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A random sphere-of-influence graph is \ created as follows. Begin by throwing ", StyleBox["m", FontSlant->"Italic"], " points randomly onto the unit cube in ", StyleBox["n", FontSlant->"Italic"], " dimentions. For each point ", Cell[BoxData[ \(TraditionalForm\`x\_i\)]], " create a sphere of influence by drawing an open ball about ", Cell[BoxData[ \(TraditionalForm\`x\_i\)]], " with radius equal to the distance to ", Cell[BoxData[ \(TraditionalForm\`x\_i\)]], "'s nearest neighbor. Then draw an edge between two points if their spheres \ intersect. The integral ", Cell[BoxData[ \(TraditionalForm\`I\_n\)]], " arose in calculating the probability that there exists an edge between \ two points." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["The solution of 1).", "Section"], Cell[TextData[{ "Setting ", Cell[BoxData[ \(TraditionalForm\`x = z\/\(z + 1\)\)]], ", we find" }], "Text"], Cell[BoxData[ \(TraditionalForm \`I\_n = \(\[Integral]\_0\%1 \( x\^\(n - 1\)\/\((\((1 - x)\)\^n + x\^n)\)\^2\) \[DifferentialD]x\ = \ \[Integral]\_0 \%\[Infinity]\(\(\( z\^\(n - 1\)\)(z + 1)\)\^\(n - 1\)\/\(( z\^n\ + \ 1)\)\^2\) \[DifferentialD]x\)\)], "Text", TextAlignment->Center], Cell[TextData[{ "Expanding ", Cell[BoxData[ \(TraditionalForm\`\((z + 1)\)\^\(n - 1\)\)]], " into the series" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\(\((z + 1)\)\^\(n - 1\)\), "=", RowBox[{\(\[Sum]\+\(k = 0\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], " ", \(z\^k\)}]}]}], TraditionalForm]], "Text",\ TextAlignment->Center], Cell["and changing the order of summation and integration, we get", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(I\_n\), " ", "=", " ", RowBox[{\(\[Sum]\+\(k = 0\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], " ", \(\[Integral]\_0 \%\[Infinity]\( z\^\(\ k\ + \ n - 1\)\/\((z\^n\ + \ 1)\)\^2 \) \[DifferentialD]z\)}]}]}], TraditionalForm]], "Text",\ TextAlignment->Center], Cell["Evaluating the inner integral ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\[Infinity]\( z\^\(k + n - 1\)\/\((z\^n + 1)\)\^2\) \[DifferentialD]z\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"If", "[", RowBox[{ \(Re(n) > 0 \[And] Re(n - k) > 0 \[And] Re(k + n) > 0\), ",", \(\(k\ \[Pi]\ \(csc(\(k\ \[Pi]\)\/n)\)\)\/n\^2\), ",", RowBox[{ SubsuperscriptBox["\[Integral]", "0", InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]], \(\(z\^\(k + n - 1\)\/\((z\^n + 1)\)\^2\) \[DifferentialD]z\)}]}], "]"}], TraditionalForm]], "Output"] }, Open ]], Cell["we finally obtain ", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(I\_n\), "=", RowBox[{\(1\/n\), "+", RowBox[{\(\(\[Pi]\ \)\/n\^2\), RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 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1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], " ", RowBox[{"(", RowBox[{"1", " ", "+", RowBox[{\(\[Sum]\+\(j = 1\)\%\[Infinity]\), RowBox[{ \((2 - 2\^\(2 - 2\ j\))\), \(k\^\(2\ j\)\/n\^\(2\ j\)\), TagBox[ RowBox[{"\[Zeta]", "(", TagBox[\(2\ j\), (Editable -> True)], ")"}], InterpretTemplate[ Zeta[ #]&]]}]}]}], ")"}]}]}]}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell["and changing the order of summation, we find that", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(I\_n\), "=", RowBox[{\(1\/n\), "+", RowBox[{\(1\/n\), RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 1\)\), TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False]}]}], " ", "+", " ", RowBox[{\(1\/n\), RowBox[{\(\[Sum]\+\(j = 1\)\%\[Infinity]\), RowBox[{\(\((2 - 2\^\(2 - 2\ j\))\)\/n\^\(2\ j\)\), TagBox[ RowBox[{"\[Zeta]", "(", TagBox[\(2\ j\), (Editable -> True)], ")"}], InterpretTemplate[ Zeta[ #]&]], RowBox[{"(", RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], \(k\^\(2 j\)\)}]}], ")"}]}]}]}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell["Taking into account that", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(1\/n + \[Sum]\+\(k = 1\)\%\(n - 1\)Binomial[n - 1, k]\/n\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(\(-2\) + 2\^n\)\/\(2\ n\) + 1\/n\)], "Output"] }, Open ]], Cell["we arrive at", "Text"], Cell[BoxData[ FormBox[ RowBox[{\(I\_n\), "=", RowBox[{\(2\^\(n - 1\)\/n\), " ", "+", " ", RowBox[{\(1\/n\), RowBox[{\(\[Sum]\+\(j = 1\)\%\[Infinity]\), RowBox[{\(\((2 - 2\^\(2 - 2\ j\))\)\/n\^\(2\ j\)\), TagBox[ RowBox[{"\[Zeta]", "(", TagBox[\(2\ j\), (Editable -> True)], ")"}], InterpretTemplate[ Zeta[ #]&]], RowBox[{"(", RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], \(k\^\(2 j\)\)}]}], ")"}]}]}]}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "To find the asymptotic expansion of ", Cell[BoxData[ \(TraditionalForm\`\(I\_n, \)\)]], " when ", Cell[BoxData[ \(TraditionalForm\`n \[Rule] \[Infinity]\)]], " we need to find an asymptotic expantion of the inner sum. First of all we \ observe that" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], \(k\^\(2 j\)\)}]}], " ", "=", \(lim\+\(z \[Rule] 1\)\[ThinSpace]\(( \(\((z d\/dz)\)\^\(2 j\)\) \((z + 1)\)\^\(n - 1\))\)\)}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Now let us find several values of this sum when ", StyleBox["j", FontSlant->"Italic"], " is changing form 1 to 4" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Table[ Factor[Nest[z*D[#1, \ z]\ &\ , \ \((z\ + \ 1)\)^\((n\ - \ 1)\), \ \n\ \ \ \ 2*j] /. \ z\ -> \ 1], \ {j, \ 1, \ 4}] // TableForm \)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[GridBox[{ {\(2\^\(n - 3\)\ \((n - 1)\)\ n\)}, {\(2\^\(n - 5\)\ \((n - 1)\)\ n\ \((n\^2 + 3\ n - 6)\)\)}, { \(2\^\(n - 7\)\ \((n - 1)\)\ n\ \((n\^4 + 10\ n\^3 - 5\ n\^2 - 70\ n + 80)\)\)}, { \(2\^\(n - 9\)\ \((n - 1)\)\ n\ \((n\^6 + 21\ n\^5 + 63\ n\^4 - 385\ n\^3 - 280\ n\^2 + 2212\ n - 1904)\)\)} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ { Times[ Power[ 2, Plus[ -3, n]], Plus[ -1, n], n], Times[ Power[ 2, Plus[ -5, n]], Plus[ -1, n], n, Plus[ -6, Times[ 3, n], Power[ n, 2]]], Times[ Power[ 2, Plus[ -7, n]], Plus[ -1, n], n, Plus[ 80, Times[ -70, n], Times[ -5, Power[ n, 2]], Times[ 10, Power[ n, 3]], Power[ n, 4]]], Times[ Power[ 2, Plus[ -9, n]], Plus[ -1, n], n, Plus[ -1904, Times[ 2212, n], Times[ -280, Power[ n, 2]], Times[ -385, Power[ n, 3]], Times[ 63, Power[ n, 4]], Times[ 21, Power[ n, 5]], Power[ n, 6]]]}]], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ It is easy to see that elements of this list can be written as\ \>", "Text"], Cell[BoxData[ \(TraditionalForm \`2\^\(n\ - \ 2 j - 1\)\ \(\(n\^\(2\ j\)\)(1\ + e\_1\/n + e\_2\/n\^2 + \ ... )\)\)], "Text", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`e\_p\)]], " are coefficients depend on ", StyleBox["j ", FontSlant->"Italic"], "only. If we divide the above list by the common multiplicator we obtain" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(TableForm[ Table[Collect[ \((Nest[z*D[#1, \ z]\ &\ , \ \((z\ + \ 1)\)^\((n\ - \ 1)\), \ \n\ \ \ \ 2*j] /. \ z\ -> \ 1)\)\ \ 1/\((2^\((n - 2 j - 1)\)\ n^\((2 j)\))\), \ n], \ {j, \ 