Solving $f\left(x\right)+f\left(1/x\right)=K$

Question

I would like to know if there's a way to solve $$ \forall x \in \mathbb{R}^{*+}, \ K \in \mathbb{R}, \ f\left(x\right)+f\left(\frac{1}{x}\right)=K $$ We know that for $\displaystyle K=\frac{\pi}{2}$ we have ( $x \in \mathbb{R}^{*+}$ ) $$ \text{arctan}\left(x\right)+\text{arctan}\left(\frac{1}{x}\right)=\frac{\pi}{2} $$ And for $K=0$ $$ \ln\left(x\right)+\ln\left(\frac{1}{x}\right)=0 $$ Only thing i've found is that it implies $$ -f'\left(\frac{1}{x}\right)+x^2f'\left(x\right)=0 \text{ and }f\left(1\right)=\frac{K}{2} $$

Answer 1

$$f(x)+f(1/x)=K$$ Change of function : $\quad f(x)=F\left(\ln|x|\right)+\frac{K}{2}$

$f(1/x)=F\left(\ln|1/x|\right)+\frac{K}{2}=F\left(-\ln|x|\right)+\frac{K}{2}$

$$F\left(\ln|x|\right)+F\left(-\ln|x|\right)=0$$ Change of variable : $\quad X=\ln|x|$ $$F(X)+F(-X)=0$$ $$F(X)=-F(-X)$$ Thus $F(X)$ is any odd function. The solution is : $$f(x)=F(\ln|x|)+\frac{K}{2}\quad\text{with any odd function }F.$$ EXAMPLES :

With $F(X)=X$ :

$\begin{cases} f(x)=\ln|x|+\frac{K}{2}\\ f(1/x)=-\ln|x|+\frac{K}{2} \end{cases} \quad f(x)+f(1/x)=\ln|x|+\frac{K}{2}+\left(-\ln|x|+\frac{K}{2}\right)=K$

With $F(X)=\sin(X)$ :

$\begin{cases} f(x)=\sin(\ln|x|)+\frac{K}{2}\\ f(1/x)= -\sin(\ln|x|)+\frac{K}{2} \end{cases} \quad f(x)+f(1/x)=\sin(\ln|x|)+\frac{K}{2}+\left( -\sin(\ln|x|)+\frac{K}{2} \right)=K$

Etc.

Answer 2

First, I should point out that your quantifiers are badly written. It should be $$\exists K,\forall x>0, f(x)+f(1/x)=K$$ or something similar.

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The map $\Phi_K:\{\text{odd functions }\Bbb R\to\Bbb R\}\to\{\text{functions }(0,\infty)\to \Bbb R\},$ $\ \Phi_K(g)=\frac K2+g\circ \ln$ establishes a one-to-one relation onto the subset of the functions that solve your equation: namely, if $f$ satisfies it, then $-\frac K2+f\circ \exp$ is odd and $\Phi_K(-\frac K2+f\circ \exp)=f$. The map $\Phi_K$ preserves regularity (namely, $\Phi_K(g)$ is $C^h$ if and only if $g$ is).

Answer 3

Sure, take $f(x) = \log(x)+\frac{K}{2}$.

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Edit: I'm interpreting the problem as

$$ \forall K\in\mathbb{R}, \ \exists f:\mathbb{R}^{*+}\to\mathbb{R} \text{ s.t. } \forall x \in\mathbb{R}^{*+}, \ f(x)+f\left(\frac{1}{x}\right) = K$$

Answer 4

As pointed out in the comments, the general soultion is too broad. But, based on The functional equation $f(y) + f\left(\frac{1}{y}\right) = 0$, we can at least write it as $$\boxed{f(x)=C\left(x,\frac{1}{x}\right)+\frac{K}{2},}$$ where $C(u,v)$ is any antisymmetric function, a function that satisfies $C(u,v)=-C(v,u)$. This is a bit of a cheat since we have just converted one functional equation to another, but at least we have turned it into bit more familiar concept (antisymmetric function), this can make it easier for finding some examples.

Indeed, trivial one is the $C(u,v)=0$, which leads to $$f(x)=\frac{K}{2}.$$

Or try $C(u,v)=\frac{1}{2}\log\left(\frac{u}{v}\right)$, then $$ f(x)=\log x+\frac{K}{2}, $$ or for $C(u,v)=\arctan\left(\sqrt{\frac{u}{v}}\right)-\frac{\pi}{4}$, then $$f(x)=\arctan x-\frac{\pi}{4}+\frac{K}{2}.$$

The above examples were designed to resemble the already touched ones, I am sure you can modify them to generate new ones.

Another one, based on What characteristics do functions have, where $f(x,y) = -f(y,x)$? we have another antisymmetric function $C(x,y)=x^2-y^2$, yielding $$f(x)=x^2-\frac{1}{x^2}+\frac{K}{2}.$$

And so on and on...