What is the Power Mean Theorem for Generalised Means?

[icystrike]

Homework Statement

Anyone have the proof or exact direction to work on?

\mathop{\lim}\limits_{\alpha \to \infty}(\frac{a_1^{\alpha}+a_2^{\alpha}+...+a_n^{\alpha} }{n})^{1/\alpha}=\sqrt[n]{a_{1}a_{2}...a_{n}}

Homework Equations

The Attempt at a Solution

I do not know how to start.

 

[Dick]

There's something wrong with your statement. It isn't true. Take n=2 and a_1=2 and a_2=3. You should be able to convince yourself the limit as alpha->infinity is 3. Not sqrt(2*3).

 

[micromass]

You probably want the limit with \alpha\rightarrow 0, don't you?

 

[Dick]

You probably want the limit with \alpha\rightarrow 0, don't you?

Ah, that's it. I would just take the log and then use asymptotic approximations, like exp(x)~1+x and log(1+x)~x for x very small.

 

[icystrike]

Yes! Thanks guys the limit tends toward 0. Dick you are right... which approximatation it gave a very simple and elegant proof. Thanks!

 

[╔(σ_σ)╝]

I believe you could also you the Jensen inequality for this question.

 

[icystrike]

Jesnsen inequality is a more generalised inequality ... more like working from top to bottom right?

 

[╔(σ_σ)╝]

Yes, sort of.:-)