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Re: Generated parameters

  • To: mathgroup at smc.vnet.net
  • Subject: [mg116265] Re: Generated parameters
  • From: Andrzej Kozlowski <akozlowski at gmail.com>
  • Date: Wed, 9 Feb 2011 02:09:30 -0500 (EST)

On 8 Feb 2011, at 19:05, olfa wrote:

> Hi Mathematica community,
> For this system:
> Reduce[v/d^(i/-2) == vP/d^(iP/-2) &&
> (iP == -1 || iP == 0), {iP, vP}, Reals, Backsubstitution -> True]
>
> the output is:
> (iP == 0 && C[1] \[Element] Integers && d < 0 && i == -2 C[1] &&
>  vP == (-d)^-C[1] v)
> ||
> (iP == 0 && C[1] \[Element] Integers &&
>  C[1] <= -1 && d < 0 && i == -2 C[1] &&
>  vP == (-d)^-C[1] v)
> ||
> (iP == 0 && C[1] \[Element] Integers &&
>  d < 0 && i == -2 C[1] && vP == -(-d)^-C[1] v)
> ||
> (iP == 0 &&
>  C[1] \[Element] Integers && C[1] <= -1 && d < 0 && i == -2 C[1] &&
>  vP == -(-d)^-C[1] v)
> ||
> (d > 0 && iP == 0 &&
>  vP == d^(i/2) v)
> ||
> (d > 0 && iP == -1 && vP == d^(1/2 + i/2) v)
>
> 1)Why Reduce generates parameters for this example?
> 2) and how to avoid them knowing that for this system the solution
> that I hope to get is
> ( iP == 0 && vP == d^(i/2) v)
> ||
> ( iP == -1 && vP == d^(1/2 + i/2) v)
>
>
> Thank you.
>


Your solution is not a correct or "full" solution in the sense that Reduce understands these terms since in your solution d could be negative and i odd making vP imaginary etc. Reduce divides the cases into the ones when d negative, in whcih case i must be even. The solution could be expressed without c1 since c1 is just i2, but its appearance is an artifact fo the way Reduce arrives at the solution.

To get the solution you seem to want, use Solve and not Reduce and Simplify under the condition that d>0:

Simplify[{iP, vP} /. Solve[v/d^(i/-2) == vP/d^(iP/-2) &&
        (iP == -1 || iP == 0), {iP, vP}, Reals], d > 0]

{{-1, d^((i + 1)/2)*v}, {0, d^(i/2)*v}}

Andrzej Kozlowski


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