Construct a probability function on the operator monotone functions, $g(t)=t g(t^{-1})$, fitting certain values

Question

To immediately pose the question of interest to us, without first expanding upon its (quantum-information-theoretic) origin—we seek a univariate function $f$, for which we have the ("two-qubit separability") probability results $f(\frac{1+t}{2})=\frac{25}{341} =0.0733138$ and $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}=0.0393251$. (Also, at least in a limiting sense, $f(\frac{2 t}{1+t}) =0.$)

The possible arguments of $f$ in which we are interested are the members of the (infinite) class of operator monotone functions (of $t$) . (Theorem 7 of https://www.sciencedirect.com/science/article/pii/0024379594002118 tells us that such functions $g(t)$ satisfy the relation $g(t)=t g(t^{-1})$.)

Other—than the three already given ($\frac{1+t}{2}$ (the minimal), $\sqrt{t}$ and $\frac{2 t}{1+t}$ (the maximal))--members of the infinite class of operator monotone functions for which we have (2004) calculations (but only of a numerical nature, accurate to at most four decimal places, we believe) are for $f(\frac{t^{(t-1)}}{e}) \approx 0.0609965$, $f(\frac{1}{4} \left(\sqrt{t}+1\right)^2) \approx 0.0503391$ and $f(\frac{(t-1)}{\log{t}}) \approx .0346801$. (Table II, p. 14 in https://arxiv.org/abs/quant-ph/0308037), and also $f(\frac{1+6 t +t^2}{4 +4 t}) \approx 0.0475438$ (Table I there).

For background on the first value ($\frac{25}{341}$) of $f$ given, see https://arxiv.org/abs/1901.09889, and for the second ($1-\frac{256}{27 \pi^2}$), see eq. (87) in https://arxiv.org/abs/1701.01973 .

Answer 1

Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial \begin{equation} f(x) =a_0 +a_1 x +a_2 x^2. \end{equation} Then, we can achieve the three target exact results $f(\frac{1+t}{2})=\frac{25}{341}$, $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}$ and $f(\frac{2 t}{1+t}) =0$, by taking \begin{equation} a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2}, \end{equation} \begin{equation} a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}}, \end{equation} and \begin{equation} a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}}, \end{equation} giving us that the second-degree polynomial $f(x)$ is obtainable by dividing \begin{equation} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x) \end{equation} by \begin{equation} 9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}. \end{equation}

Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Rationalizing one of them, 0.0346801, we added $f(\frac{t-1}{\log{t}})=\frac{347}{10000}$ to our set of equations, and obtained a further (larger) solution.

In this proof-of-principle exercise, one could take $f(x)$ to be other than polynomial in nature--rational functions,...

However, perhaps in my original conception of the problem I was thinking that $t$ would not be present in the expression of the desired function $f(x)$, as it certainly is in the coefficients of the second-degree polynomial given above.