let $\lceil a \rfloor$ be the nearest integer function. Is there a nice lower bound on the expression: \begin{align} | \lceil a \rfloor b- a\lceil b \rfloor| \end{align} Thank you
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What is the source of the question? – Bill Dubuque Sep 01 '14 at 15:50
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I am doing the some research and stumbled on this – Boby Sep 01 '14 at 15:52
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actually I am trying to solve a problem I posted earlier and this is one of the solutions. I am just trying to find a nice lower bound. The previous post can be found here: http://math.stackexchange.com/questions/915168/minimum-of-az-x-bz-y – Boby Sep 01 '14 at 15:54
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The obvious lower bound $0$ is sharp, for example $a = 3+\frac{1}{4}$ and $b = 4+\frac{1}{3}$. Are you maybe looking for an upper bound? – Daniel Fischer Sep 01 '14 at 16:01
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But, it's not always 0. Yes, I would like to bound it from above and below. – Boby Sep 01 '14 at 16:05
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1^A much easier example is found by just using integers. – Nishant Sep 01 '14 at 16:05
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1@Nishant one of the few examples where integers make the problem easier^^ – flawr Sep 01 '14 at 16:10
1 Answers
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Sorry for messing things up!
EDIT: The following is not the requride solution, I think the a bound could rather be achieved by finding $$|\lceil a \rfloor b - a \lceil b \rfloor | \geq \min\limits_{(x,y)\in [-0.5,0.5]^2} |(a+x)b-a(b+y)| $$
EDIT2: No this is not the term we search for, since it will result in $0$ in any way.
EDIT3:
One of the following must be true:
$$|\lceil a\rfloor b-a \lceil b \rfloor | \geq | \lfloor a \rfloor b - a \lceil b \rceil | $$
$$|\lceil a\rfloor b-a \lceil b \rfloor | \geq | \lfloor a \rfloor b - a \lfloor b \rfloor| $$
$$|\lceil a\rfloor b-a \lceil b \rfloor | \geq | \lceil a \rceil b - a \lceil b \rceil| $$
$$|\lceil a\rfloor b-a \lceil b \rfloor | \geq | \lceil a \rceil b - a \lfloor b \rfloor| $$
So you can just take the minimum of those as a lower bound (if you are lazy) or think about under which assumptions which ones of those hold. (because I am lazy=) I think you can use the same tactics for finding an upper bound (by just writing $\geq$ insted of $\leq$ or vice versa)
flawr
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I guess not since it comes from the definition of nearest integer function? – Boby Sep 01 '14 at 16:16
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I think this lower bound is wrong. Consider, $a=5.2$ and $b=3.3$. Then $| \lceil 5.2 \rfloor 3.3-\lceil 3.3 \rfloor 5.2|=|53.3-35.2|=|15.6-16.5|=0.9$. But $0.5|a-b|=0.5|5.2-3.3|=0.5|1.9|=0.95$. – Boby Sep 01 '14 at 17:15
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I am sorry, you are right! I think the correct version of my approach would be $|\lceil a \rfloor b - a \lceil b \rfloor | \geq \min\limits_{(x,y)\in [-0.5,0.5]^2} |(a+x)b-a(b+y)|$ but I am not yet sure how to find that. – flawr Sep 01 '14 at 20:39
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No problem. This is still better than what I had before. Could you please let me know if you have any ideas. Thank you very much – Boby Sep 02 '14 at 02:13
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Did you already see the ideas with $\lfloor \rfloor$ and $\lceil \rceil$? It is basically just reformulating your problem but not a solution. The idea with $\min$ does not work since it will always yield 0 as lower bound. – flawr Sep 02 '14 at 08:26
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It is basically the same argument again, but this time it is correct=) BUT you do not get such a nice expression. – flawr Sep 02 '14 at 18:06
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