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Instead of using the instruction FullSimplify, one may \ write the appropriate replacement:" }], "MathCaption"], Cell[CellGroupData[{ Cell[BoxData[ \(LaplaceCoefficient[1\/2, 0, 0, \[Alpha]]\)], "Input", CellLabel->"In[4]:="], Cell[BoxData[ \(\(4\ EllipticK[\[Alpha]\^2]\)\/\[Pi]\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(LaplaceCoefficient[3\/2, 2, 0, \[Alpha]]\)], "Input", CellLabel->"In[5]:="], Cell[BoxData[ \(15\/4\ \[Alpha]\^2\ Hypergeometric2F1[3\/2, 7\/2, 3, \[Alpha]\^2]\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(lap = LaplaceCoefficient[3\/2, 1, 1, \[Alpha]]\)], "Input", CellLabel->"In[6]:="], Cell[BoxData[ \(3\/4\ \((4\ HypergeometricPFQ[{3\/2, 5\/2}, {2}, \[Alpha]\^2] + 15\ \[Alpha]\^2\ HypergeometricPFQ[{5\/2, 7\/2}, {3}, \[Alpha]\^2])\)\)], "Output", CellLabel->"Out[6]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(lap // FullSimplify\)], "Input", CellLabel->"In[7]:="], Cell[BoxData[ \(3\ Hypergeometric2F1[3\/2, 5\/2, 2, \[Alpha]\^2] + 45\/4\ \[Alpha]\^2\ Hypergeometric2F1[5\/2, 7\/2, 3, \[Alpha]\^2]\)], "Output", CellLabel->"Out[7]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(lap /. HypergeometricPFQ[{a_, b_}, {c_}, \[Alpha]\^2] :> Hypergeometric2F1[a, b, c, \[Alpha]\^2]\)], "Input", CellLabel->"In[8]:="], Cell[BoxData[ \(3\/4\ \((4\ Hypergeometric2F1[3\/2, 5\/2, 2, \[Alpha]\^2] + 15\ \[Alpha]\^2\ Hypergeometric2F1[5\/2, 7\/2, 3, \[Alpha]\^2])\)\)], "Output", CellLabel->"Out[8]="] }, Open ]], Cell["NumericalValue", "Subsubsection", CellTags->"NumericalValue"], Cell["\<\ NumericalValue[expr, alpha] carries out the replacement expr / . \ \[Alpha] -> alpha. For example, considering the expression: \ \>", \ "MathCaption"], Cell[BoxData[ \(\(ex = \(1\/8\) \((17 b[1\/2, 1, 0] + 10\ \[Alpha]\ b[1\/2, 1, 1] + \(\[Alpha]\^2\) b[1\/2, 1, 2])\) - \(27\/8\) \[Alpha];\)\)], "Input", CellLabel->"In[9]:="], Cell["Here is the numerical value for a=0.480597:", "MathCaption"], Cell[CellGroupData[{ Cell[BoxData[ \(NumericalValue[ex, 0.480597]\)], "Input", CellLabel->"In[10]:="], Cell[BoxData[ \(0.36295417297095134`\)], "Output", CellLabel->"Out[10]="] }, Open ]], Cell["\<\ NumericalValue[expr, alpha] carries out the replacement expr /. \ \[Alpha]->alpha /. rule, where rule represents one or more replacements of \ the following type: e->number, e1->number, etc. For example, considering the expression: \ \>", "MathCaption"], Cell[BoxData[ \(\(ex = \(-\(\(\(e\^5\) \[Sigma]\^6\)\/1228\)\) \((4731447 \( \ \[Alpha]\^3\) b[7\/2, 15, 0] + 1163365 \( \[Alpha]\^4\) b[7\/2, 15, 1] + 110950 \( \[Alpha]\^5\) b[7\/2, 15, 2] + 5130 \( \[Alpha]\^6\) b[7\/2, 15, 3] + 115 \( \[Alpha]\^7\) b[7\/2, 15, 4] + \(\[Alpha]\^8\) b[7\/2, 15, 5])\);\)\)], "Input", CellLabel->"In[11]:="], Cell["\<\ Let us replace \[Alpha]=0.5331, e=0.2323, \[Sigma]=0.1025:\ \>", \ "MathCaption"], Cell[CellGroupData[{ Cell[BoxData[ \(NumericalValue[ex, 0.5331, {e \[Rule] 0.2323, \[Sigma] \[Rule] 0.1025}]\)], "Input", CellLabel->"In[12]:="], Cell[BoxData[ \(\(-2.228318550038465`*^-6\)\)], "Output", CellLabel->"Out[12]="] }, Open ]] }, FrontEndVersion->"4.1 for Macintosh", ScreenRectangle->{{0, 832}, {0, 602}}, WindowSize->{737, 575}, WindowMargins->{{4, Automatic}, {Automatic, 1}}, ShowCellBracket->True, Magnification->1, StyleDefinitions -> "HelpBrowser.nb" ] (******************************************************************* Cached data follows. 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