bad performance of Reduce (5.2)
- To: mathgroup at smc.vnet.net
- Subject: [mg77773] bad performance of Reduce (5.2)
- From: dimitris <dimmechan at yahoo.com>
- Date: Sat, 16 Jun 2007 03:34:05 -0400 (EDT)
Hello. $VersionNumer 5.2 o=Sin[ArcTan[z] + ArcTan[2*z]] -1/Sqrt[2] ; The equation o=0 has two roots in the real positive axis Plot[Sin[ArcTan[z] + ArcTan[2*z]] - 1/Sqrt[2], {z, 0, 5}] In[1]:= Needs["NumericalMath`IntervalRoots`"] In[21]:= IntervalBisection[Sin[ArcTan[z] + ArcTan[2*z]] - 1/Sqrt[2], z, Interval[{0, 5}], 0.1] List @@ % (FindRoot[Sin[ArcTan[z] + ArcTan[2*z]] - 1/Sqrt[2] == 0, {z, #1[[1]], #1[[2]]}, WorkingPrecision -> 30, PrecisionGoal -> 20] & ) /@ % Out[21]= Interval[{0.23437500000000014, 0.31250000000000033}, {1.71875, 1.7968750000000009}] Out[22]= {{0.23437500000000014, 0.31250000000000033}, {1.71875, 1.7968750000000009}} Out[23]= {{z -> 0.28077640640441513745535298432248087547755357014`30.}, {z -> 1.780776406404415137455352463993519256286808059355`30.}} Solve returns the correct roots. In[21]:= Off[N::meprec] In[22]:= Solve[Sin[ArcTan[z] + ArcTan[2*z]] == 1/Sqrt[2], z] FullSimplify[Sin[ArcTan[z] + ArcTan[2*z]] == 1/Sqrt[2] /. %] Out[22]= {{z -> (1/4)*(-3 + Sqrt[17])}, {z -> (1/4)*(3 + Sqrt[17])}} Out[23]= {True, True} However, it is well known that Solve is not proper for equations like o=0. Let's use Reduce In[24]:= Reduce[Sin[ArcTan[z] + ArcTan[2*z]] == 1/Sqrt[2], z, Reals] (*ommited output*) Let's be more specifically! In[25]:= Reduce[Sin[ArcTan[z] + ArcTan[2*z]] == 1/Sqrt[2] && z > 0, z, Reals] Out[25]= ((-Pi + 4*ArcTan[(1/4)*(-3 + Sqrt[17])] + 4*ArcTan[(1/2)*(-3 + Sqrt[17])])/(8*Pi) Integers && z == (1/4)*(-3 + Sqrt[17])) || ((-3*Pi + 4*ArcTan[(1/4)*(3 + Sqrt[17])] + 4*ArcTan[(1/2)*(3 + Sqrt[17])])/(8*Pi) Integers && z == (1/4)*(3 + Sqrt[17])) The roots are correct but it appears a strange condition ((-Pi + 4*ArcTan[(1/4)*(-3 + Sqrt[17])] + 4*ArcTan[(1/2)*(-3 + Sqrt[17])])/(8*Pi) Integers Any comments? Dimitris
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