DanA> When I try to evaluate that exact sum in Mma, it gives me only numbers that are to > 20 decimal places equal to 0. (Adding only 1000 or 10000 terms, asking for 10 digits' precision.) I'd guess only a set of measure 0 of real numbers x would have their Sum[Round[x^n]-x^n,{n,0,∞}] <> 0. But it's easy to believe that many algebraic numbers don't. What's your secret? --Dan On 2013-09-10, at 8:11 PM, Bill Gosper wrote: Let t:=1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)), the tribonacci constant. Then, empirically, Sum[Round[x^n]-x^n,{n,0,∞}] Gaa, I meant Round[t^n]-t^n (Mac OS crashed as I prepared to send.) ==-0.80851211604688125... -2 + ((3 Sqrt[3] - Sqrt[11])^(1/3) + (3 Sqrt[3] + Sqrt[11])^(1/3))/(2^(2/3) Sqrt[3]) In[2]:= NSum[Round[GoldenRatio^k] - GoldenRatio^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 99] Out[2]= 0.618033988749894848204586834365638117720309179806089199107257598931385927496544945583310480117151672 In[3]:= (% + 1/2)^2 Out[3]= 1.2500000000000000000000000000000000000000000000007297116525363033564609 5964404522000200488900285933 This can't be new. Anybody? --rwg
PVS numbers? http://en.wikipedia.org/wiki/Pisot–Vijayaraghavan_number http://en.wikipedia.org/wiki/Salem_number WFL On 9/11/13, Bill Gosper <billgosper@gmail.com> wrote:
DanA>
When I try to evaluate that exact sum in Mma, it gives me only numbers that are to > 20 decimal places equal to 0. (Adding only 1000 or 10000 terms, asking for 10 digits' precision.)
I'd guess only a set of measure 0 of real numbers x would have their Sum[Round[x^n]-x^n,{n,0,∞}] <> 0.
But it's easy to believe that many algebraic numbers don't.
What's your secret?
--Dan
On 2013-09-10, at 8:11 PM, Bill Gosper wrote:
Let t:=1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)), the tribonacci constant. Then, empirically, Sum[Round[x^n]-x^n,{n,0,∞}]
Gaa, I meant Round[t^n]-t^n (Mac OS crashed as I prepared to send.)
==-0.80851211604688125... -2 + ((3 Sqrt[3] - Sqrt[11])^(1/3) + (3 Sqrt[3] + Sqrt[11])^(1/3))/(2^(2/3) Sqrt[3])
In[2]:= NSum[Round[GoldenRatio^k] - GoldenRatio^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 99]
Out[2]= 0.618033988749894848204586834365638117720309179806089199107257598931385927496544945583310480117151672
In[3]:= (% + 1/2)^2
Out[3]= 1.2500000000000000000000000000000000000000000000007297116525363033564609 5964404522000200488900285933
This can't be new. Anybody?
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Re: Pisot numbers When the third edition of the Mathematica Book came out in 1996, it included this neat example in its Formula Gallery: http://reference.wolfram.com/legacy/v4/MainBook/G.1.2.html For the most recent version of Mathematica, a Pisot sampler is relegated to a "neat example" for Root and an application for FractionalPart. And for the function AlgebraicIntegerQ we learn how to recognize a Pisot number: In[1]:= pisotNumberQ[a_] := AlgebraicIntegerQ[a] && Element[a, Reals] && (a > 1) && (Count[ List @@ (Last /@ Roots[MinimalPolynomial[a, x] == 0, x]), _?(Abs[#] > 1 &)] == 1) So, to borrow Bill's tribonacci constant… In[2]:= pisotNumberQ[1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3))] Out[2]= True
participants (3)
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Bill Gosper -
Fred Lunnon -
Hans Havermann