Mutually odd functions

The floor of a real number x is the largest integer n ≤ x, written ⌊x⌋.

The ceiling of a real number x is the smallest integer n ≥ x, written ⌈x⌉.

The floor and ceiling have the following symmetric relationship:

⌊-x⌋ = -⌈x
⌈-x⌉ = -⌊x

The floor and ceiling functions are not odd, but as a pair they satisfy a generalized parity condition:

f(-x) = -g(x)
g(-x) = -f(x)

If the functions f and g are equal, then each is an odd function. But in general f and g could be different, as with floor and ceiling.

Is there an established name for this sort of relation? I thought of “mutually odd” because it reminds me of mutual recursion.

Can you think of other examples of mutually odd functions?

Related posts:

Saved by symmetry
Odd numbers in odd bases
The power of parity

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9 comments on “Mutually odd functions
  1. Richard Holmes says:

    Am I seriously misunderstanding something, or does _every_ function have a function with which it’s mutually odd? In general any function on the reals f(x) can be written as o(x)+e(x) where o(x) is an odd function and e(x) is an even function. But then if g(x) = o(x)–e(x), we have f(–x)=o(–x)+e(–x)=–o(x)+e(x)=–g(x), and likewise g(–x)=–f(x).

  2. John says:

    Good observation. I didn’t realize that it’s that simple.

  3. Kyle says:

    I feel like there should be a category way of thinking about this.

  4. It’s even a bit easier than that. The mop (mutually odd partner) of f is g defined by

    g(x) = -f(-x)

  5. John Moeller says:

    In fact g(x) = -f(-x).

  6. Benjamin says:

    Btw, these two operations have an interesting commutator

    [⌊, ⌈]x = [⌈, ⌊]x = 1

  7. Dan says:

    Amplifying what John Moeller pointed out: for any real function on a symmetrically defined domain g(x)=-f(-x) defines such a pair, and similarly g(x)=+f(-x) defines a “mutually even pair”.

  8. neizod says:

    one thing that remind me of this mutually-relation is differential over trigonometry function:

    d/dx sin(x) = cos(x)
    d/dx cos(x) = -sin(x)
    d/dx -sin(x) = -cos(x)
    d/dx -cos(x) = sin(x)

  9. Thomas says:

    f(-x) = -g(x) implies g(-x) = -f(x):

    f(-x) = -g (x);
    y = -x;
    => f(y) = -g(-y);
    -f(y) = g(-y);