-1
$\begingroup$

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 .

$\endgroup$
  • 1
    $\begingroup$ 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 $f(t)$. $\endgroup$ – Iosif Pinelis Jul 28 at 15:30
  • $\begingroup$ Per comment of Iosif Pinelis, changed $f(t)$ to $f$ at outset of question. Hopefully, the intent of the question is clear. $\endgroup$ – Paul B. Slater Jul 28 at 16:00
  • 2
    $\begingroup$ 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. $\endgroup$ – Iosif Pinelis Jul 28 at 16:16
  • 2
    $\begingroup$ 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!) $\endgroup$ – Pietro Majer Jul 28 at 17:18
  • $\begingroup$ 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 $f$.) $\endgroup$ – Paul B. Slater Jul 28 at 17:21
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.