Re: algebraic solutions
- To: mathgroup@smc.vnet.net
- Subject: [mg11320] Re: algebraic solutions
- From: weber@math.uni-bonn.de (Matthias Weber)
- Date: Fri, 6 Mar 1998 00:40:32 -0500
- Organization: RHRZ - University of Bonn (Germany)
- References: <6dj2ag$a32@smc.vnet.net>
In article <6dj2ag$a32@smc.vnet.net>, Ersek_Ted%PAX1A@mr.nawcad.navy.mil wrote: > Daniel wrote: > ---------- > |I want to find the roots of the following equation: | > | z^5 + 2z^3 - p + 1 = 0 > | > |I want to solve for z in terms of p. Now, if I pick some random number > |for p, I can get mathematica to solve for z, but I cant get a solution > |in terms of p. Alternatively, I would like to be able to plot this > What you are trying to get is a closed form solution for a fifth order > polynomial. > This can only be done if you are satisfied with horribly complicated > expressions in terms of Elliptic functions. Mathematica will not do it > for you. I think the designers figured such a solution is of no > practical use. It is a question of taste what you consider to be horrible. There is a Mathematica package at mathsource for solving the quintic in various ways. > For degrees higher than five it gets even worse. This is a famous fact > in mathematics. > The famous fact is that it is impossible to solve the general quintic (or higher degree equations) in terms of radicals This doesn't say i) that Daniel's equation has a "horrible" solution ii) that the solutions in whatever form are of no practical use If you are doing number theory, you might be very well interested in (say) the prime factors of "explicit horrible solutions". I would suggest first to compute the Galois group of the equation, and if it is solvable, have a look at the above mentioned MathSource package. There were routines which could solve the solvable quintics, if I remember correctly. Matthias