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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 $$ -f'\left(\frac{1}{x}\right)+x^2f'\left(x\right)=0 \text{ and }f\left(1\right)=\frac{K}{2} $$

B. Mehta
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Atmos
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  • What does $\mathbb{R}^{*+}$ denote? – B. Mehta Jun 01 '18 at 22:58
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    This is nearly trivial, I'm afraid; since $x\leftrightarrow 1/x$ is a bijection between $(0,1)$ and $(1, \infty)$, you can define an arbitrary $f$ on $(0,1)$ and then define $f(x)=K-f(1/x)$ for all $x\gt 1$. Continuity/differentiability conditions might constrain the value and/or derivative(s) of your function at $x=1$, but not enough to make solutions at all unique. – Steven Stadnicki Jun 01 '18 at 22:59
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    (Indeed, you can see this even with your examples; $f(x) = \arctan(x)-\pi/4$ looks nothing like $\ln(x)$ but still satisfies $f(x)+f(1/x)=0$ by dint of the first identity.) – Steven Stadnicki Jun 01 '18 at 23:05

4 Answers4

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$$f(x)+f(1/x)=K$$ 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.

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

Edit: I'm interpreting the problem as

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

Alex Jones
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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...

Sil
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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.

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).

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    While I agree that OP question is not really clear, I'm not sure that you rewriting is better (at least to me). – Surb Jun 01 '18 at 23:23
  • @Surb There must be something that establishes the fact that $K$ is a parameter of the equation: namely, that we aren't asking for a function such that for all $x$ there is $K$ such that $f(x)+f(1/x)=K$; or, even worse, for a function such that for all $x$ and $K$ et cetera. Asking for a function such that there is $K$ such that for all $x$ et cetera appears to be the best we can make out of sheer quantifiers, as far a clarity of exposition is concerned. The second part of the discussion just explains the fact that the problem is tantamount to finding all the odd functions. –  Jun 01 '18 at 23:28
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    I agree that $K$ should be a parameter and the way I understand the question, the variable is $f$. – Surb Jun 01 '18 at 23:37