Mentioned in the wikipedia article, the $0$th power mean is defined to be the geometric mean. Why is this? I understand that a convenient consequence is that the means are ordered by their exponent. But is there an intuitive reason why the $0$th mean should be the geometric mean? Is it true that

$$\lim_{r\downarrow 0} \left(\frac{a_1^r + \dotsb + a_n^r}{n}\right)^{1/r}= \left( a_1 \dotsb a_n\right)^{1/n}?$$

If I take the logarithm and use L'Hospital's rule I get

$$\ln L = \lim_{r\downarrow 0} \frac{a_1^r + \dotsb + a_n^r}{n\left( a_1^r \ln{a_1} + \dotsb + a_n^r \ln{a_n} \right)}.$$

How might we evaluate this?

  • $\begingroup$ Writing "$\log$" in the LHS and "$\ln$" in the RHS is certainly a slip-up. Though I think that $\ln$ is the right one, but somehow mathematicians prefer to consider the inverse function of $\exp$ to be written $\log$ rather than the shorter $\ln$ (see for instance the answers given so far). $\endgroup$ – Marc van Leeuwen Oct 19 '13 at 7:24

Use the fact that as $r \rightarrow 0$, for all $a>0$: $$a^r = 1 + r \log a + o(r)$$ So the sum becomes: $$\frac{1}{n}\sum_{i=1}^n a_i^r=\frac{1}{n}\sum_{i=1}^n(1 + r \log a_i + o(r))=1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)$$

Furthermore $\lim_{r \rightarrow 0} (1+rx + o(r))^{1/r}=e^x$.

$$\lim_{r \rightarrow 0} \left(\frac{1}{n}\sum_{i=1}^n a_i^r\right)^{1/r}=\lim_{r \rightarrow 0} \left(1+r\frac{1}{n}\sum_{i=1}^n\log a_i + o(r)\right)^{1/r}=e^{\left(\frac{1}{n}\sum_{i=1}^n\log a_i\right)}= \left( \prod_{i=1}^n a_i\right)^{1/n}.$$

  • $\begingroup$ Wow, that is very beautiful $\endgroup$ – Eric Auld Oct 19 '13 at 8:03

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