Limit 3

**Posted:** Sat Oct 31, 2009 6:58 pm

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Kent Merryfield

The full result is this:

Fix $a,b>0$ with $a\ne b.$ Define $M(p)=\left(\frac{a^p+b^p}2\right)^{\frac1p}$ for $p\ne0$ and define $p(0)=\sqrt{ab}.$ Then $M(p)$ is continuous on $\mathbb{R}$ (which notably includes the statement that $\lim_{p\to0}M(p)=M(0),$ which is what this problem asks for). Also: $M(p)$ is strictly increasing, $\lim_{p\to\infty}M(p)=\max(a,b),$ and $\lim_{p\to-\infty}M(p)=\min(a,b).$

This is a pretty famous collection of results. After all, $M(0)<M(1)$ is AM-GM and $M(-1)<M(1)$ is AM-HM.

Just to look at the part asked for here:

$\ln M(p)=\frac1p\ln\left(\frac{a^p+b^p}2\right)=\frac1p\ln\left(\frac{e^{p\ln a}+e^{p\ln b}}2\right)$

For $p$ near zero, $e^{p\ln a}=1+p\ln a+O(p^2)$ and $e^{p\ln b}=1+p\ln b+O(p^2).$

Add those together to get that

$\frac{e^{p\ln a}+e^{p\ln b}}2=1+p\left(\frac{\ln a+\ln b}2\right)+O(p^2)$

$\ln\left(\frac{e^{p\ln a}+e^{p\ln b}}2\right)=p\left(\frac{\ln a+\ln b}2\right)+O(p^2).$

Divide that by $p$ and take the limit at $p\to0$ to get that

$\lim_{p\to0}\ln M(p)=\frac{\ln a+\ln b}2$

Exponentiate that to get that $\lim_{p\to 0}M(p)=\sqrt{ab},$ which is the geometric mean of $a$ and $b.$

**Posted:** Sat Oct 31, 2009 8:12 pm

Hong Quy · Yang-Mills Theory Offline Joined: 20 Feb 2005 Posts: 542

Thanks Kent Merryfield!!!
I like this solution...

**Posted:** Sat Oct 31, 2009 9:48 pm

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WWW · Hodge Conjecture Online Joined: 19 Oct 2009 Posts: 95

Letting $M(p) = \left(\frac {a^p + b^p}2\right)^{\frac1p}$ as Kent did, we could also set $f(p) = \ln {\frac{a^p+b^p}{2}}$ and note that $\lim_{p->0} \ln{M(p)} = \lim_{p->0}\frac{f(p)-f(0)}{p} = f'(0)$ by definition of the derivative. We have $f'(p) = \frac {2}{a^p+b^p}\frac{a^p\ln{a} +b^p\ln{b}}{2}$ , and so $f'(0)=(\ln{a} +\ln{b})/2$ as desired.

**Posted:** Sun Nov 01, 2009 4:24 pm