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What is the period of the function $f(x)=x-[x]$?

(Here, $[\,.]$ represents the greatest integer function)

On a tangential note: does that affect the periodicity of $e^{x-[x]}$?

HK Lee
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Shivanandan
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3 Answers3

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The period is 1, because $f(x+1) = (x+1) - [x+1] = x+1-([x]+1) = x-[x] = f(x)$. Common sense suggest that it cannot be less than 1, but to prove this rigorously you only need to see that for $x \in [0,1)$, $f(x) = x$ so it's injective on an interval of length 1.

This also influences the other function: if $f(x)$ is periodic with a period $a$, $g(f(x))$ is also periodic with period $a$ (or smaller).

Glorfindel
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  • Just a warning to the OP, This only proves that the period of $f$ is at most 1. – 5xum Mar 15 '17 at 07:59
  • @5xum you're right. 'Common sense' suggests that it's not less than 1, but 'common sense' also has problems with the Banach-Tarski paradox. – Glorfindel Mar 15 '17 at 08:04
  • Yup. I mean, it's fairly easy to prove that if the period is not smaller than $1$ (which I see you did), but it's good to konw for the OP that he still has to show that. – 5xum Mar 15 '17 at 08:08
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We know that, for all $x\in\mathbb{R}$ :

$$[x]\le x<[x]+1$$

Hence :

$$[x]+1\le x+1<([x]+1)+1$$

Which proves that $[x]+1$ is the greatest integer less that $x+1$. In other words :

$$[x+1]=[x]+1$$

which can be written :

$$x+1-[x+1]=x-[x]$$

As a conclusion, $f:x\mapsto x-[x]$ is $1$-periodic.

Adren
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For any periodic function: $f(x) = f(x+T)$ , where T is the fundamental period or a multiple of it

So, $x -[T]= (x +T) - [x+T] $ $[x+T]=[x]+T $

Which implies that , $ T \in Z$ And we want the fundamental period to be the smallest and positive : $T=1$

daksh
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