1, \ 5}] /. \ n^s_\ :> \ 0\ /; \ s\ < \(-2\)]\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[GridBox[{ {\(1 - 1\/n\)}, {\(1 + 2\/n - 9\/n\^2\)}, {\(1 + 9\/n - 15\/n\^2\)}, {\(1 + 20\/n + 42\/n\^2\)}, {\(1 + 35\/n + 270\/n\^2\)} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ { Plus[ 1, Times[ -1, Power[ n, -1]]], Plus[ 1, Times[ -9, Power[ n, -2]], Times[ 2, Power[ n, -1]]], Plus[ 1, Times[ -15, Power[ n, -2]], Times[ 9, Power[ n, -1]]], Plus[ 1, Times[ 42, Power[ n, -2]], Times[ 20, Power[ n, -1]]], Plus[ 1, Times[ 270, Power[ n, -2]], Times[ 35, Power[ n, -1]]]}]], TraditionalForm]], "Output"] }, Open ]], Cell["We can guess that", "Text"], Cell[BoxData[ \(e\_1[j_] := a\_1\ j\^2 + a\_2\ j + a\_3\)], "Input"], Cell[BoxData[ \(e\_2[j_] := b\_1\ j\^4 + b\_2\ j\^3 + b\_3\ j\^2 + b\_4\ j + b\_5\)], "Input"], Cell[TextData[{ "when ", Cell[BoxData[ \(TraditionalForm\`j\ = 1, \ 2, \ \( ... . \)\)]], " Then from the second raw it follows" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{e\_1[1] == \(-1\), e\_1[2] == 2, e\_1[3] == 9}, {a\_1, a\_2, a\_3}]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`{{a\_1 \[Rule] 2, a\_2 \[Rule] \(-3\), a\_3 \[Rule] 0}}\)], "Output"] }, Open ]], Cell["Thus", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Factor[e\_1[j] /. %\[LeftDoubleBracket]1\[RightDoubleBracket]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`j\ \((2\ j - 3)\)\)], "Output"] }, Open ]], Cell[TextData[{ "Similar for ", Cell[BoxData[ \(TraditionalForm\`e\_\(2 : \)\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[{e\_2[1] == 0, e\_2[2] == \(-9\), e\_2[3] == \(-15\), e\_2[4] == 42, e\_2[5] == 270}, {b\_1, b\_2, b\_3, b\_4, b\_5}]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`{{b\_1 \[Rule] 2, b\_2 \[Rule] \(-10\), b\_3 \[Rule] 23\/2, b\_4 \[Rule] \(-\(7\/2\)\), b\_5 \[Rule] 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Factor[e\_2[j] /. %\[LeftDoubleBracket]1\[RightDoubleBracket]]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/2\ \((j - 1)\)\ j\ \((2\ j - 7)\)\ \((2\ j - 1)\)\)], "Output"] }, Open ]], Cell["\<\ Finally, we derive the following asymptoric expansion for the inner \ sum\ \>", "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(k = 1\)\%\(n - 1\)\), RowBox[{ TagBox[ RowBox[{"(", GridBox[{ {\(n - 1\)}, {"k"} }], ")"}], Binomial, Editable->False], \(k\^\(2 j\)\)}]}], " ", "\[TildeEqual]", \(2\^\(n\ - \ 2 j - 1\)\ \(\(n\^\(2\ j\)\)( 1\ + \(j\ \((2\ j - 3)\)\)\/n + \(\((j - 1)\)\ j\ \((2\ j - 7)\)\ \((2\ j - 1)\)\)\/\(2 n\^2\) + \ ... )\)\)}], TraditionalForm]], "Text", TextAlignment->Center], Cell[TextData[{ "Thus, the integral ", Cell[BoxData[ \(TraditionalForm\`I\_n\)]], " can be rewritten as" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{\(I\_n\), "\[TildeEqual]", RowBox[{\(2\^\(n - 1\)\/n\), " ", "+", " ", RowBox[{\(1\/n\), RowBox[{\(\[Sum]\+\(j = 1\)\%\[Infinity]\), RowBox[{\(\((2 - 2\^\(2 - 2\ j\))\)\/n\^\(2\ j\)\), TagBox[ RowBox[{ RowBox[{"\[Zeta]", "(", TagBox[\(2\ j\), (Editable -> True)], ")"}], \(2\^\(n\ - \ 2 j - 1\)\), " ", \(n\^\(2\ j\)\)}], InterpretTemplate[ Zeta[ #]&]], \((1\ + \(j\ \((2\ j - 3)\)\)\/n + \(\((j - 1)\)\ j\ \((2\ j - 7)\)\ \((2\ j - 1)\)\)\/\(2 n\^2\))\)}]}]}]}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell["The infinite sum is doable:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2\^\(n - 1\)\/n + \(\(2\^\(n - 1\)\ \)\/n\) \(\[Sum]\+\(j = 1\)\%\[Infinity]\(( \((2 - 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