Re: [math-fun] tribia
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
Incidentally Salem's book on "Algebraic Numbers and Fourier Analysis" (despite its title) includes a wealth material of general interest which --- as far as I know --- is found nowhere else: for example, a neat elementary proof using only GCD's that a linear recurring sequence over the integers has finitely many zeros not lying in regularly-spaced zero subsequences. WFL On 9/12/13, 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
* Fred Lunnon <fred.lunnon@gmail.com> [Sep 12. 2013 08:43]:
Incidentally Salem's book on "Algebraic Numbers and Fourier Analysis" (despite its title) includes a wealth material of general interest which --- as far as I know --- is found nowhere else: for example, a neat elementary proof using only GCD's that a linear recurring sequence over the integers has finitely many zeros not lying in regularly-spaced zero subsequences.
WFL
[...]
Yves Meyer: Algebraic Numbers and Harmonic Analysis, North-Holland Publishing Company, (1972).
From the Preface: "This book is dedicated to the memory of Raphael Salem: it contains most of his beautiful discoveries acd the proof of his conjecture about the role played by Pisot numbers in the problem of spectral synthesis."
Best, jj
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
As close as I can tell, it's vee-juh-yuh-RAH-guh-vun. (The fact that the third A is long can't be seen in English spelling, but is visible in the original Tamil.) On Thu, Sep 12, 2013 at 3:59 AM, Bill Gosper <billgosper@gmail.com> wrote:
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
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Re Vijayaraghavan: Maybe it's Tamil, but it sure looks like a not-unusual name in Hindi with a typical transliteration into English. In which case I'd venture that the second vowel sound is like a long I. (But where is the long A sound?) --Dan On 2013-09-12, at 10:29 AM, Allan Wechsler wrote:
As close as I can tell, it's vee-juh-yuh-RAH-guh-vun. (The fact that the third A is long can't be seen in English spelling, but is visible in the original Tamil.)
The long A is after the R. In Tamil (and Hindi) short A sounds like "uh", and long A like "ah". The syllabification is Vi-ja-ya-ra-gha-van, so there isn't much diphthong (long I) effect. For an English example, I can make up a word, "a-yearning". There's *sort of* a long I audible there, but it's not salient. On Thu, Sep 12, 2013 at 1:41 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Re Vijayaraghavan:
Maybe it's Tamil, but it sure looks like a not-unusual name in Hindi with a typical transliteration into English.
In which case I'd venture that the second vowel sound is like a long I.
(But where is the long A sound?)
--Dan
On 2013-09-12, at 10:29 AM, Allan Wechsler wrote:
As close as I can tell, it's vee-juh-yuh-RAH-guh-vun. (The fact that the third A is long can't be seen in English spelling, but is visible in the original Tamil.)
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Oh, so *that's* what you mean by a "long A". On 2013-09-12, at 11:03 AM, Allan Wechsler wrote:
The long A is after the R. In Tamil (and Hindi) short A sounds like "uh", and long A like "ah".
The syllabification is Vi-ja-ya-ra-gha-van, so there isn't much diphthong (long I) effect. For an English example, I can make up a word, "a-yearning". There's *sort of* a long I audible there, but it's not salient.
Actually, when a short a (as in ah) is followed by a y sound, a long I automatically occurs. (Just say ja-ya not too slow.) --Dan On 2013-09-12, at 11:03 AM, Allan Wechsler wrote:
The syllabification is Vi-ja-ya-ra-gha-van, so there isn't much diphthong (long I) effect.
Eesh, let Neil try the guy's first name: Tirukkannapuram. (tee-rook-kun-NUP-oo-rum). On Thu, Sep 12, 2013 at 1:29 PM, Allan Wechsler <acwacw@gmail.com> wrote:
As close as I can tell, it's vee-juh-yuh-RAH-guh-vun. (The fact that the third A is long can't be seen in English spelling, but is visible in the original Tamil.)
On Thu, Sep 12, 2013 at 3:59 AM, Bill Gosper <billgosper@gmail.com> wrote:
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
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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
participants (5)
-
Allan Wechsler -
Bill Gosper -
Dan Asimov -
Fred Lunnon -
Joerg Arndt