# What floor function identity makes this true: $(-1)^{\lfloor x\rfloor} = -2\lfloor x\rfloor + 4\left\lfloor\frac {\lfloor x\rfloor}2\right\rfloor +1$?

I know that the graph of these two functions is the same:

$$(-1)^{\lfloor x\rfloor} = -2\lfloor x\rfloor + 4\left\lfloor\frac {\lfloor x\rfloor}2\right\rfloor + 1$$

Both of them interchange sign at integer points in the same manner. I'm trying to figure out what identity allows them to be equal, though. I know that I cannot apply a logarithm as that wouldn't do any good. I'm just trying to figure out how I could relate this to floor as an identity rather than just two alternate forms for the exact same function.

• But there is no broader subject area that this function belongs to at this given time... – The Great Duck Feb 24 '16 at 19:30
• Just write $n$ instead of $\lfloor x\rfloor$, and you're halfway there. – user147263 Feb 24 '16 at 19:32
• If I may ask, why did you ass Algebra-Precalculus? Is that really what floor falls under? – The Great Duck Feb 24 '16 at 19:35

The only property of the floor function you need to use is, for $n$ integer: $$\left\lfloor\frac n2\right\rfloor= \begin{cases} \frac{n-1}2&\mbox{if n is odd}\\ \frac n2 &\mbox{if n is even}\\ \end{cases}$$ Prove the result by writing $n:=\lfloor x\rfloor$; then argue by cases.
• Yeah, that's the identity. If there's a generalization to be made of the result you found, it might be that $f(n):= n-k\lfloor\frac nk\rfloor$ takes $k$ possible values, and will generate a periodic sequence of those values. – grand_chat Feb 24 '16 at 20:06