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From: AES on 13 Aug 2010 06:53
Some time back there was a post in this group about how to convert the
expression

(Eq_1) � a*b + a*c - 2*a*d + b*c - 2*b*d - 2*c*d + 3*d^2

into

(Eq_2) � (a-d)*(b-d) + (a-d)*(c-d) + (b-d)*(c-d)

to which I made an absurd response. My meds are working better now, so
I hope the following will make sense.

Although I'm not attempting to answer the original question, I did want
to understand better the symmetry of these expressions. So, suppose we
scale a, b and c to d, and rename them as x, y, z. The expression then
obviously has a zero at x=y=z=1, so shift the origin of coordinates to
that point (suffix 's') and get

f=(x-1)(y-1)+(x-1)(z-1)+(y-1)(z-1);
fs=f/.{x->xs+1,y->ys+1,z->zs+1};
fs//Expand

xs ys+xs zs+ys zs

This has an obvious axis of symmetry, so rotate the axes into alignment
with it (suffix 'r') and get

{xs,ys,zs}=RotationMatrix[{{0,0,1},{1,1,1}}] .{xr,yr,zr};
fr=fs;fr//Expand

-xr^2/2 - yr^2/2 + zr^2

Note sure what a geometer would call that, but it's obviously a figure
of rotation about an axis (ellipsoid of rotation about an imaginary
axis?). Confirm this by making a couple of plots:

x=Sin[theta]Sin[phi];y=Sin[theta]Cos[phi];z=Cos[theta];
fs=x y +x z+y z;
fr=z^2 - 0.5(x^2+y^2);
ps=Graphics[SphericalPlot3D[fs,theta,phi]];
pr=Graphics[SphericalPlot3D[fr,theta,phi]];
GraphicsRow[{ps,pr}]

or by mapping f onto the surface or spheres of varying diameter:

Manipulate[
x=a Sin[theta]Sin[phi];y=a Sin[theta]Cos[phi];z=a Cos[theta];
f=z^2-0.5(x^2+y^2);
Show[Graphics[SphericalPlot3D[a+ b f,theta,phi]]],
{{a,1},0,5},{{b,0.1},0,1}]

[and in the process learn that Mathematica allows spheres to have
negative radii.]

Follow-up question: Suppose you want to plot a unit sphere with
SphericalPlot3D in which different areas or patches on the surface have
different colors. depending on the theta, phi (or x,y,z) values. Or, a
SphericalPlot3D[r[theta,phi], theta, phi] for which positive values of r
give a red surface and negative values a blue surface. How might one do
that in some simple way?

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