       Re: Finding a continuous solution of a cubic

• To: mathgroup at smc.vnet.net
• Subject: [mg85945] Re: [mg85899] Finding a continuous solution of a cubic
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Thu, 28 Feb 2008 02:53:12 -0500 (EST)
• References: <200802270925.EAA16050@smc.vnet.net>

```Hugh Goyder wrote:
> I am investigating the maxima and minima of the expression, e, below.
> The expression is a quartic in x and has two parameters a and b which,
> for my problem, are restricted to the range -1 < a < 1 and 0 < b (the
> interesting part is 0<b<1).
>
> I start with a simple approach and take the derivative with respect to
> x and solve to find the three turning points. The solution is
> complicated and not easy to examine. I work out the discriminant to
> see when I will have three real roots or one and plot the region. I
> work out three functions for the three roots r1, r2 and r3, and three
> functions for the value of e at the roots, f1, f2, and f3. I am
> expecting three real functions when within the region defined by the
> discriminant and one outside this region. I plot the three roots and
> the three functions (there are some artifacts where the functions turn
> from real to complex, this is not a problem).
>
> Now my problem. The function f3 is continuous in the region where
> there are three solutions and this I like. Functions f1 and f2 are not
> continuous but contain holes. If functions f1 and f2 are plotted
> together then they each fill in the others holes. The holes makes my
> life difficult because I do not have continuous functions.
>
> Is it possible to make Solve find roots that will give rise to three
> functions one corresponding to the maxima and two corresponding to
> each minima without the functions crossing over and jumping between
> the turning points?
>
> I also notice, from the plots when shown together,  that my desired
> functions are even with respect to the parameter a. Is this correct
> and how would I show this if I can't assemble them as continuous
> functions?
>
> Thanks
>
> Hugh Goyder
>
> e = ((1 - x)^2 + b*(-1 + a)^2)*((1 + x)^2 +
>           b*(1 + a)^2);
>
> d = Simplify[D[e, x]];
>
> dis = Discriminant[d, x];
>
> ContourPlot[dis, {a, -1, 1}, {b, 0, 1},
>    ContourShading -> False, Contours -> {0},
>    FrameLabel -> {"a", "b"}]
>
> sol = Solve[d == 0, x];
>
> ClearAll[x1, x2, x3, f1, f2, f3];
> x1[a_, b_] := Evaluate[x /. sol[]];
> x2[a_, b_] := Evaluate[x /. sol[]];
> x3[a_, b_] := Evaluate[x /. sol[]];
> f1[a_, b_] := Evaluate[e /. sol[]];
> f2[a_, b_] := Evaluate[e /. sol[]];
> f3[a_, b_] := Evaluate[e /. sol[]];
>
> r1 = Plot3D[x1[a, b], {a, -1, 1}, {b, 0, 1}]
>
> r2 = Plot3D[x2[a, b], {a, -1, 1}, {b, 0, 1}]
>
> r3 = Plot3D[x3[a, b], {a, -1, 1}, {b, 0, 1}]
>
> Show[r1, r2, r3, PlotRange -> All]
>
> s1 = Plot3D[f1[a, b], {a, -1, 1}, {b, 0, 1}]
>
> s2 = Plot3D[f2[a, b], {a, -1, 1}, {b, 0, 1}]
>
> s3 = Plot3D[f3[a, b], {a, -1, 1}, {b, 0, 2}]

Before using Solve, do

SetOptions[Roots, Cubics -> False];

This will force Solve to use Root[] objects instead of Cardano formulas.
The latter can jump around due to branch cut structure of (nested)
radicals. Also I think Plot might prefer the former because it always
knows when they are real valued. With the radicals there might be
spurious imaginary part artifacts upon numerical evaluation.

Daniel Lichtblau
Wolfram Research

```

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