2
$\begingroup$

How do I prove that $f(x) = x - {\lfloor}x{\rfloor}$ is periodic and find its minimal period?

I've taken the following steps:

Let $x = x_0 + \Delta{x}$ where $x_0 \in \mathbb Z$ and $\Delta{x} \in [0;1)$. I need to prove that for given $x$ and $T$: $f(x) = f(x+T)$ where $T$ is some period to be defined. Let $T = n + \Delta{T}$ where $n \in \mathbb N$ and $\Delta{T} \in [0, 1)$

$$ f(x) = f(x+T) \\ x- {\lfloor}x{\rfloor} = x+T - {\lfloor}x+T{\rfloor} \\ x_0+\Delta{x} - {\lfloor}x_0+\Delta{x}{\rfloor} = x_0 + \Delta{x} + n +\Delta{T} - {\lfloor}x_0 + \Delta{x} + n + \Delta{T}{\rfloor} $$

So since $\Delta{x} \in [0;1)$ and $x_0 \in \mathbb Z$ then ${\lfloor}x_0+\Delta{x}{\rfloor} = x_0 +{\lfloor}\Delta{x}{\rfloor}$. Based on that LHS may be rewritten as:

$$ x_0+\Delta{x} - {\lfloor}x_0+\Delta{x}{\rfloor} = \Delta{x} - {\lfloor}\Delta{x}{\rfloor} = \Delta{x} $$

At the same time:

$$ x_0 + \Delta{x} + n +\Delta{T} - {\lfloor}x_0 + \Delta{x} + n + \Delta{T}{\rfloor} = \\ = (x_0 + n) + \Delta{x} + \Delta{T} - (x_0 + n) - {\lfloor}\Delta{x} + \Delta{T}{\rfloor} = \\ = \Delta{x} + \Delta{T} - {\lfloor}\Delta{x} + \Delta{T}{\rfloor} $$ That means that for LHS and RHS to be equal $\Delta{T}$ must be equal to $0$, hence $T=n+\Delta{T} \in \mathbb N$. And the smallest natural number is $1$, which gives that the function is indeed periodic and its minimal period is $1$.

Is the proof above valid?

$\endgroup$
3
  • 2
    $\begingroup$ It's valid, but overkill. $f(x)=f(y)$ iff they have the same fractional part, which happens iff they differ by an integer. Since the integer with smallest nonzero absolute value is 1, it follows that 1 is the minimal period. $\endgroup$ Jun 27, 2018 at 18:03
  • $\begingroup$ Before trying to prove it's periodic, just ask how in general this function behaves. After you answer that it will be clear what the period is. $\endgroup$ Jun 27, 2018 at 19:20
  • $\begingroup$ Too long. I didn't read it. $[x+n]=[x]+n;n\in\mathbb Z$ is easy to prove so $f(x+1)=(x+1)-[x+1]=(x+1)-[x]-1=x-[x]=f(x)$ so $f(x)$ has a period of $1$. If $k<1$; $n-1< x <n-k$ for some integer $n$ then $f(x)=x-n+1$ but $f(x+k)=x+k-n+1$. So $f$ does not have period $k$. So $1$ is minimum period. $\endgroup$
    – fleablood
    Jun 27, 2018 at 21:26

2 Answers 2

7
$\begingroup$

Note that $$f(x+T) = f(x) \iff x+T - \lfloor x+T\rfloor = x -\lfloor x\rfloor \iff T=\lfloor x+T\rfloor-\lfloor x\rfloor.$$ Since the RHS is the difference of two integers, $T$ must be an integer. Also, $T=\lfloor x+T\rfloor-\lfloor x\rfloor$ for all integers. As such, the minimal period is $T=1$.

$\endgroup$
2
  • $\begingroup$ This one is much more elegant than mine, thanks $\endgroup$
    – roman
    Jun 27, 2018 at 19:13
  • $\begingroup$ I do have to admire the eloquence. $\endgroup$
    – fleablood
    Jun 27, 2018 at 21:28
1
$\begingroup$

Proof:

$x= \text{floor}(x) + \{x\}$ where ${x}$ denotes the fractional part function.

$\implies f(x)= \{x\}$

$\implies f(x+1)= \{x+1\}$

$\implies f(x+1)=\{x\} $

$\implies f(x)=f(x+1)$.

Thus, $f(x)$ is periodic with period $1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.