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Re: How to Eliminate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg75076] Re: How to Eliminate
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Mon, 16 Apr 2007 20:10:27 -0400 (EDT)
*References*: <evsr25$14h$1@smc.vnet.net>
Note that returning the results as a list of rules is a very useful
setting of Solve for
further actions. I don't see many reasons why someone would not want
it.
E g
In[20]:=
Solve[x^4 + a^4 == 0, x]
x^4 + a^4 /. %
Out[20]=
{{x -> (-(-1)^(1/4))*a}, {x -> (-1)^(1/4)*a}, {x -> (-(-1)^(3/4))*a},
{x -> (-1)^(3/4)*a}}
Out[21]=
{0, 0, 0, 0}
Anyway...you can define your function that will use Solve in order to
get the solutions
and in turn returning the results as you want.
In[21]:=
norulSolve[eq_Equal, x_] := Solve[eq, {x}] /. {a_ -> b_} :> b
An example follows now
In[24]:=
norulSolve[(x^4 + x^2 + x)*(x^3 + a^3) == 0, x]
Out[24]=
{0, -(2/(3*(-9 + Sqrt[93])))^(1/3) + ((1/2)*(-9 + Sqrt[93]))^(1/3)/
3^(2/3),
-(((1 + I*Sqrt[3])*((1/2)*(-9 + Sqrt[93]))^(1/3))/(2*3^(2/3))) + (1
- I*Sqrt[3])/(2^(2/3)*(3*(-9 + Sqrt[93]))^(1/3)),
-(((1 - I*Sqrt[3])*((1/2)*(-9 + Sqrt[93]))^(1/3))/(2*3^(2/3))) + (1
+ I*Sqrt[3])/(2^(2/3)*(3*(-9 + Sqrt[93]))^(1/3)), -a,
(-1)^(1/3)*a, (-(-1)^(2/3))*a}
And here is a nice example of application
In[32]:=
norulSolve[x^4 + 16 == 0, x]
Factor[1/(x^4 + 16), Extension -> %]
Apart[%]
Out[32]=
{-2*(-1)^(1/4), 2*(-1)^(1/4), -2*(-1)^(3/4), 2*(-1)^(3/4)}
Out[33]=
1/((2*(-1)^(1/4) - x)*(2*(-1)^(3/4) - x)*(2*(-1)^(1/4) +
x)*(2*(-1)^(3/4) + x))
Out[34]=
(-1)^(1/4)/(32*(2*(-1)^(1/4) - x)) + (-1)^(3/4)/(32*(2*(-1)^(3/4) -
x)) + (-1)^(1/4)/(32*(2*(-1)^(1/4) + x)) +
(-1)^(3/4)/(32*(2*(-1)^(3/4) + x))
Regards
Dimitris
=CF/=C7 JikaiRF at aol.com =DD=E3=F1=E1=F8=E5:
> Dear Members:
> When I use "Solve" in order to solve polynomials with high orders, its
> solutions are usually reterned in terms of a pair of nested braces,
> like {{x -> 0.1}, {x -> 0.5}}. In this situation, I would like to
> obtain a pair of braces such as {0.1, 0.5}. In other words, I would
> like to know how to eliminate x ->.
> How can I obtain my result?
>
> Sincerely
> F. Takata.
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