NeilB just completely nuked the problem. Defining roundoff[x_]:=Round[x]-x, he finds the simple Sum[roundoff[Pisot^k],{k,∞}] as roundoffsum[a_] := If[ ! pisotNumberQ[a] || InexactNumberQ[a], Print["roundoffsum::nonpisot - This isn't a Pisot number!"], Block[{minpoly = MinimalPolynomial[a], conjugates, k = 0}, conjugates = Select[(#1[[1, 2]] & ) /@ Solve[minpoly[x] == 0, x], Abs[#1] < 1 & ]; While[Sum[Abs[conjugates[[i]]]^k, {i, 1, Length[conjugates]}] >= 1/2, k++]; Print["minpoly degree " ~~ ToString[Length[conjugates] + 1] ~~ ", k=" ~~ ToString[k]]; Sum[Round[a^n] - a^n, {n, 0, k - 1}] + Sum[conjugates[[i]]^k/ (1 - conjugates[[i]]), {i, 1, Length[conjugates]}]]] Testing with the atypical sextic in http://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number#Small_Pisot... roundoffsum[Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 2]] minpoly degree 6, k=50 8532175911 + Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 1]^50/( 1 - Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 1]) - Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 2] - Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 2]^2 - ... +(Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 6]^50/( 1 - Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 6])) In[145]:= RootReduce[%] Out[145]= Root[-13 - 17 #1 + 7 #1^2 + 26 #1^3 + 20 #1^4 + 7 #1^5 + #1^6 &, 2] Numerically: In[163]:= RootApproximant[ SetPrecision[ NSum[roundoff[Root[-1 + #1 - #1^2 + #1^4 - 2 #1^5 + #1^6 &, 2]^k], {k, 3333}, NSumTerms -> 3333, WorkingPrecision -> 1111], 33]] Out[163]= Root[-13 - 17 #1 + 7 #1^2 + 26 #1^3 + 20 #1^4 + 7 #1^5 + #1^6 &, 2] (*SLOW* convergence.) 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. On Wed, Sep 11, 2013 at 4:26 PM, Bill Gosper <billgosper@gmail.com> wrote:
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 _________________________________ Thanks, Fred! So, defining roundoff[x]:=FractionalPart[x+1/2]-1/2 (or Round[x]-x), why has nobody thought to Sum[roundoff[Pisot^k]]? At first I thought it's because of the modern habit of norming everything. But Sum[Abs[roundoff[Pisot^k]]] is quadratic when Pisot is, so somebody should have noticed.
This is and endless source of identities. It seems that, if it converges, Sum[roundoff[Pisot1^k]^p1*roundoff[Pisot2^k]^p2.../r^k] is algebraic with the same degree as Pisot1 Pisot2 ... . E.g., NSum[(FractionalPart[(1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)))^k + 1/2] - 1/2)^2/2^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 288];
In[31]:= RootApproximant[%]
Out[31]= Root[-215317 + 2949880 #1 + 1899072 #1^2 + 326144 #1^3 &, 1]
In[29]:= RootApproximant[NSum[(FractionalPart[(1/ 3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)))^ k + 1/2] - 1/2)*(FractionalPart[GoldenRatio^k + 1/2] - 1/2), {k, 288}, NSumTerms -> 288, WorkingPrecision -> 105]]
Out[29]= Root[ 41 + 567 #1 - 144 #1^2 - 313 #1^3 - 8 #1^4 + 55 #1^5 + 11 #1^6 &, 2] --rwg