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Thread: reflection imagine

  1. #1
    Junior Member
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    Jun 2011
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    reflection imagine

    Hi!

    I got the following:
    A function $\displaystyle f_{i} \left(x \right) = \int_{0}^{x} (-1)^{\lfloor t \cdot 2^{i}\rfloor} \ dt$ and the binary expansions with $\displaystyle \tilde{b_{i}} = 1 - b_{i}$ which are given by
    $\displaystyle x_{0}= 0.....0b_{n+1} b_{n+2} \dots$ and
    $\displaystyle x_{1}= 0.....0\tilde{b_{n+1}} \tilde{b_{n+2}} \dots$

    I guess that the function $\displaystyle f_{i} \left(x \right)$ is reflection symmetric about the vertical line $\displaystyle x=2^{-(n+1)}, n \geq 1$.

    Is the reflection imagine about that line of $\displaystyle x_{0}$ $\displaystyle x_{1}$ because I can verify that
    if $\displaystyle x_{0} < x_{1}$ then I get $\displaystyle x_{0} + 2 (x - x_{0}) = x_{1}$ and
    if $\displaystyle x_{1} < x_{0}$ then I get $\displaystyle x_{1} + 2 (x - x_{1}) = x_{0}$ ?

    Thanks.
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  2. #2
    Junior Member
    Joined
    Jun 2011
    Posts
    36

    Re: reflection imagine

    Quote Originally Posted by Tahoe View Post

    Is the reflection imagine about that line of $\displaystyle x_{0}$ $\displaystyle x_{1}$ because I can verify that
    if $\displaystyle x_{0} < x_{1}$ then I get $\displaystyle x_{0} + 2 (x - x_{0}) = x_{1}$ and
    if $\displaystyle x_{1} < x_{0}$ then I get $\displaystyle x_{1} + 2 (x - x_{1}) = x_{0}$ ?

    Thanks.
    My question here is if that is the way on how to show that the reflection image of $\displaystyle x_0$ is $\displaystyle x_1$ over the line $\displaystyle x=2^{-(n+1)}$? Or do I have to show it in a different manner?
    If I calculate it that way it is all correct.

    Thanks.
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