0
$\begingroup$

How to get the following limit: $$\lim_{p \to 0}\sqrt[p]{\frac{1}{n}\sum_{i=1}^{n}x_i^p}=\sqrt[n]{\prod_{i=1}^n x_i}$$ I.e. how to get geometric mean from definition of generalized means?

I attempted to apply exponentiation/derivation/logarithmization but all these gives nothing to me.

$\endgroup$
  • $\begingroup$ For ease of notation, consider the $n=2$ case. You can write $\sqrt[p]{\frac{x_1^p + x_2^p}{2}}$ as $\exp\left(\frac{\ln\left(\frac{x_1^p + x_2^p}{2}\right)}{p}\right)$. You could then try using L'Hôpital's rule. $\endgroup$ – Minus One-Twelfth Feb 23 at 12:02
  • $\begingroup$ An equivalent question here $\endgroup$ – Fabio Lucchini Feb 23 at 14:33
1
$\begingroup$

Take logarithm on the LHS to obtain $$ \lim_{p\to 0}\frac{\log\left(\frac1{n}\sum_{i=1}^n x_i^p\right)}p=\lim_{p\to 0}\frac{\log\left(\frac1{n}\sum_{i=1}^n x_i^p\right)}{\frac1{n}\sum_{i=1}^n x_i^p-1}\cdot\lim_{p\to 0}\frac{\frac1{n}\sum_{i=1}^n (x_i^p-1)}{p}. $$ Note that the first term tends to $1$ since $\lim_{t\to 1}\frac{\log t}{t-1}=(\log t)'|_{t=1}=1$. The second term is equal to $$ \frac1 n\sum_{i=1}^n\lim_{p\to 0}\frac{x_i^p-1}{p}=\frac1 n\sum_{i=1}^n\log(x_i)\lim_{p\to 0}\frac{e^{\log(x_i)\cdot p}-1}{\log(x_i)\cdot p}=\frac1 n\sum_{i=1}^n\log(x_i) $$ since $\lim_{t\to 0}\frac{e^t-1}t=1$. So, we have that the LHS converges to $\exp\left(\frac1 n\sum_{i=1}^n\log(x_i)\right)=\left(\prod_{i=1}^n x_i\right)^{\frac1 n}$ as wanted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.