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Math Help - Unbounded sequence?

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation Unbounded sequence?

    I have a solution for the following problem but I really wonder what kind of ways will be provided for this.

    Problem. Let
    a(m,n)=<br />
\begin{cases}<br />
    67/256,&\mathrm{mod}(m,4)=0\text{ and }\mathrm{mod}(n,2)=0\\ %\delta<br />
    27/256,&\text{ otherwise}.<br />
\end{cases}
    Prove that
    <br />
\limsup\limits_{m,n\to\infty}A(m,n)=\infty,<br />
    where
    <br />
A(m,n):=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}  a(i,j)\Bigg\{\frac{4}{3}\Bigg(\sum\limits_{u=i+1}^  {i+3}a(u,j+1)\Bigg)^{1/4}-1\Bigg\}<br />
    ................. -\frac{1}{3}\Bigg[\sum\limits_{i=m-3}^{m}\sum\limits_{j=1}^{n}\sum\limits_{u=i+1}^{i+  3}a(u,j+1)+\sum\limits_{i=1}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)\Bigg].
    Last edited by bkarpuz; October 4th 2008 at 03:20 AM.
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  2. #2
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    The sequence a(m,n) is "almost" constant. Let's bound A(m,n) from below by replacing a(m,n) by a=\frac{27}{256} in the first term and by a'=\frac{67}{256} in the second one (after the minus sign): we get A(m,n)\geq mna(\frac{4}{3}(3a)^{1/4}-1)-(12n+6m)a'. However, we have \frac{4}{3}(3a)^{1/4}=\left(\frac{256 a}{27}\right)^{1/4}=1 so that this lower bound is slightly too strong, but if we had a positive factor instead in front of mn (no matter how small), this would be right. Then, to bound the first term, it suffices to consider the indices where 4|i+1 and 2|j+1 (you can bound the terms with other indices by 0 thanks to the previous computation). There are about \frac{mn}{8} such couples (i,j), and for each of them the summand is greater than a(\frac{4}{3}(2a+a')^{1/4}-1), which is strictly greater than a(\frac{4}{3}(3a)^{1/4}-1)=0. We get that A(m,n)\geq \lambda mn- \alpha m - \beta n for some positive \lambda, \alpha,\beta. Choosing m=n for instance, you easily see that this is unbounded from above.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Cool

    Here is my solution.
    Solution. Clearly, we have a(m+4,n)=a(m,n) and a(m,n+2)=a(m,n) for all m,n\in\mathbb{N}.
    Therefore, we have
    \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}a(i,j)\Bigg\{\frac{4}{3}\Bigg(\sum\limits_{u  =i+1}^{i+3}a(u,j+1)\Bigg)^{1/4}-1\Bigg\}=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)>0........(1)
    and
    \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)=3.................................................. ...............(2)
    for all m,n\in\mathbb{N}.
    Therefore, using (1) and (2), we have the following inequalities:
    <br />
\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}a(i,j)\B  igg\{\frac{4}{3}\Bigg(\sum\limits_{u=i+1}^{i+3}a(u  ,j+1)\Bigg)^{1/4}-1\Bigg\}\geq\frac{81}{256}\bigg(\frac{\sqrt{11}}{3  }-1\bigg)\bigg\lfloor\frac{m}{4}\bigg\rfloor\bigg\lf  loor\frac{n}{2}\bigg\rfloor<br />
......(3)
    \sum\limits_{i=m-3}^{m}\sum\limits_{j=1}^{n}\sum\limits_{u=i+1}^{i+  3}a(u,j+1)\leq3\bigg\lceil\frac{n}{2}\bigg\rceil.................................................. .........._(4)
    \sum\limits_{i=1}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)\leq3\bigg\  lceil\frac{m}{4}\bigg\rceil.................................................. ..........,(5)
    for all m,n\in\mathbb{N}.
    Hence, from (3)--(5), we learn that
    A(m,n)\geq B(m,n)
    holds for all m,n\in\mathbb{N}, where
    B(m,n):=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)\bigg\lfloor\frac{m}{4}\bigg\rfloor\bigg\lf  loor\frac{n}{2}\bigg\rfloor-\Bigg\{\bigg\lceil\frac{n}{2}\bigg\rceil+\bigg\lce  il\frac{m}{4}\bigg\rceil\Bigg\}
    for m,n\in\mathbb{N}.
    It suffices to prove that B is unbounded.
    As k\to\infty, we have
    B(4k,2k)=\frac{81}{256}\bigg(\frac{\sqrt{11}}{3}-1\bigg)k^{2}-2k\to\infty.
    Thus, the proof is done. \rule{0.3cm}{0.3cm}

    Notation. \lfloor\cdot\rfloor denotes the least integer function and \lceil\cdot\rceil denotes the greatest integer function.

    Appendix. Here, (2) is shown.
    By the periodicity, we obtain
    \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}a(i,j+1)=1.................................................. .......................(6)
    for all m,n\in\mathbb{N}.
    Using (6), we get
    \sum\limits_{i=m-3}^{m}\sum\limits_{j=n-1}^{n}\sum\limits_{u=i+1}^{i+3}a(u,j+1)=\sum\limit  s_{i=m-3}^{m}\sum\limits_{u=i+1}^{i+3}\sum\limits_{j=n-1}^{n}a(u,j+1)
    ......................................_ =\sum\limits_{u=1}^{3}\sum\limits_{i=m+u-3}^{m+u}\sum\limits_{j=n-1}^{n}a(i,j+1)
    ......................................_ =\sum\limits_{u=1}^{3}1=3
    for all m,n\in\mathbb{N}.
    Last edited by bkarpuz; October 4th 2008 at 06:44 AM. Reason: Appendix is added.
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