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Re: Big Memory Needed Again

  • To: mathgroup at smc.vnet.net
  • Subject: [mg111183] Re: Big Memory Needed Again
  • From: Artur <grafix at csl.pl>
  • Date: Fri, 23 Jul 2010 07:09:46 -0400 (EDT)
  • References: <4C4860C4.8010903@csl.pl>
  • Reply-to: grafix at csl.pl

All sign ^ was eaten in previous email, sorry for cofusion!

Solve[1/2 Sqrt[
  3] ((-1 - 3 p + 2 z^2 + 2 z Sqrt[-1 + z^2])/(
   2 (-z + Sqrt[-1 + z^2]))) z +
 1/2 Sqrt[
  1 - ((-1 - 3 p + 2 z^2 + 2 z Sqrt[-1 + z^2])/(
    2 (-z + Sqrt[-1 + z^2])))2] z -
 1/2 I ((-1 - 3 p + 2 z^2 + 2 z Sqrt[-1 + z^2])/(
   2 (-z + Sqrt[-1 + z^2]))) Sqrt[-1 + z^2] +
 1/2 I Sqrt[3] Sqrt[
  1 - ((-1 - 3 p + 2 z^2 + 2 z Sqrt[-1 + z^2])/(
    2 (-z + Sqrt[-1 + z^2])))2] Sqrt[-1 + z^2] == q, z]

Artur pisze:
> Dear Mathematica Gurus,
> If sombody have memory bigger as 8MB try to help me solve equation 
> (don't try with smallest memory).
> Also if will be out of memory inform me how much memory do you use.
> Solve[1/2 Sqrt[
>   3] ((-1 - 3 p + 2 z2 + 2 z Sqrt[-1 + z2])/(
>    2 (-z + Sqrt[-1 + z2]))) z +
>  1/2 Sqrt[
>   1 - ((-1 - 3 p + 2 z2 + 2 z Sqrt[-1 + z2])/(
>     2 (-z + Sqrt[-1 + z2])))2] z -
>  1/2 I ((-1 - 3 p + 2 z2 + 2 z Sqrt[-1 + z2])/(
>    2 (-z + Sqrt[-1 + z2]))) Sqrt[-1 + z2] +
>  1/2 I Sqrt[3] Sqrt[
>   1 - ((-1 - 3 p + 2 z2 + 2 z Sqrt[-1 + z2])/(
>     2 (-z + Sqrt[-1 + z2])))2] Sqrt[-1 + z2] == q, z]
>
> Or if you have idea how simplify or do somethink inspite let me know!
>
> Best wishes
> Artur
>


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