solving third order equation: bad result
- To: mathgroup at smc.vnet.net
- Subject: [mg22152] solving third order equation: bad result
- From: "Wolter Kaper" <kaper at chem.uva.nl>
- Date: Wed, 16 Feb 2000 02:35:18 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Dear Mathgroup members, When I solve the following equation for V, in terms of symbolic P and then substitute a value for P, I get a result that is wrong. (The right result is returned when I solve the equation with P substituted beforehand.) P == RT /(V - b) - a/V^2 /. {a -> 0.142666, b -> 3.913*10^(-5), RT -> 2494}) The following notebook demonstrates the problem and compares results after the equation has been rewritten in standard (third order polynomial ==0) form. In that case the right result is returned. My questions are: does anybody know why this behaviour occurs, has anybody seen it before, should I report a bug to Wolfram research? Thanks beforehand, Wolter Kaper Univ. Of Amsterdam, dept. of Chemistry Notebook follows. (*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info at 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[ 5727, 190]*) (*NotebookOutlinePosition[ 6380, 213]*) (* CellTagsIndexPosition[ 6336, 209]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Faulty / inaccurate solution", "Title"], Cell["W.H. Kaper (kaper at chem.uva.nl)", "Subtitle"], Cell[TextData[{ StyleBox["Problem:", FontWeight->"Bold"], " When I solve the following equation for V, in terms of symbolic P and \ then substitute a value for P, I get a result that is wrong. The right result \ is returned when I solve the equation with P substituted beforehand." }], "Text"], Cell[BoxData[ \(Remove["\<Global`*\>"]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(eq1 = P == T\/\(V - b\) - a\/V\^2 /. {a \[Rule] \ 0.142666, b \[Rule] \ 3.913*10^\(-5\), T \[Rule] \ 2494}\)], "Input"], Cell[BoxData[ \(P == 2494\/\(\(-0.00003913`\) + V\) - 0.142666`\/V\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(OplA = Solve[eq1, \ V]\)], "Input"], Cell[BoxData[ \({{V \[Rule] \(-\(\(6.6666666666666664`*^-15\ \((\(-1.247`*^17\) - 1.9565`*^9\ P)\)\)\/P\)\)}, {V \[Rule] \ \(-\(\(6.6666666666666664`*^-15\ \((\(-1.247`*^17\) - 1.9565`*^9\ P)\)\)\/P\)\)}, {V \[Rule] \ \(-\(\(6.6666666666666664`*^-15\ \((\(-1.247`*^17\) - 1.9565`*^9\ P)\)\)\/P\)\)}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(OplA /. P \[Rule] 4*10^8\)], "Input"], Cell[BoxData[ \({{V \[Rule] 0.000015121666666666666`}, {V \[Rule] 0.000015121666666666666`}, {V \[Rule] 0.000015121666666666666`}}\)], "Output"] }, Open ]], Cell["This is wrong. (physical considerations)", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eq2 = eq1\ /. P \[Rule] 4*10^8\)], "Input"], Cell[BoxData[ \(400000000 == 2494\/\(\(-0.00003913`\) + V\) - 0.142666`\/V\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[eq2, \ V]\)], "Input"], Cell[BoxData[ \({{V \[Rule] 4.774318802122206`*^-7 - 0.000017720931948096697`\ \[ImaginaryI]}, {V \[Rule] 4.774318802122206`*^-7 + 0.000017720931948096697`\ \[ImaginaryI]}, {V \[Rule] 0.00004441013623957556`}}\)], "Output"] }, Open ]], Cell["\<\ These solutions are OK. None of them is equal to the faulty 0.0000151217.\ \>", "Text"], Cell["\<\ As a comparison, I can also rewrite eq1 in standard \"third order \ polynomial\" form:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(eq3 = P*V^3 - \((T + P*b)\)*V^2\ + \ a*V\ - \ a*b \[Equal] 0 /. {a \[Rule] \ 0.142666, b \[Rule] \ 3.913*10^\(-5\), T \[Rule] \ 2494}\)], "Input"], Cell[BoxData[ \(\(-5.582520579999999`*^-6\) + 0.142666`\ V - \((2494 + 0.00003913`\ P)\)\ V\^2 + P\ V\^3 == 0\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(OplB = Solve[eq3, \ V];\)\)], "Input"], Cell[BoxData[ \(General::"spell1" \(\(:\)\(\ \)\) "Possible spelling error: new symbol name \"\!\(OplB\)\" is similar to \ existing symbol \"\!\(OplA\)\"."\)], "Message"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(OplB /. P \[Rule] 4*10^8\)], "Input"], Cell[BoxData[ \({{V \[Rule] 4.774318802122236`*^-7 - 0.000017720931948096676`\ \[ImaginaryI]}, {V \[Rule] 4.774318802122236`*^-7 + 0.000017720931948096676`\ \[ImaginaryI]}, {V \[Rule] 0.000044410136239575546`}}\)], "Output"] }, Open ]], Cell[TextData[{ "Now it works allright, even when I leave P symbolic and only substitute \ afterwards.\n", StyleBox["So my question is", FontWeight->"Bold"], ": why does the deviating solution \"0.0000151217\" appear? Is it a matter \ of limited numerical accuracy? \nAnd why does \"Solve\" come up with three \ solutions that are identical? I know that a cubic equation has three \ (complex) roots, but apparently \"Solve\" has not used the general routine \ for solving a cubic equation, because then it should have arrived at the \ result presented last." }], "Text"] }, Open ]] }, FrontEndVersion->"4.0 for Microsoft Windows", ScreenRectangle->{{0, 800}, {0, 527}}, WindowSize->{497, 431}, WindowMargins->{{19, Automatic}, {Automatic, 0}} ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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