(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 23281, 656]*) (*NotebookOutlinePosition[ 23952, 680]*) (* CellTagsIndexPosition[ 23908, 676]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["once at a time ...", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ \(Product[\((1\ + \ 1/Prime[k]^w)\), \ {k, \[Infinity]}] \[Equal] HoldForm[Sum[MoebiusMu[k]^2/k^w, {k, \[Infinity]}]] == Zeta[w]\/Zeta[2\ w]\)], "Input"], Cell[BoxData[ RowBox[{ RowBox[{ UnderoverscriptBox["\[Product]", \(k = 1\), InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]], \((1 + Prime[k]\^\(-w\))\)}], "==", 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Cell[BoxData[ \(Sum[\((\(-1\))\)^\((p)\)/\((2 p - 1)\)^w, {p, 1, \[Infinity]}] // FullSimplify\)], "Input"], Cell[BoxData[ \(4\^\(-w\)\ \((\(-Zeta[w, 1\/4]\) + Zeta[w, 3\/4])\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(% /. \ w \[Rule] 2\)], "Input"], Cell[BoxData[ \(\(-Catalan\)\)], "Output"] }, Open ]], Cell[" http://mathworld.wolfram.com/RiemannZetaFunction.html", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Sum[\ 2\ ^\[Omega][k]/k^s, {k, \[Infinity]}]\ == Zeta[s]^2/\ Zeta[2 s]\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%\[Infinity] 2\^\[Omega][k]\/k\^s == Zeta[s]\^2\/Zeta[2\ s]\)], "Output"] }, Open ]], Cell["\<\ where \[Omega][k] is the number of distinct prime factors of k (Hardy and \ Wright 1979,p.254).\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Sum[\ \[Lambda][k]/k^s, {k, \[Infinity]}]\ == Zeta[2 s]/\ Zeta[s]\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%\[Infinity] \[Lambda][k]\/k\^s == Zeta[2\ s]\/Zeta[s]\)], "Output"] }, Open ]], Cell["\<\ where \[Lambda][k]is the Liouville function (-1)^r[k],with r[k] the number \ of not necessarily distinct prime factors of k.\ \>", "Text"], Cell["\<\ Sum[1(1-\[Lambda][k])/2/(k^2)^s,{k,1,\[Infinity]}]==(Zeta[2s]^2-Zeta[4s])/2/\ Zeta[2s]\ \>", "Input"], Cell["\<\ according to http://mathworld.wolfram.com/PrimeSums.html (20), giving:\ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ Table[(Zeta[2s]^2-Zeta[4s])/2/Zeta[2s],{s,4}]\ \>", "Input"], Cell[BoxData[ \({\[Pi]\^2\/20, \[Pi]\^4\/1260, \(4\ \[Pi]\^6\)\/225225, \(59\ \ \[Pi]\^8\)\/137837700}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Zeta[2 n] \[Equal] 2^\((2 n - 1)\)\ Pi^\((2 n)\)/\(\((2 n)\)!\)\ Abs[ BernoulliB[2 n]]\)], "Input"], Cell[BoxData[ \(Zeta[ 2\ n] == \(2\^\(\(-1\) + 2\ n\)\ \[Pi]\^\(2\ n\)\ Abs[BernoulliB[2\ \ n]]\)\/\(\((2\ n)\)!\)\)], "Output"] }, Open ]], Cell[BoxData[ \(Zeta[ 1 - s] \[Equal] \((\((Pi^\((1/2 - s)\)\ Gamma[s/2])\)/ Gamma[\((1 - s)\)/2])\)\ Zeta[s]\)], "Input"], Cell[BoxData[ \(Sum[ Zeta[2\ k]\ Zeta[2\ n - 2\ k], {k, 1, n - 1}] \[Equal] \((n + 1/2)\)\ Zeta[2\ n] /; n \[Element] Integers && n > 1\)], "Input"], Cell[BoxData[ \(Zeta[s] \[Equal] Limit[\((1/\((2^\((1 - s)\) - 1)\))\)\ Sum[\((\((\(-1\))\)^k\ Binomial[2\ n, n - k])\)/\((k^s\ Binomial[2\ n, n])\), {k, 1, n}], n \[Rule] Infinity]\)], "Input"], Cell[TextData[StyleBox[" special case of a more general form\n\n / 1 \ \\ / 1 \\ / 1 \\\n PROD( ---------- ) = \ PROD( ------------ ) PROD( ------------ )\n \\1 - 1/p^2 / \\ 1 - \ a(p)/p / \\ 1 - b(p)/p /\n\nwhere we've replaced the unit numerators \ with the functions a(p) \nand b(p), which for each prime p are the roots of \ x^2 - 1, but any \npermutation of those roots is allowable for any given \ prime. For \nexample, we are free to define a(p) and b(p) in terms of the \n\ Legendre symbol\n /-1 \\ / +1 if p=1,2 mod 4\n \ a(p) = -b(p) = ( ) = (\n \\ p / \\ -1 \ if p=3 mod 4\n\nin which case we have decomposed zeta(2) into the factors\n\n \ / 1 \\ / 1 \\ pi pi\n PROD( \ ------------ ) PROD( ------------ ) = -- --\n \\ 1 - a(p)/p / \ \\ 1 - b(p)/p / 2 3\n\nwhere the first factor is simply 1/(1 - \ 1/2) times the arctan series \nfor pi/4, and the second factor is 1/(1 + 1/2) \ times the sign-reversal \nof the Leibniz series (product) as discussed in the \ previous note.", FontFamily->"Courier", FontSize->10]], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["mail to rcs", "Subsubsection"], Cell[TextData[{ "From: ", StyleBox["Wouter Meeussen", FontColor->RGBColor[0, 0, 0.996109], FontVariations->{"Underline"->True}], " \nTo: ", StyleBox["rcs@cs.arizona.edu", FontColor->RGBColor[0, 0, 0.996109], FontVariations->{"Underline"->True}], " \nSent: Thursday, December 20, 2001 11:01 PM \nSubject: good enough for \ math-fun? \n", StyleBox["dear Rich,\n\nits maybe a good thing I can't get onto mathfun \ from my home address,\nsaving you all from the confused meandrings of an \ untrained mind, but...\nCould you glance at this, and decide if there's \ anything to it?\nIf not, a mild rebuke would be nice.\n\n\ -----------------------\ngiven that \nProduct[1-1/Prime[p]^2,{p,inf}] = \ 6/Pi^2 (* nega...*)\nand\nProduct[1+1/Prime[p]^2,{p,inf}] = 15/Pi^2 (* \ posi... *)\n-----------------------\nI looked at the contributions to both \ products by the\nprimes p with p+1 either =0 mod 4 or =2 mod 4\n(call 'm even \ & odd)\n--------------------\nno surprise: the approximations with \ {10,100,1000,10^4, up to 10^5} primes gives :\nnegaeven= Table[\nN[ \ Product[p=Prime[q];w=If[Mod[p+1,4]===0,p^2,Infinity];1-1/w,{q,1,10^w}] ,20], \ {w,5}]\n{0.85953201993726147609, 0.85621898617553122609, \ 0.85611445092405188430, \n0.85610930908832459737, 0.85610900364841359348}\n\n\ negaodd= \nTable[\nN[ Product[p=Prime[q];w=If[Mod[p+1,4]===2,p^2, \ Infinity];1-1/w,{q,1,10^w}] ,20], {w,5}]\n{0.94988656559647151288, \ 0.94692587750730265684, 0.94681238397007787775,\n0.94680676782855386613, \ 0.94680643141970876075}\n\nbothnega=3/4 negaeven negaodd/(6/Pi^2)\n\ {1.0072645831404900822, 1.0002546921538941542, 1.0000127010605822276, \n\ 1.0000007632993523593, 1.0000000512132967145}\n\nstill no surprise, the \"3/4\ \" takes care of the funny prime p=2,\nand both parts should come together to \ 6/Pi^2, as in the first line above.