Vrftsjtryk

I believe that I have found a trigonometric expression for both the ceiling and floor function, and I seek confirmation that it is, indeed, correct.

Update

Note that you have the identity $tanleft(frac{pi}{2}-xright)=cot(x)$. Using this, your formula for the floor function is:

$$
begin{split} lfloor x rfloor &= x-frac{1}{2}+frac{arctanleft(tanleft(frac{pi}{2}-picdot xright)right)}{pi} \ &=x-frac{1}{2}+frac{frac{pi}{2}-picdot x+npi}{pi}, text{ for }ninmathbb{Z} \ &=x-frac{1}{2}+frac{1}{2}-x+n \ &=n end{split}
$$

Then there is some $nin mathbb{Z}$ such that $n=lfloor x rfloor$. If, as usual, you insist that $-pile arctan(x) le pi$ this then forces:

$$
begin{split} && -1le frac{1}{2}-x+n le 1 \ &implies& -frac{3}{2}le n-x le frac{1}{2} \ &implies& x-frac{3}{2} le n le x+frac{1}{2} end{split}
$$

So the value for $n$ is close to $lfloor x rfloor$. I'm not sure how computers choose which value to pick for $arctan$ but it appears to always pick the right one for your formula to work. I'm not sure it would be a good idea to use this formula in your work unless you know which value of $arctan$ to pick to ensure that you get the right answer.

If you're using the single-valued $arctan$, i.e. $-frac{pi}{2} leq arctan(x) leq frac{pi}{2}$, which is what all computer languages I've seen use, this works perfectly fine.

Proof:

Let $x = s(a + b)$, where $a in mathbb{N}$, $0 leq b < 1$ and $s in {-1,1}$. We can call $a$ the absolute integer part, $b$ the absolute fractional part, and $s$ is $x$'s sign.

Then:
$$
begin{align*}
frac{arctan(cot(pi x))}{pi}
&= frac{arctan(cot(pi s(a + b)))}{pi}\[1em]
&= frac{arctan(scdotcot(pi a + pi b)))}{pi}\[1em]
&= frac{arctan(scdotcot(pi b))}{pi}\[1em]
&= frac{arctan(scdottan(frac{pi}{2} - pi b))}{pi}\[1em]
&= frac{arctan(scdottan(pi(frac{1}{2} - b)))}{pi}\[1em]
&= frac{arctan(tan(pi s(frac{1}{2} - b)))}{pi}\[1em]
&= frac{pi s(frac{1}{2} - b)}{pi}\[1em]
&= sbigg(frac{1}{2} - bbigg)\[1em]
end{align*}
$$

Note that we can assert that $arctan(tan(pi s(frac{1}{2} - b)) = pi s(frac{1}{2} - b)$ without issues only because

$$
begin{alignat*}{5}
0 && quad leq && b quad && < quad && 1 &&implies\[0.5em]
0 && quad geq && -b quad && > quad && -1 &&implies\[0.5em]
frac{1}{2} && quad geq && quad frac{1}{2} - b quad && > quad && -frac{1}{2} &&implies\[0.5em]
frac{pi}{2} && quad geq && piBigg(frac{1}{2} - bBigg) quad && > quad && -frac{pi}{2} &&\
end{alignat*}
$$

$arctan(tan(x)) = x$ only holds for $x$ in that range. For example, $arctan(tan(x)) = x - pi$ if $frac{pi}{2} < x < pi$.

With that said...
$$
begin{align*}
lfloor x rfloor &= x - frac{1}{2} + sbigg(frac{1}{2} - bbigg)\[0.5em]
&= s(a + b) - frac{1 - s}{2} - sb\[0.5em]
&= s(a + b - b) - frac{1 - s}{2}\[0.5em]
&= sa - frac{1 - s}{2}\[0.5em]
end{align*}
$$

If $s=1$, $lfloor xrfloor = a - frac{1 - 1}{2} = a$.

If $s=-1$, $lfloor xrfloor = -a - frac{1 + 1}{2} = -a - 1 = -(a + 1)$.

This works because $lfloor pm y.uwv rfloor$ is y if the number is positive, and $-(y + 1)$ if the number is negative. e.g. $lfloor 1.2rfloor = 1$ but $lfloor -1.2rfloor = -2$.

Proving that your $ceil$ function works should go about the same way.

I realize I am three years late to this but I initially was going to post my own trigonometric floor:

$$
lfloor x rfloor = x - text{frac}(x)\[1.4em]
%
text{frac}(x) = text{sgn}Big(sin(2pi x)Big)Bigg( frac{cos^{-1}(cos(2pi x))}{2pi} - frac{1}{2} Bigg) + frac{1}{2}left| text{sgn}Big(sin(2pi x)Big)right|\[1.4em]
%
text{sgn}(x) = frac{1}{2}Bigg(frac{cot^{-1}(x) - cot^{-1}(-x)}{big|cot^{-1}(x)big|}Bigg)
$$

But yours seems a bit simpler.