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.

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

  • $\begingroup$ I was more thinking if there was any kind of identity I could pull from the equation. I'm satisfied that they are equal. I was more just thinking if there was any kind of general relation having to do with exponential functions. I built the two equations myself, so I know why they oscillate. I was just trying to see if I could pull an identity from it. I guess... that is the identity. $\endgroup$ – The Great Duck Feb 24 '16 at 19:49
  • $\begingroup$ 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. $\endgroup$ – grand_chat Feb 24 '16 at 20:06
  • $\begingroup$ Well that's not real special. All you've done is write n % k. I was thinking more of how to remove that pesky (-1)^floor(x). Like I said, there probably isn't really any identity to be garnered from this. Thank you, though. $\endgroup$ – The Great Duck Feb 24 '16 at 21:33
  • $\begingroup$ To think back on this old question I think I was thinking that a series of sub-steps existed each one bringing one equation closer to the other. I've realized since then that such identities are actually the rarer variant and that in reality most of lower mathematics is the rarest math in the grand spectrum of all math. I'm addressing this here in a comment as i believe this is the actual flaw in my thinking back then and others might fall into that same pitfall. My reasoning wasn't wrong. I worked by observation. It was the set of observations that was biased. $\endgroup$ – The Great Duck Oct 31 '18 at 4:06

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