Thread: proving a formula

1. proving a formula

Hello!

I want to prove the following:
$<< k x >>= << k (1-x) >>$ where $<>$ is the distance to the nearest integer. $x \in $0,1$$ and k is an integer.

Here is what I did so far:
Since $x \in $0,1$$ I know there exists a c $\in \{0, 1,..., 2^{n} -1 \}$ with $x \in \left[\frac{c}{2^{n}}, \frac{c+1}{2^{n}} \right] , n \geq 1.$. Thus I can conclude that $x = \frac{s}{2^{n}}$ where $c \leq s \leq c+1 (*)$

Now I got $<< k x >> = << k \frac{s}{2^{n}}>>$

I know because of $(*)$ that $\frac{c \cdot k}{2^{n}} \leq k \frac{s}{2^{n}} \leq \frac{c \cdot k}{2^{n}} + \frac{k}{2^{n}}$

I should now look at the cases that k is even and odd and greater or less than zero. But I don´t get to anything when I look at the last inequality to calculate $<< k \frac{s}{2^{n}}>>$ The case k=0 is trivial.

Plan is to look at k greater and less 0 and k even and odd for this case and then also do the same thing for the right hand side and compare those values and it should be the same result. Then I am done. Can anyone give me a hint on how to progress from here?

Thanks.

2. Re: proving a formula

Consider: kx + k(1-x)

3. Re: proving a formula

Hello SammyS! I don´t really get the point right now. What do you want to do with that term?

4. Re: proving a formula

Let k be some integer, and x be some real number. Then kx is a real number.

The max distance any real number is from the nearest integer is (1/2) .

∴ there is some integer M such that M-(1/2) ≤ kx ≤ M+(1/2).
Case 1: M-(1/2) ≤ kx < M

Then $\ll kx\gg\,=M-kx \,.$

M-(1/2) ≤ kx < M → -M < -kx ≤ (1/2) - M → k - M < k - kx ≤ (1/2) + k - M

Since k - M is the closest integer to k - kx = k(1 - x), we have $\ll k(1-x)\gg\,=k-kx-(k-M)=M-kx \,.$

For Case 1: $\ll kx\gg\,=\ll k(1-x)\gg \,.$

Case 2: kx = M (This case should be easy.)

Case 3: M < kx ≤ M+(1/2) (This case proceeds much like Case 1.

5. Re: proving a formula

Thanks for your answer. I guess I got it.
In the 3rd case I get $<< kx>> = << k(1-x)>> = kx - M$