Question

Is floor(x) periodic?

Answer 1

No

Explanation:

A periodic function has a period T such that, for every x, we have

`f(x+T)=f(x)`

In this case, it would mean that there exists a number T such that, for every x,

`\lfloor x+T\rfloor = \lfloor x \rfloor`

but we can always choose a value for x that is near enough (i.e. less than T) an integer. For example, if T=0.2, we can choose x=0.9 (which is 0.1 away from the next integer, which is less than 0.2).

This choice implies that
`x+T` and
`x` don't share the same integer part, and thus their floor is different:

`\lfloor x+T\rfloor = \lfloor 0.9+0.2\rfloor =\lfloor 1.1\rfloor=1`

whereas

`\lfloor x \rfloor=\lfloor 0.9 \rfloor=0`

Answer 2

If you look at the graph of y = floor(x), you'll see a stairstep pattern that climbs up as you read from left to right. There are no vertical components to the graph. There are only horizontal components.

The graph is not periodic because the y values do not repeat themselves after a certain x value is passed. For instance, start at x = 0 and go to x = 3. You'll see the y values dont repeat themselves as if a sine function would. If you wanted the function to be periodic, the steps would have to go downhill at some point; however, this does not happen.

Conclusion: The function floor(x) is not periodic.