Maximum, Minimum, Supremum and Infimum

kubebm · #1 $Old$ November 6th, 2009, 07:21 AM

please explain me how i'm solving it and how i should present the solution :

thanks!

TriKri · #2 $Old$ November 6th, 2009, 01:17 PM

In a), I think you could solve it by looking at the function

$F(k)=\prod_{n=1}^k\left(1+\frac{1}{n}\right)^{1/k} = e^{f(k)}$

where

$f(k)=\ln\left(\prod_{n=1}^k\left(1+\frac{1}{n}\right)^{1/k}\right) = \sum_{n=1}^k\frac{\ln\left(1+\cfrac{1}{n}\right)}{k}$

and by realizing that a k that gives a minimal or a maximal value for f, will also give a minimal or a maximal value, respectively, for F.

Also note that

$\ln\left(1+\cfrac{1}{n}\right) \leq\ \frac{1}{n}$

Drexel28 · #3 $Old$ November 6th, 2009, 03:39 PM

I'm going to assume that I am interpreting these questions correctly.

Problem: Let $\mathcal{I}=\left\{x: x=\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k}\right)\right]^{\frac{1}{k}}\quad k\in\mathbb{N}\right\}$ . Find $\max\text{ }\mathcal{I},\min\text{ }\mathcal{I},\inf\text{ }\mathcal{I},\sup\text{ }\mathcal{I}$

Solution: There is no "minimum" of this set. Those concepts (at least for how I have them defined) act only on members of the set. So $\max\text{ }\mathcal{I}$ existing would imply $\exists \delta\in\mathcal{I}$ such that $\forall x\in\mathcal{I}\quad x\le\delta$ . Before we proceed we need a lemma.

Lemma: Let $\sigma_k=\ln\left\{\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k}\right)\right]^{\frac{1}{k}}\right\}=\frac{1}{k}\sum_{\ell=1}^k \ln\left[1+\frac{1}{\ell}\right]$ , then $\sigma_{k+1}<\sigma_{k}\quad\forall k\in\mathbb{N}$

Proof: This is just true by a series of observations. Namely:

$\ln\left[1+\frac{1}{k+1}\right]\le \frac{1}{k}\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ . Adding $\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ to both sides gives $\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]+\ln\left[1+\frac{1}{k+1}\right]=\sum_{\ell=1}^{k+1}\ln\left[1+\frac{1}{\ell}\right]\le\left(1+\frac{1}{k}\right)\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ . Dividing both sides by $k+1$ gives $\frac{1}{k+1}\sum_{\ell=1}^{k+1}\ln\left[1+\frac{1}{\ell}\right]\le\frac{1}{k+1}\left(1+\frac{1}{k}\right)\sum_{\ell=1}^k\ln\left[1+\frac{1}{\ell}\right]=\frac{1}{k}\sum_{\ell=1}^{k}\ln\left[1+\frac{1}{\ell}\right]$ The conclusion follows. $\blacksquare$

Since $\ln(x)$ is an increasing function this implies that $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k}\right)\right]^{\frac{1}{k}}$ is a decreasing function.

Thus we can show that $\min\text{ }\mathcal{I}$ does not exist. We do this by contradiction. Suppose $\max\text{ }\mathcal{I}=\delta$ . Then $\delta=\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k}\right)\right]^{\frac{1}{k}}$ for some $k\in\mathbb{N}$ . But $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k+1}\right)\right]^{\frac{1}{k+1}}\le \delta$ and $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k+1}\right)\right]^{\frac{1}{k+1}}\in\mathcal{I}$ contradicting our choice of $\delta$ . Therefore $\min\text{ }\mathcal{I}$ does not exist.

From our lemma we can easily deduce that $2\in\mathcal{I}$ and $\delta\le2\quad\forall \delta\in\mathcal{I}$ . Thus, $\max\text{ }\mathcal{I}=2$ . Consequently, $\sup\text{ }\mathcal{I}=2$

Lastly we must find $\inf\text{ }\mathcal{I}$ . We claim that it is.......I leave the rest for you.

kubebm · #4 $Old$ November 6th, 2009, 06:11 PM

wow!! i can't even understand what you wrote, it suppose to be more simple, we didn't learn that, i'm sure.

HallsofIvy · #5 $Old$ November 6th, 2009, 09:03 PM

Well, you should have learned to at least calculate some of those numbers!

In the first problem, you have numbers of the form $\left[\left(1+\frac{1}{1}\right)\cdots\left(1+\frac{1}{k}\right)\right]^{\frac{1}{k}}$ .

When k= 1, that is $(1+ 1)^{1/1}= 2$ .

When k= 2, that is $[(1+1/1)(1+1/2)]^{1/2}= [(2)(3/2)]^{1/2}= 3^{1/2}$ .

When k= 3, that is $[(1+1/1)((1+ 1/2)(1+ 1/3)]^{1/3}= [(3)(4/3)]^{1/3}= 4^{1/3}$ .

That is, $(1+ 1/1)(1+ 1/2)(1+1/3)\cdot\cdot\cdot(1+ 1/(k-1))(1+1/k)$ $= (2)(3/2)(4/3)\cdot\cdot\cdot k/(k-1))((k+1)/k= k+1$ so your sequence reduces to $(k+1)^{1/k}$ . It should be easy to see that that is a decreasing sequence.

TriKri · #6 $Old$ November 7th, 2009, 05:54 AM

Quote:

Originally Posted by HallsofIvy $View Post$

That is, $(1+ 1/1)(1+ 1/2)(1+1/3)\cdot\cdot\cdot(1+ 1/(k-1))(1+1/k)=$ $(2)(3/2)(4/3)\cdot\cdot\cdot(k/(k-1))((k+1)/k)= k+1$ so you sequence reduces to $(k+1)^{1/k}$ . It should be easy to see that that is a decreasing sequence.

Nice! I got to try harder on seeing those things.