Jtdylktuy

I am trying to use floor and ceiling functions to construct a 2-periodic extension of the function $f(x) = |x|, -1 leq x leq 1$.

Through trial an error I have been able to show that:

$f(x) = 1 - bigg( lfloor x rfloor - 2 lfloor frac{lfloor x rfloor}{2} rfloor)(x - lfloor x rfloor) + (lfloor x-1 rfloor - 2lfloor{frac{lfloor x-1 rfloor}{2}}rfloor)(lceil x-1 rceil -(x-1)bigg)$

However, this formula does not work when $x$ is an even integer since it gives 1 instead of 0.

Is there an easier way to do this?

If you can also use trigonometric functions, then there are several ways to do it.

If we can construct a graph like the following, which I will call $r(x)$ then $f(x)=r(x)+r(-x)$ will almost be what you want depending whether or not $f(2k)=1$ and $f(2k+1)=0$ for all integers $k$.

First, let us create a periodic "cutting" function

$$ c(x)=lfloor1+sinpi xrfloor $$

Now look at the graph of

$$ g(x)=lceil xrceil-x $$

If we multiply $g(x)$ and $c(x)$ we will get a function whose graph looks like the first graph above. So we define

begin{eqnarray}
r(x)&=&g(x)c(x)\
&=&(lceil xrceil-x)lfloor1+sinpi xrfloor
end{eqnarray}

And let

begin{eqnarray}
f_0(x)&=&r(x)+r(-x)\
&=&(lceil xrceil-x)lfloor1+sinpi xrfloor+(lceil -xrceil+x)lfloor1-sinpi xrfloor
end{eqnarray}

This function has the following graph

This is almost what we want, but we need to correct the missing points by adding a function which equals one for all even integers and equals zero elsewhere. A function such as

$$ c(x)=leftlfloorfrac{1+cospi x}{2}rightrfloor $$

So we define

$$ f(x)=(lceil xrceil-x)lfloor1+sinpi xrfloor+(lceil -xrceil+x)lfloor1-sinpi xrfloor+leftlfloorfrac{1+cospi x}{2}rightrfloor $$