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Brillhart's Cubic and e^(Sqrt[163] π)

Posted 3 years ago

The Brillhart matrix has eigenvalues that give the solutions of b^3-8 b-10=0. In 1965, Brillhart noticed this led to an exotic continued fraction, with numerous large terms early in the expansion.

bc = Eigenvalues[{{0, 2, 2}, {1, 0, 2}, {2, 1, 0}}][[1]]
ContinuedFraction[N[bc, 240]] 
{3,3,7,4,2,30,1,8,3,1,1,1,9,2,2,1,3,22986,2,1,32,8,2,1,8,55,1,5,2,28,1,5,1,1501790,1,2,1,7,6,1,1,5,2,1,6,2,2,1,2,1,1,3,1,3,1,2,4,3,1,35657,1,17,2,15,1,1,2,1,1,5,3,2,1,1,7,2,1,7,1,3,25,49405,1,1,3,1,1,4,1,2,15,1,2,83,1,162,2,1,1,1,2,2,1,53460,1,6,4,3,4,13,5,15,6,1,4,1,4,1,1,2,1,16467250,1,3,1,7,2,6,1,95,20,1,2,1,6,1,1,8,1,48120,1,2,17,2,1,2,1,4,2,3,1,2,23,3,2,1,1,1,2,1,27,325927,1,60,1}

Amazingly, that leads to the following three values, all created with different methods.

 Column[{
   AccountingForm[N[E^(Sqrt[163] \[Pi]), 32]],
   AccountingForm[N[(2 + bc)^24 - 24, 32]],
   AccountingForm[N[ 640320^3 + 744 - 1/54^7, 32]]}]

262537412640768743.99999999999925
262537412640768743.99999999999925
262537412640768743.99999999999925

Try it yourself!

POSTED BY: Ed Pegg
2 Replies

Another trick

Brillhart = {{0, 2, 2}, {1, 0, 2}, {2, 1, 0}};
bc = Eigenvalues[Brillhart][[1]]; 
N[Log[(2 + bc)^24 - 24]/Sqrt[163], 34]

3.141592653589793238462643383279503

N[Pi, 34]

3.141592653589793238462643383279503

POSTED BY: Ed Pegg

Ed, you might what to add this to the entry in the OEIS in the PROG section at least.

https://oeis.org/A002937

The 163 number is treated in

https://www.youtube.com/watch?v=DRxAVA6gYMM

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