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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