The power-mean inequality states that for positive real numbers $a_i$ and all real numbers $k_1,k_2$: $$(\frac{1}{n}\sum_{i=1}^{n}a_i^{k_1})^{\frac{1}{k_1}} \ge (\frac{1}{n}\sum_{i=1}^{n}a_i^{k_2})^{\frac{1}{k_2}}$$ if $k_1>k_2$. https://artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality
When $k_1=2,k_2=1$, one can prove the RMS-AM inequality (RMS is the root mean square). And when $k_1=1,k_2=-1$, one can prove the AM-HM inequality (HM is the harmonic mean).
This site(https://brilliant.org/wiki/power-mean-qagh/) states that by convention when $k_1$ or $k_2$ is $0$, then that side of the inequality is equal to the geometric mean.
So, in some sense could one say that the statement below is true?
$$\sqrt[0]{(\frac{1}{n}\sum_{i=1}^{n}a_i^{0})} = \sqrt[n]{\prod_{i=1}^na_i}$$
Though this may simply be a complete coincidence, it seems to fit too perfectly for me not to wonder about it. I am asking, besides simply being a nice notational trick, is it useful for mathematicians to define the zeroeth root as a way to invert operations or something similar? For example, in the example above, the zeroeth root seems to turn $\frac{1}{n}$ into the nth root and it turns the summation into a product.
Also, are there any other situations or theorems in math where a zeroeth root could mean something?
There is a reason for that. The limit of the $k$-th power mean as $k\to 0$ is precisely the geometric mean. here is a proof with $n=2$:
Call $$ f(k)=\ln \left(\frac{a_1^k+a_2^k}{2}\right)^{1/k} $$. Then $$ \lim_{k\to 0}f(k)=\lim_{k\to 0}\frac{\ln\left(1+\frac{a_1^k+a_2^k}{2}-1\right)}{k}=\lim_{k\to 0}\frac{a_1^k+a_2^k-2}{2k}. $$ By L'Hopitals rule then
$$ \lim_{k\to 0}f(k)=\lim_{k\to 0}\frac{a_1^k\ln a_1+a_2^k\ln a_2}{2}=\frac{\ln a_1a_2}{2} $$ Therefore, the limit of the power means is $$ e^{\frac{\ln a_1a_2}{2}}=\sqrt{a_1a_2}. $$ A similar proof works for any $n$.