Re: Forcing surds into the numerator

*To*: mathgroup at smc.vnet.net*Subject*: [mg64680] Re: [mg64656] Forcing surds into the numerator*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 26 Feb 2006 05:08:02 -0500 (EST)*References*: <200602250753.CAA13968@smc.vnet.net> <9761453B-CD02-4FFD-BA7E-C011DA040EFF@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I forgot of course to add that my second function RationalizeDenominator2 will only work if you load in the package << Algebra`PolynomialExtendedGCD` Also, I should add that the reason why RationalizeDenominator1[Sin[Pi/ 12]] returns an answer with Sqrt[2] in the denominator is precisely the fact that Mathematica will always convert Sqrt[2]/2 to 1/Sqrt[2]. Andrzej Kozlowski On 25 Feb 2006, at 18:46, Andrzej Kozlowski wrote: > > On 25 Feb 2006, at 08:53, Tony King wrote: > >> Does anyone know how I can force Mathematica to display surds in the >> numerator of an expression, or a function that can be applied to >> do the job? >> >> For example, FullSimplify[1/(3+Sqrt[2])] returns itself and not 1/7 >> (3-root2) >> >> Similarly, Sin[Pi/12] returns (-1+root3)/(2root2) and not 1/4 >> (root6-root2) >> >> Many thanks >> >> Tony >> > > I answered this question in 1999, so below I am just copying the > functions I defined in that post without any further thinking about > it. > > > You can do it with FullSimplify and a suitably chosen complexity > function that will penalise the presence of radicals in the > denominator. Here is such a function: > > RationalizeDenominator1[expr_] := > > FullSimplify[expr, ComplexityFunction -> > ( > Count[#, _? > (MatchQ[Denominator[#], Power[_, _Rational] _. + _.] &), > {0, Infinity} > ] + If[FreeQ[#, Root], 0, 1] & > ) > ] > > This will work on your first example: > > > RationalizeDenominator1[1/(3 + Sqrt[2])] > > > (1/7)*(3 - Sqrt[2]) > > but not quite on yours second > > In[14]:= > RationalizeDenominator1[Sin[Pi/12]] > > Out[14]= > (-1 + Sqrt[3])/(2*Sqrt[2]) > > > To deal with such and more complicated cases I wrote a function > that will remove radicals from the denominator. It has quite clumsy > interface so one has to specify the root you want to remove but it > is not difficult to make a function that will remove all radicals. > Here is the function: > > RationalizeDenominator2[f_, a_] := > Module[{t, > MinimalPoly}, MinimalPoly[x_, t_] := RootReduce[ > x][[1]][t]; MinimalPoly[Sqrt[b_], t_] := t^2 - b; > Numerator[f]* > PolynomialExtendedGCD[Denominator[f] /. {a -> t}, > MinimalPoly[a, t]][[2, 1]] /. t -> a // Expand] > > > To use it you have to specify which radical in the denominator you > want move to the numerator: > > > RationalizeDenominator2[1/(3 + Sqrt[2]), Sqrt[2]] > > > 3/7 - Sqrt[2]/7 > > > > RationalizeDenominator2[RationalizeDenominator2[Sin[Pi/12], Sqrt > [3]], Sqrt[2]] > > > Sqrt[3/2]/2 - 1/(2*Sqrt[2]) > > > with complicated expressions RationalizeDenominator1 and > RationalizeDenominator2 will give you different answers: > > > expr = (3 - Sqrt[5])/(1 + 5^(1/7)); > > > RationalizeDenominator1[expr] > > > (-(-3 + Sqrt[5]))*Root[6*#1^7 - 7*#1^6 + 21*#1^5 - 35*#1^4 + > 35*#1^3 - 21*#1^2 + 7*#1 - > 1 & , 1] > > > RationalizeDenominator2[expr, 5^(1/7)] > > > 1/2 - (5*5^(1/14))/6 - 5^(1/7)/2 + (5*5^(3/14))/6 + 5^(2/7)/2 - > (5*5^(5/14))/6 - > 5^(3/7)/2 - Sqrt[5]/6 + 5^(4/7)/2 + 5^(9/14)/6 - 5^(5/7)/2 - 5^ > (11/14)/6 + > 5^(6/7)/2 + 5^(13/14)/6 > > > Note also, however, that Mathematica will always do this: > > > Sqrt[2]/2 > > > 1/Sqrt[2] > > which means that no matter what you do if Mathematica encounters > something like the above you will get square roots in the > denominator. There is no way of preventing this without using Hold. > > > Andrzej Kozlowski > > > > >

**Follow-Ups**:**Re: Re: Forcing surds into the numerator***From:*"Kai Gauer" <kai.g.gauer@gmail.com>

**References**:**Forcing surds into the numerator***From:*"Tony King" <mathstutoring@ntlworld.com>