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December 18th, 2015, 11:11 AM   #1
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Explain

Explain that $\displaystyle \sum \limits_{i=1}^{n} \text{int}\!\left(\!\frac{n}{x_i}\!\right)\leq \text{int}\!\left(\sum \limits_{i=1}^{n} \frac{n}{x_i}\right)$; $\displaystyle x_i \in \mathbb{N}$

Last edited by skipjack; December 18th, 2015 at 04:42 PM.
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December 18th, 2015, 12:11 PM   #2
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Your explanation should talk about how on the left hand side every (non-integer) term is rounded down, while on the right only the total of the sum is rounded. You can think about the fractional parts that are thrown away and how each of them is a positive number. This tells us something about their sum.
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December 18th, 2015, 12:27 PM   #3
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No, I was trying to say show that it is true mathematically.

Last edited by skipjack; December 18th, 2015 at 04:34 PM.
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December 18th, 2015, 12:54 PM   #4
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I've just outlined how you'd do it. I don't intend to do your homework for you.
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December 18th, 2015, 05:20 PM   #5
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$\displaystyle \text{int}(\frac{n}{x_i})=p_i$
$\displaystyle \text{int}(\sum \limits_{i=1}^{n} \frac{n}{x_i})=\text{int}(\sum \limits_{i=1}^{n}L(\frac{n}{x_i}))+\sum p_i \geq \sum p_i$
$\displaystyle \text{int}(\sum L(\frac{n}{x_i}))\geq 0$
And we cannot use this : $\displaystyle \text{int}(a)=\text{int}(b) $ ;$\displaystyle \text{int}(a-\text{int}(b))\neq 0$
It is solved using remainders of $\displaystyle \frac{n}{x_i}=L_i +p_i$

Last edited by skipjack; December 18th, 2015 at 06:24 PM.
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December 18th, 2015, 05:58 PM   #6
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I would suggest:
$$\newcommand{\int}[1]{\text{int}\left({#1}\right)}{n \over x_i}=\int{n \over x_i} + f_i$$
where $f_i \gt 0$ is the fractional part. So then
$$\sum_{i=1}^n {n \over x_i} = \sum_{i=1}^n \left(\int{n \over x_i} + f_i\right) = \sum_{i=1}^n \int{n \over x_i} + \sum_{i=1}^n f_i$$
And then, since the first term on the right is an integer, we have
$$\int{\sum_{i=1}^n {n \over x_i} }= \sum_{i=1}^n \int{n \over x_i} + \int{\sum_{i=1}^n f_i}$$
And the last term is non-negative.

It may be necessary to justify more completely the last equation there.
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December 18th, 2015, 06:18 PM   #7
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$\displaystyle p=kx+r$
It is like saying : $\displaystyle A$mod$\displaystyle B$$\displaystyle \geq 0$
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