5
$\begingroup$

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?

  • 1
    For all $x>0$, we have $\arctan(x)+\arctan(1/x)=\pi/2$$ – 2017-01-15
  • 1
    If $f(x)$ is one such function, then $g(x)=af(x)+b$ is another. – 2017-01-15
  • 1
    If $f(x)$ is one such function, then so is $f(1/x)$. – 2017-01-15
  • 0
    @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. – 2017-01-15
  • 1
    @Veliko Right; or you can consider $f(x)=sgn(x)\,\arctan(x)$ where $sgn$ denotes the sign function. Now you have : $\forall x\neq0,\,f(x)+f(1/x)=\pi/2$ – 2017-01-15

1 Answers 1

5

For any function $h(x)$, let $f(x)=A[h(x)-h(1/x)]+B$. Then, $$ f(1/x)=A[h(1/x)-h(x)]+B\implies f(x)+f(1/x)=2B. $$

  • 0
    So, for example, $x-\frac1x$ works? – 2017-01-15
  • 0
    @AkivaWeinberger Yes, it does. – 2017-01-15
  • 1
    This is probably the most general answer possible for continuous functions. Caution is needed when working with the Domain. But thank you very much. – 2017-01-15