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From: Artur on 23 Jul 2010 07:09 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
From: Artur on 23 Jul 2010 07:09
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 > |