\n-----------------\n\nSame dull stuff for \ the positive sign : Product[1+1/Prime[p]^2,{p,inf}] = 15/Pi^2\n\n\ posieven=Table[\nN[ \ Product[p=Prime[q];w=If[Mod[p+1,4]===0,p^2,Infinity];1+1/w,{q,1,10^w}] ,20]\n\ , {w,5}]\n\n{1.1484905905452709371, 1.1529324174297167506, \ 1.1530731952448214522, \n1.1530801206605984133, 1.1530805320529420541}\n\n\n\ posiodd=Table[\nN[ \ Product[p=Prime[q];w=If[Mod[p+1,4]===2,p^2,Infinity];1+1/w,{q,1,10^w}] ,20]\n\ , {w,5}]\n{1.0510220003120611640, 1.0543069099627281654, \ 1.0544332886068237382, \n1.0544395431528854488, 1.0544399178047436977}\n\nand \ again, the two bits combined give a nice approximation to the true (15/Pi^2) \ :\n\nbothposi=5/4 posieven posiodd/(15/Pi^2)\n{0.9927908084153098869, \ 0.9997453730135355179, 0.9999872991008042526,\n0.9999992367012302905, \ 0.9999999487867059083}\n---------------------", FontFamily->"Arial", FontSize->10], " \n", StyleBox["\nAnd now for the (or at least my) surprise:", FontFamily->"Arial", FontSize->10], "\n", StyleBox["---------------------\nMultiply both approximations together:\n\n\ bothposi * bothnega\n\n{1.0000030197841572664, 1.0000000003159340499, \ 1.0000000000000715900, \n1.0000000000000000243, 1.0000000000000000000}\n\n\ Question:\nwhy is it so good?\n\nSame as:\nwhy is the residual error factor \ of \"bothposi\" so 'uncomfortably' close to the reciproce\nof the residual \ error factor of \"bothnega\"?\n\nI feel confident that a number theorist (or \ good pro mathematecian) can make this quite clear to me.\n\n\nSorry to bother \ you with this,\n\nWouter\n(de amatoribus nihil sed bonum....)", FontFamily->"Arial", FontSize->10] }], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["other stuff with catalan", "Subsubsection"], Cell[TextData[{ " ----- Original Message ----- \nFrom: \"wouter meeussen\" <", StyleBox["wouter.meeussen@pandora.be", FontColor->RGBColor[0, 0, 0.996109], FontVariations->{"Underline"->True}], "> \nTo: \"math-fun\" <", StyleBox["math-fun@mailman.xmission.com", FontColor->RGBColor[0, 0, 0.996109], FontVariations->{"Underline"->True}], "> \nSent: Sunday, November 18, 2007 5:20 PM \nSubject: near identity for \ Catalan \n\n> or possibly a true identity (?)\n> negaodd posieven Catalan ==\n\ > {0.9992596423851822204, 0.9999976893644401925, 0.9999999241377495063,\n> \ 0.9999999985217359313, 0.9999999999896304167}\n> \n> with \"negaodd\" defined \ as product(p prime, p mod 4=1, 1-1/p^2)\n> and \"posieven\" defined as \ product(p prime, p mod 4=3, 1+1/p^2)\n> \n> negaodd= Table[N[ \ Product[p=Prime[q];w=If[Mod[p+1,4]===2,p^2,\n> Infinity];1-1/w,{q,1,10^w}] \ ,20], {w,5}]\n> posieven= Table[N[ \ Product[p=Prime[q];w=If[Mod[p+1,4]===0,p^2,\n> Infinity];1+1/w,{q,1,10^w}] \ ,20], {w,5}]\n> \n> nothing under ", StyleBox["http://mathworld.wolfram.com/CatalansConstant.html", FontColor->RGBColor[0, 0, 0.996109], FontVariations->{"Underline"->True}], "\n> \n> Wouter." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Finch makes my day", "Subsubsection"], Cell["\<\ LandauRamanujan[n_] := With[{K = Ceiling[Log[2, n*Log[3, 10]]]}, N[1/Sqrt[2]*Product[(((1 - 2^(-2^k))*4^2^k*Zeta[2^k])/ (Zeta[2^k, 1/4] - Zeta[2^k, 3/4]))^2^(-k - 1), {k, 1, K}], n]]\ \>", "Input"], Cell[CellGroupData[{ Cell["\<\ kijknowes= Solve[{3/4 ne no==6/Pi^2, 5/4 pe po==15/Pi^2,no pe Catalan ==1, 16 K^2/Pi^2==no},{ne,no,pe,po}]\ \>", "Input"], Cell[BoxData[ \({{ne \[Rule] 1\/\(2\ K\^2\), po \[Rule] \(192\ Catalan\ 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