Like without using if even or odd. Like how you can define a bijection $f\colon\mathbb{N}\to\mathbb{Z}$ by is $f(n)=\lfloor n/2\rfloor\cdot(-1)^n$.
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1What exactly do you mean by "piecewise"? – MJD Nov 19 '14 at 19:29
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@MJD Usually that means the function has multiple definitions, one for each portion of the domain. – Dunka Nov 19 '14 at 19:40
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The floor function is defined piecewise, so I'm not sure that "counts". – D Wiggles Nov 19 '14 at 20:11
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What's the problem with “piecewise functions”? – egreg Nov 19 '14 at 21:04
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I was just curious if such a function existed. It seems all of them are piecewise. – srujzs Nov 20 '14 at 02:36
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If $0\notin \Bbb N$, then $$\left|2x-\frac12\right|+\frac12$$ works. If not, then subtract $1$ to get $\left|2x-\frac12\right|-\frac12$. To be honest, absolute values have an implicit poecewise-ness to it, but it's hidden away.
Arthur
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2Replace it by $\sqrt{x^2}$ then. Then it's hidden even better, in the convention for the square root. – Peter Franek Nov 19 '14 at 19:47
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@PeterFranek That's a good idea. I would not say that $\sqrt{x^2}$ is piecewise. – Eff Nov 19 '14 at 19:48
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Let $f(x) = \max(2x+1,-2x) = {1\over 2} + \max(2x+{1\over 2},-{1 \over 2}-2x) = {1 \over 2} + |2x+{1\over 2}|$.
Then $f: \mathbb{Z} \to \mathbb{N}$ is a bijection.
copper.hat
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IBWiglin's example, written another way: $$ A(n) := \frac{(-1)^n(-2n-1)+1}{4} $$ Then $$ A(0)=0, \\ A(1)=1, \\ A(2)=-1, \\ A(3)=2, \\ A(4 )=-2 $$ The inverse of this function can presumably be written in terms of the Lambert W function...
GEdgar
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1I tried that Lambert thing, but you need to use a different branch of W for each value ... sort of the ultimate in piecewise writing. – GEdgar Nov 19 '14 at 21:38
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If I can use the absolute value, then, for example, $$2|x|+\frac{1}{2}\frac{x+1/2}{|x+1/2|}+\frac{1}{2}.$$
Peter Franek
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It doesn't exactly answer your question, but I made a function going the opposite direction $f\colon\mathbb{N}\to\mathbb{Z}$ which is bijective and not implicitly defined piecewise like the example you gave. $$ f(n)=\frac{1}{4}-\frac{1+2n}{4}\cos(\pi n) $$
This is the restriction to $\mathbb{N}$ of the spiral $f\colon \mathbb{R}\to\mathbb{C}$ defined in the complex plane by $$ f(x)=\frac{1}{4}-\frac{1+2x}{4}e^{ix}$$ which can be extended to a complex analytic function on all of $\mathbb{C}$.
D Wiggles
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