1. ## reflection imagine

Hi!

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

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

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

Thanks.

2. ## Re: reflection imagine

Originally Posted by Tahoe

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

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

Thanks.