgroup$ Oct 30, 2014 at 14:54
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    $\begingroup$ @Ant: No. Think of how the exact value of a definite integral is determined by a limit of Riemann sums; that is basically what is going on here. I'm giving the limit of Riemann sums version. Right-hand sums are used, which is a good idea because the integral is improper at the left end-point. $\endgroup$ Oct 31, 2014 at 14:24

3 Answers 3


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 $4e^{-1}$.

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    $\begingroup$ Actually $0^0$ is defined, but in the context of anything looking like a limit it is an indeterminate form $\endgroup$ Oct 30, 2014 at 13:03
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    $\begingroup$ @HagenvonEitzen The way I've seen it, is that in the natural numbers $\mathbb N$, there's nothing stopping us from defining $0^0=1$ (and it actually is the most logical choice for combinatorial and set-theoretical reasons), but in the real numbers $\mathbb R$ it makes more sense to call it indeterminate. $\endgroup$ Oct 30, 2014 at 22:08

It would be

$$\exp\left(\int_{0}^{1}\ln x \, dx\right)=e^{-1}$$

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    $\begingroup$ $\exp(-\infty)=0$ !! $\endgroup$
    – scineram
    Oct 30, 2014 at 7:17
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    $\begingroup$ @scineram: $\int_{0}^{1} \ln x dx = (x \ln x - x) |_{0}^{1}= -1 - \lim_{x \rightarrow 0} x \ln x = -1$ $\endgroup$
    – RRL
    Oct 30, 2014 at 7:30

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.


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