In a comment to


user JimB provided an answer to the question posed there of finding the "two-quater[nionic]bit" absolute separability Hilbert-Schmidt probability. (Previously, in TwoQubit, he had obtained the "two-qubit" counterpart.)

The answer now took the form

327574875999612773528659/95105071448064 - 2951081236201839/(524288 Sqrt[2]) - (15390446918294583135 \[Pi])/(17179869184 Sqrt[2]) + (15390446918294583135 ArcCsc[18/Sqrt[50 + 17 Sqrt[2]]])/(2147483648 Sqrt[2])

An earlier answer,

-((13 (-216449750678398795533760757497856 + 
   176860737736399592490919645937664 Sqrt[2] + 
   279292548969739228073088142369304501839785 Sqrt[2] \[Pi] - 
   558572941247617043110461841280869072896000 Sqrt[2]
     ArcCot[Sqrt[2]] + 
   23637916932187025487103667523337320 Sqrt[2]
     ArcCot[2 Sqrt[2]] - 
   16178155879591789043088455851252390200 Sqrt[2]
     ArcCot[3 + Sqrt[2]] - 
   558589165778586158484606527963549721006600 Sqrt[2]

of somewhat different form, to the very same question had been provided in eq. (36) in


Both formulas above evaluate (N[,50]) to


However, the FullSimplify command does not reveal the formulas' evident equivalence.

It might seem that some inverse trigonometric function transformations would be required to accomplish that.

So, can the evident equivalence of the two formulas be formally established employing Mathematica?

  • $\begingroup$ Block[{$MaxExtraPrecision = 1000}, ans1 - ans2 // N[#, 1000] &] is close enough for me. $\endgroup$
    – Bob Hanlon
    Oct 3, 2020 at 19:25

1 Answer 1


FindIntegerNullVector is your friend. Notice that this method allows you to obtain an even simpler expression.

A. Find relations between sets of different functions

v[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]],ArcTan[Sqrt[2]],ArcCot[3+Sqrt[2]]};
v[2]={ArcCot[Sqrt[2]],ArcCot[2 Sqrt[2]],ArcCot[3+Sqrt[2]]};
v[3]={ArcCot[2 Sqrt[2]],ArcCot[Sqrt[2]],ArcTan[Sqrt[2]]};

B. This is the main part

Out[1]= {{-3,-1,11},{-2,1,4},{1,1,-1},{1,1,-1}}

C. Make a list of replacement rules

r[1]={ArcCsc[18/Sqrt[50+17 Sqrt[2]]]->(-ArcTan[Sqrt[2]]+11ArcCot[3+Sqrt[2]])/3};
r[2]={ArcCot[3+Sqrt[2]]->1/4(2ArcCot[Sqrt[2]]-ArcCot[2 Sqrt[2]])};
r[3]={ArcCot[2 Sqrt[2]]->ArcTan[Sqrt[2]]-ArcCot[Sqrt[2]]};

D. Apply the rules

Out[2]= 0


In principle, all but one identities found by the LLL method are easy to prove by hand. Only the first one, namely $$11\,\mathrm{arccot}\big(3+\sqrt{2}\big) = \mathrm{arctan}\big(\sqrt{2}\big)+ 3\,\mathrm{arccsc}\!\Bigg\{\frac{18}{\sqrt{50+17 \sqrt{2}}}\Bigg\}$$ is somewhat unexpected.

  • $\begingroup$ Interesting and impressive! Didn't know of the FindIntegerNullVector command. To what does "LLL method" refer? $\endgroup$ Oct 3, 2020 at 20:38
  • 1
    $\begingroup$ @PaulB.Slater LLL is the Lenstra–Lenstra–Lovász lattice basis reduction algorithm. In Mathematica it is implemented in the LatticeReduce command, and FindIntegerNullVector is using it. $\endgroup$
    – yarchik
    Oct 3, 2020 at 20:41
  • 1
    $\begingroup$ Very nice! In a related question of @PaulB.Slater the quantity $\phi=\cos ^{-1}\left(\frac{1}{3}\right)$ held a sacred place. Using FindIntegerNullVector[N[{ArcTan[Sqrt[2]], \[Pi], ArcCos[1/3]}, 1000]] one can construct the rule ArcTan[Sqrt[2]] -> (\[Pi] - \[Phi])/2 resulting in another equal representation: $\frac{15390446918294583135 \phi }{1073741824 \sqrt{2}}+\frac{327574875999612773528659}{95105071448064}-\frac{2951081236201839}{524288 \sqrt{2}}-\frac{46171340754883749405 \pi }{8589934592 \sqrt{2}}$. $\endgroup$
    – JimB
    Oct 4, 2020 at 2:50
  • $\begingroup$ @JimB That is probably the most economical way of writing it, nice! $\endgroup$
    – yarchik
    Oct 4, 2020 at 8:07
  • $\begingroup$ What can be done with this formula -((3840 (-5358569267936 + 33756573946095 Sqrt[2] [Pi] - 270052591568760 Sqrt[2] ArcCot[Sqrt[2]] + 11149704525960 Sqrt[2] ArcCot[2 Sqrt[2]] + 270052591568760 Sqrt[2] ArcCot[3 + Sqrt[2]]))/(-1959684729929728 + 1601255307608064 Sqrt[2] + 1529087492782080 Sqrt[2] [Pi] - 45247615492565918250 Sqrt[2] ArcCot[Sqrt[2]] + 22619730179635540245 Sqrt[2] ArcSec[3]))? It's from eq. (35) in the 2009 paper and is claimed there to be the Hilbert-Schmidt area/volume ratio for the absolutely separable two-qubit states. $\endgroup$ Oct 4, 2020 at 13:07

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