# Functions $f$ that satisfy $f(x) + f(1/x) = \rm constant$ for each $x$.

Hello I am thinking about all the functions that satisfy $$f(x)+f(1/x) = C$$ for each $x$. The constant $C$ is the same for all $x$ in the Domain.

It is clear that $\log_a(x)$ works for all possible bases. Can you think of another function?

• For all $x>0$, we have $\arctan(x)+\arctan(1/x)=\pi/2$$Commented Jan 15, 2017 at 16:52 • If f(x) is one such function, then g(x)=af(x)+b is another. Commented Jan 15, 2017 at 16:55 • If f(x) is one such function, then so is f(1/x). – user384138 Commented Jan 15, 2017 at 16:59 • @Adren, good example, though. If I make the correction f(x) = arctan(\sqrt{x}) + arctan(1/\sqrt{x}) = \pi/2 for all x>0 and this is everything in the domain now. Commented Jan 15, 2017 at 17:00 • @Veliko Right; or you can consider f(x)=sgn(x)\,\arctan(x) where sgn denotes the sign function. Now you have : \forall x\neq0,\,f(x)+f(1/x)=\pi/2 Commented Jan 15, 2017 at 17:08 ## 1 Answer For any function h(x), let f(x)=A[h(x)-h(1/x)]+B. Then,$$ f(1/x)=A[h(1/x)-h(x)]+B\implies f(x)+f(1/x)=2B.$$• So, for example,$x-\frac1x\$ works? Commented Jan 15, 2017 at 17:04
• @AkivaWeinberger Yes, it does. Commented Jan 15, 2017 at 17:05
• This is probably the most general answer possible for continuous functions. Caution is needed when working with the Domain. But thank you very much. Commented Jan 15, 2017 at 17:08