Re: Finding a continuous solution of a cubic
- To: mathgroup at smc.vnet.net
- Subject: [mg86310] Re: Finding a continuous solution of a cubic
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Sat, 8 Mar 2008 05:40:51 -0500 (EST)
- References: <fq3agu$fq0$1@smc.vnet.net> <fqb92v$n6s$1@smc.vnet.net> <fqlmnf$jcm$1@smc.vnet.net>
Hugh Goyder <h.g.d.goyder at cranfield.ac.uk> wrote: > Thanks to Andrzei and Daniel for their helpful comments -they have > enabled me to simplify the problem. The response from David appears > to do exactly what I want but I am completely mystified as to what he > did to get these region boundaries. Please could you reveal the magic > you used to find the regions? I'm glad that I was able to give you exactly what you wanted. Namely, "...the three functions you want, the solutions of d == 0, can be expressed as x -> Root[-2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 1] applicable when 27 a^2 b^2 <= (1 - b - a^2 b)^3 Or a <= 0 x -> Root[-2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 2] applicable when 27 a^2 b^2 <= (1 - b - a^2 b)^3 x -> -Root[2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 1] applicable when 27 a^2 b^2 <= (1 - b - a^2 b)^3 Or a >= 0 " As to the boundaries, I hate to say that I merely graphed the solutions, thought about the discriminant, and then wrote down the appropriate conditions for applicability. In a private email, you said that you had other similar problems and therefore hoped that I would disclose my method. I've done so now, but it's not really a "method" in the sense you wanted. Perhaps one of our more algebraically inclined cognoscenti could describe a true "method". David