On Thu, Sep 12, 2013 at 12:59 AM, Bill Gosper <billgosper@gmail.com> wrote:
NeilB just completely nuked the problem. Defining roundoff[x_]:=Round[x]-x,
[...]
Amazingly, under intense pressure of looming bedtime, he then derived the general case, Sum[roundoff[Pisot1^k]*roundoff[Pisot2^k].../q^k,{k,∞}], retiring before having the chance to email me the formula. --rwg But he still can't pronounce Vijayaraghavan.
To sum this with Neil's general case, roundoffsum[{Pisot1,Pisot2,...},q]. E.g. In[14]:= pisotNumberQ[GoldenRatio] Out[14]= True In[15]:= pisotNumberQ[PlasticConstant = Root[#^3 - # - 1 &, 1]] Out[15]= True In[8]:= roundoffsum[{GoldenRatio}, q] ks are {2} {Subscript[i, 1],1,1} Out[8]= (2 - GoldenRatio)/q + (1 - Sqrt[5])^2/(4 q (1/2 (-1 + Sqrt[5]) + q)) In[9]:= FullSimplify[%] Out[9]= 1/q - 2/(2 + q + Sqrt[5] q) Simplify[roundoffsum[{PlasticConstant, GoldenRatio}, 2^(1/5)]] ks are {10, 2} {Subscript[i, 1],1,2}{Subscript[i, 2],1,1} ((-1 + Sqrt[ 5])^10 (((1 - I Sqrt[3]) (9 - Sqrt[69])^( 1/3) + (1 + I Sqrt[3]) (9 + Sqrt[69])^(1/3))^10/( 2^(1/5) + ( I (-1 + Sqrt[ 5]) ((I + Sqrt[3]) (9 - Sqrt[69])^( 1/3) - (-I + Sqrt[3]) (9 + Sqrt[69])^(1/3)))/( 4 2^(1/3) 3^( 2/3))) + ((1 + I Sqrt[3]) (9 - Sqrt[69])^( 1/3) + (1 - I Sqrt[3]) (9 + Sqrt[69])^(1/3))^10/( 2^(1/5) - ( I (-1 + Sqrt[ 5]) ((-I + Sqrt[3]) (9 - Sqrt[69])^( 1/3) - (I + Sqrt[3]) (9 + Sqrt[69])^(1/3)))/( 4 2^(1/3) 3^(2/3)))))/(24461180928 2^(2/15) 3^(2/3)) + ((2 - GoldenRatio) (1 - Root[-1 - #1 + #1^3 &, 1]))/2^(1/5) + [...] (Too hairy for RootReduce.) In[20]:= N[%, 33] Out[20]= -0.00178817291753598836056721461690086 + 0.*10^-40 I In[24]:= SetPrecision[ NSum[roundoff[GoldenRatio^k]*roundoffPlasticConstant^k]/2^(k/5), {k, 333}, NSumTerms -> 333, WorkingPrecision -> 333], 33] Out[24]= -0.00178817291753598836056721461690086 Neil made us a nice .nb and .pdf with the functions, formula, derivation, and proof sketch: http://gosper.org/pisotresults4.pdf http://gosper.org/pisotresults4.nb --rwg