Inversion of an Equation with Floor

Whilst working with the equations for the previous post on the fusc function, I encountered one I wished to invert. That equation was essentially

y = x - 2 \lfloor x \rfloor

but I had not noticed that this equation is special, and that the ‘2’ is significant.

It turns out that the inverse is identical:

x = y - 2 \lfloor y \rfloor

Similarly, the inverse of

y_2 = 2 \lfloor x_2 \rfloor - x_2

is given by

x_2 = 2 \lceil y_2 \rceil - y_2

but note the switch from floor to ceiling in this case.

I wondered if equations slightly more general than this had a solution, and it seems that the answer is essentially ‘no’. To be more specific, consider the equation

y = f(x) = x + c \lfloor x \rfloor

Following is a table of some properties of this function of x, for different values of c.

c inj surj cts mono bound
 (−∞,−2)  ✔  ✗  ✗  ✗  ✗
 −2  ✔  ✔  ✗  ✗  ✗
 (−2,−1)  ✗  ✔  ✗  ✗  ✗
 −1  ✗  ✗  ✗  ✗  ✔ to [0,1)
 (−1,0)  ✗  ✔  ✗  ✗  ✗
 0  ✔  ✔  ✔  ✔ (<)  ✗
 (0,+∞)  ✔  ✗  ✗  ✔ (<)  ✗


inj = injective or 1-to-1 (f(y_1) = f(y_2) \implies y_1 = y_2 )
surj = surjective or onto (\forall y \exists x \bullet f(x) = y)
cts = continuous
mono = monotonic:
(<) for strictly increasing (x_1 < x_2 \implies f(x_1) < f(x_2))
bound = bounded range

For f to be invertible (i.e. for f to have an inverse), f must be bijective, that is, both injective and surjective. There are only two values for which that is the case: c \in \{-2,0\}. When c=0, f is just the identity function, and thus the case c=−2 is the only interesting case.

Please see also the question I raised on MathOverflow prior to writing this entry.


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