Someone asked me :what's the geometric mean for (0,1)
then I wrote this:
$$\mu=\sqrt[n]{x_0x_1x_2...x_n}\\ \\n \to \infty \\\mu=\sqrt[n]{\frac{n-1}{n}.\frac{n-2}{n}...\frac{n-(n-1)}{n}}\\ \ln \mu=\frac{1}{n} \ln(\frac{n-1}{n}.\frac{n-2}{n}...\frac{n-(n-1)}{n})=\\ \frac{1}{n}\sum_{i=1}^{n-1}\ln \frac{n-i}{n}=\\ \frac{1}{n}\sum_{i=1}^{n-1}\ln(1-\frac{i}{n})=\\ \int_{0}^{1}\ln(1-x)dx=\\(1-x)\ln(1-x)-x=-1\\ \to \mu=e^{-1}=\frac{1}{e}$$ Am I right ?
Is it possible to find geometric mean for [0,1]
thanks in advaced .
Yes, though I would say the geometric mean across the interval $(0,1]$ was $$\exp\left(\int_0^1 \log_e(x)\,dx \right) =\exp\left(\lim_{b\to 0+}\left[x\log_e(x)-x \right]_b^1 \right) = \exp\left(-1 \right) = \dfrac{1}{e}$$