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Math Help - Maximum, Minimum, Supremum and Infimum

  1. #1
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    Maximum, Minimum, Supremum and Infimum

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


    thanks!
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  2. #2
    Senior Member TriKri's Avatar
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    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}\rig  ht)^{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}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    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_{\e  ll=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.
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  4. #4
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    wow!! i can't even understand what you wrote, it suppose to be more simple, we didn't learn that, i'm sure.
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  5. #5
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    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.
    Last edited by HallsofIvy; November 7th 2009 at 06:43 AM.
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  6. #6
    Senior Member TriKri's Avatar
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    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.
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