# How do you call functions that fulfill $f(x)=\pm f(\pm 1/x)$?

A function $f(x)$ that fulfills $f(x)=\pm f(-x)$ is called (a)symmetric even/odd.

How do you call functions that fulfill $f(x)=\color{blue}\pm f(\color{red}\pm 1/x)$?

"$\color{red}{\text{Positive/Negative}}$ Reciprocal $\color{blue}{\text{(anti)symmetric}}$"..?

edited according to the comments...

• Usually a function satisfying $f(x) = \pm f(-x)$ is called even or odd, not (a)symmetric. Maybe (anti)symmetric with respect to $x \to -x$. Certainly it should be (anti) rather than (a), because "asymmetric" just means "not symmetric". And then if $s$ is an involution, you can call a function satisfying $f(x) = \pm f(s(x))$ (anti)symmetric with respect to the involution $s$. – Robert Israel Jan 23 '13 at 0:11
• If you only consider $f(x)=\pm f(1/x)$ on $\mathbb R^+$, then $g=f\circ\exp$ is even/odd. Equivalently, $f$ is the composition of an even/odd function and the logarithm function... – user856 Jan 23 '13 at 0:19
• '-1', I did not find your question useful to me. – user103816 Jun 3 '14 at 17:17
• @user31782 when you think I downvoted your other question, you are wrong. I upvoted your answer instead. Very sad that you feel the need for revenge. Where is the love... – draks ... Jun 3 '14 at 19:31
• Hi @draks... the downvote is not due to revenge. Let's talk about this on meta: meta.math.stackexchange.com/q/13921/103816 – user103816 Jun 4 '14 at 1:50