reflection imagine

**Tahoe** · July 9th 2011, 11:55 AM

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.

**Tahoe** · July 10th 2011, 08:49 AM

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.

Thread: reflection imagine

LinkBack

Thread Tools

Display

reflection imagine

Re: reflection imagine

Similar Math Help Forum Discussions

Imagine that an impact occurred on the continental plate millions and millions of yea

Reflection

Reflection

imagine and real part of inner product?

Reflection

Search Tags