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MathGroup Archive 2006

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Re: Re: Forcing surds into the numerator

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64685] Re: [mg64680] Re: [mg64656] Forcing surds into the numerator
  • From: "Kai Gauer" <kai.g.gauer at gmail.com>
  • Date: Mon, 27 Feb 2006 00:18:33 -0500 (EST)
  • References: <200602250753.CAA13968@smc.vnet.net> <9761453B-CD02-4FFD-BA7E-C011DA040EFF@mimuw.edu.pl> <200602261008.FAA22170@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Apparently, none of the 3 solutions seem to fully
RationalizeDenominator for this particular example:

x:=Sqrt[2]*Sqrt[(-(a*b*c) + a*f^2 + b*g^2 - 2*f*g*h + c*h^2)/
    ((-(a*b) + h^2)*(-a - b + (-a + b)*Sqrt[1 + (4*h^2)/(a - b)^2]))]

RationalizeDenominator1[x] - doesn't fully try all that it might be wanted to

RationalizeDenominator2[x, Sqrt[(a - b)^2 + 4*h^2]] (** The 2nd
argument is chosen in this form because it makes sense to eliminate
the denomintator of the less-simplified Sqrt[1 + (4*h^2)/(a - b)^2]
form **) - returns the warning: PolynomialExtendedGCD::"onevar":
    "PolynomialExtendedGCD is defined only for polynomials in one
variable." (Mathematica 3.0)

SurdFunction[x] - also doesn't tackle it successfully

My particular functions also have the related quantity of x, call it y:

y:=Sqrt[2]*Sqrt[(-(a*b*c) + a*f^2 + b*g^2 - 2*f*g*h + c*h^2)/
    ((-(a*b) + h^2)*(-a - b + (a - b)*Sqrt[1 + (4*h^2)/(a - b)^2]))]

Is there any way to have all the surd solutions return as a single
list, but also for symbolic forms? Can I also instead attach a (-1)^k
value term here to help pick up the sign changes of doing operations
such as FullSimplify (usually k for my purposes will be an Integer
type, expecting at least a factor of (-(a*b*c) + a*f^2 + b*g^2 -
2*f*g*h + c*h^2) above the -1) or have a return value in the form

(<esc> +- <esc>), or (<esc> -+ <esc>) operators? I do not know how to
define extra rules which will allow for Simplify, etc to recognize
these in the same format as Plus, Minus.

Certain TargetFunctions -> {Abs, Arg}, allow for tries with other
functions, but seem to be limited for the forms which they can return.
I do not necessarily need to keep the parity of the list returned (ie
the list can be re-ordered if all the solutions are returned) as it
more necessary in my application to run a Simplify on 4 values at a
time and get 4 returns (with, in certain cases, possibly double-root
solutions).

Any ideas?

Kai G. Gauer

On 2/26/06, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> 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
> >
> >
> >
> >
> >
>
>


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