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

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 .

• What is $t$ in (say) $f(\frac{1+t}{2})=\frac{25}{341}$? Also, what do you mean by "function/functions $f(t)$"? Is $t$ the argument of a function $f$? Then the function is $f$, not $f(t)$. – Iosif Pinelis Jul 28 at 15:30
• Per comment of Iosif Pinelis, changed $f(t)$ to $f$ at outset of question. Hopefully, the intent of the question is clear. – Paul B. Slater Jul 28 at 16:00
• It's still unclear what $t$ is. It is not specified in your post by quantifiers "for all" or "there exist(s)" or in any other way. – Iosif Pinelis Jul 28 at 16:16
• Actually it is only clear that you are looking for a function $f$. You should kindly add a definition of its domain, a definition of its co-domain, and a list of properties you want it to have. (Each point in unambiguous way, otherwise the there will be a bunch of possible interpretations!) – Pietro Majer Jul 28 at 17:18
• Theorem 7 of sciencedirect.com/science/article/pii/0024379594002118 tells us (changing $f$ there to $g$ here) that operator monotone functions $g(t)$ satisfy the relation $g(t)=t g(t^{-1})$--as can be checked with $\frac{1+t}{2}$, and the other examples. (So, maybe I should have set up the whole problem using $g$, not $f$.) – Paul B. Slater Jul 28 at 17:21

Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial $$$$f(x) =a_0 +a_1 x +a_2 x^2.$$$$ Then, we can achieve the three target exact results $$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$$, by taking $$$$a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2},$$$$ $$$$a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}},$$$$ and $$$$a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}},$$$$ giving us that the second-degree polynomial $$f(x)$$ is obtainable by dividing $$$$\left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x)$$$$ by $$$$9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}.$$$$
Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Rationalizing one of them, 0.0346801, we added $$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.