       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 in the denominator is precisely
the fact that Mathematica will always convert Sqrt/2 to 1/Sqrt.

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)] 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)]
>
>
> (1/7)*(3 - Sqrt)
>
> but not quite on yours second
>
> In:=
> RationalizeDenominator1[Sin[Pi/12]]
>
> Out=
> (-1 + Sqrt)/(2*Sqrt)
>
>
> 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][][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), Sqrt]
>
>
> 3/7 - Sqrt/7
>
>
>
> RationalizeDenominator2[RationalizeDenominator2[Sin[Pi/12], Sqrt
> ], Sqrt]
>
>
> Sqrt[3/2]/2 - 1/(2*Sqrt)
>
>
> with complicated expressions RationalizeDenominator1 and
> RationalizeDenominator2 will give you different answers:
>
>
> expr = (3 - Sqrt)/(1 + 5^(1/7));
>
>
> RationalizeDenominator1[expr]
>
>
> (-(-3 + Sqrt))*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/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
>
>
> 1/Sqrt
>
> 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
>
>
>
>
>

```

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