If $\{x\}:=x-\lfloor x\rfloor$, then are these the same? $\{l+m\}$, $\left\{l+\{m\}\right\}$, $\left\{\left\{l\right\}+\{m\}\right\}$

Question

I am currently working on some signal processing project and come across this particular problem:

Define $\{x\} := x - \lfloor x\rfloor$ and consider $$y_1 = \{l+m\}$$ $$y_2=\left\{l+\{m\}\right\}$$ $$y_3=\left\{\left\{l\right\}+\{m\}\right\}$$ Are $y_1$, $y_2$ and $y_3$ the same?

My gut feeling is that they are all the same, since

\begin{align} \exp\left(2\pi j l\right) &= \exp\left(2\pi j\{l\}\right)\\ \exp\left(2\pi j m\right) &= \exp\left(2\pi j\{m\}\right)\\ \exp\left(2\pi j l\right)\exp\left(2\pi j m\right) &= \exp\left(2\pi j\{l\}\right)\exp\left(2\pi j\{m\}\right)\\ &=\exp\left(2\pi j(\{l\}+\{m\})\right)\\ &=\exp\left(2\pi j\{\{l\}+\{m\}\}\right)\\ \exp\left(2\pi j l\right)\exp\left(2\pi j m\right) &=\exp\left(2\pi j (l+m)\right) \\ &=\exp\left(2\pi j \{l+m\}\right) \\ \exp\left(2\pi j l\right)\exp\left(2\pi j m\right) &=\exp\left(2\pi j l\right)\exp\left(2\pi j \{m\}\right)\\ &=\exp\left(2\pi j (l+\{m\})\right) \\ &=\exp\left(2\pi j \{l+\{m\}\}\right) \end{align}

I would greatly appreciate any help to verify the proof.

Answer 1

$y_1 = \{l+m\} = \{\lfloor l \rfloor\ + \{l\} + \lfloor m \rfloor\ + \{m\}\} = \{\{l\} + \{m\}\}$

$y_2 = \{l+\{m\}\} = \{\{l\} + \lfloor m \rfloor\ + \{m\}\} = \{\{l\} + \{m\}\}$

$y_3 = \{\{l\} + \{m\}\}$

So yes, they are the same.

Answer 2

Consider $l = n_1+f_1$ and $m=n_2+f_2$, where $n_1, n_2$ are integers and $f_1, f_2$ are the fractional parts. We then have that $$y_1=\{n_1+n_2+f_1+f_2\} = \{f_1+f_2\}$$ $$y_2 = \{ n_1+f_1+f_2\} = \{ f_1+f_2\}$$ $$y_3 = \{ f_1+f_2\}$$

Therefore all are equal to $\{f_1+f_2\}$

Answer 3

While your computations involving the exponential might help you to build some intuition (particularly in the context of signal processing), if you really want to determine whether or not the expressions are equivalent, then you need to work with the definition of the the fractional part, rather than with examples. Examples build intuition, but don't give proof. More generally, when you don't know what else to do, write down the definition and see where it gets you.

Gabe has provided these computations, but I will reiterate them (with some additional commentary) so that this answer remains self-contained. The two key insights are that

1. $\{ x + m \} = \{ x \}$ for any integer $m$ and any $x\in\mathbb{R}$ (this, really, is the more important observation‐it lets us put integers into braces and pull them out again all willy-nilly), and
2. $x = \lfloor x \rfloor + \{ x \}$ for any $x\in\mathbb{R}$ (by algebraic manipulation of the definition of $\{x\}$).

Starting with the first expression \begin{align} y_1 &= \{ l + m \} \\ &= \left\{ \left( \lfloor l \rfloor + \{l\} \right) + \left( \lfloor m \rfloor + \{m\} \right) \right\} && \text{by (2), applied to $l$ and $m$} \\ &= \left\{ \left( \lfloor l \rfloor + \lfloor m \rfloor \right) + \left( \{l\} + \{m\} \right) \right\} && \text{addition is commutative} \\ &= \left\{ \{ l \} + \{ m \} \right\} && \text{by (1), since $\lfloor l \rfloor + \lfloor m \rfloor$ is an integer} \\ &= y_3. \end{align} By a similar computation, \begin{align} y_2 &= \left\{ l + \{m\} \right\} \\ &= \left\{ \left( \lfloor l \rfloor + \{ l \} \right) + \{m\} \right\} && \text{by (2), applied to $l$} \\ &= \left\{ \{l\} + \{m\} \right\} && \text{by (1), since $\lfloor l\rfloor$ is an integer} \\ &= y_3. \end{align} Therefore $y_1 = y_2 = y_3$, which gives us the desired result.