closest integer value

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July 12th 2011, 10:05 AM
Tahoe

closest integer value

Hello!
Let {x} be the distance of x to the nearest integer value and let $x=0.\overline{b_{1}b_{2} \cdots b_{m}}$ be a binary expansion with repetition.

I want to prove that for all integers k and $k_{0}$ the following is valid:
$\{2^{mk + k_{0} } x\} = \{2^{k_{0} } x \}$

Well, in my opinion it is only valid for positive integers k and $k_{0}$ .
Anyway, here is what I did:

$\{2^{mk + k_{0} } x\} = \{2^{mk + k_{0} } \cdot 0.\overline{b_{1}b_{2} \cdots b_{m}}\} = \{ 2^{k_{0}} \cdot 2^{mk} \cdot } 0.\overline{b_{1}b_{2} \cdots b_{m}} \} = \{ 2^{k_{0}} \cdot \cdot \sum_{i=1}^{m} \overline{b_{i}} 2^{-i + mk} \}$

Now some of these values in the sum must be integer values.
Or do I just have to say that there is going to be a k-times shift of m digits and because of the repetition of x it will be always the same value. I just don´t know how to represent that integer value in the sum which will then disappear because of the definition of {}.

Does anyone have any advice? Thanks!

Regards
July 13th 2011, 07:20 AM
Tahoe

Re: closest integer value

Here is my solution: If there are mistakes could you please let me know? Thank you so much (Clapping)

If we look at $x=0.b_{1} b_{2} \cdots b_{m}$ then
$2^{m} \cdot x = 2^{m} \sum_{i=1}^{m} b_{i} 2^{-i} = \sum_{i=1}^{m} b_{i} 2^{m-i} = b_{1} 2^{m-1} + b_{2} 2^{m-2} \cdots + b_{m} 2^{m-m}$
will be an integer.

When we look at $x=\overline{0.b_{1} b_{2} \cdots b_{m}}$ we get
$2^{m} \cdot \overline{0.b_{1} b_{2} \cdots b_{m}} = 2^{m} \cdot \left(\sum_{i=1}^{m} b_{i} 2^{-i} + \underbrace{0.0 \cdots 0}_{m \ zeros} \overline{b_{1} \cdots b_{m}}\right) = \\ 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + 2^{m} \cdot \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i-m} = 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i}$

Thus
$\{ 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \} = \{\sum_{i=1}^{m} \overline{b_{i}} \}$

We can repeat such a procedure k times and that is why we get the result.

I want to ask: What do you think of that solution? Thanks! :)
July 14th 2011, 02:03 AM
Tahoe

Re: closest integer value

Hello guys!

I don´t want to bother but I really wonder what you think of my solution? I just want to make sure I don´t have any big mistakes in there :)

Thank you!