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 >