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

Question

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.

Answer 1

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.