Re: Re: Bounds for Trig expression
- To: mathgroup at smc.vnet.net
- Subject: [mg54751] Re: [mg54683] Re: Bounds for Trig expression
- From: Chris Chiasson <chris.chiasson at gmail.com>
- Date: Mon, 28 Feb 2005 03:29:01 -0500 (EST)
- References: <cvk3hc$d6p$1@smc.vnet.net> <200502270629.BAA25356@smc.vnet.net>
- Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
- Sender: owner-wri-mathgroup at wolfram.com
I was messing around with this the other day, but I don't remember if
I sent out the associated email - I think my computer borked on me
when I was writing it (which is unfortunate because the previous email
was more coherent...).
Anyway, I changed the step size on the table functions and played
around with the limits as well. Some of the plotting commands I issued
had over two million points in them.
k seems to be *roughly* analogous to theta in the Cartesian to complex
coordinate system maps. Try its limits from one pi to two pi (with
steps of 0.01 pi). a seems analogous to r (but in a kind of rmax way
-- it sets the furthest distance a point can be from the origin). Try
its limits from 0 to 1.0, with a step size of 0.01. Sr also seems
analogous to r, but each chosen value of Sr seems to plot a new
"ribbon" of points. For this reason, set the limits of Sr to be from 1
to 1 in steps of 1 (essentially just sets Sr to 1, but you don't have
to modify the other commands).
Finally, instead of wrapping the table command in flatten and partition, try:
Flatten[Table[blah],2]
BTW, the reason for the 2 is different than the reason for the 2 in
your partition command.
You should get some kind of lightly dense ellipse (or circular) cloud
that is off center from the origin.
One note, AspectRatio sets the aspect ratio of the plot area itself,
not the aspect ratio of the coordinate system inside the plot (if that
makes sense).
Okay, time to not be lazy and actually post some code:
ee = (-1 +
Cos[a*k] + I*Sin[a*k])*(6 - 3*Sr + 3*(2 + Sr)*Cos[(-2 +
a)*k] - (2 + Sr)*Cos[
2*(-1 + a)*k] - 2*Cos[a*k] + Sr*Cos[a*k] - 6*I*Sin[(-2 + a)*
k] - 3*I*Sr*Sin[(-2 + a)*k] + 2*I*Sin[2*(-1 + a)*k] + I*Sr*
Sin[2*(-1 + a)*k] - 2*I*Sin[a*k] + I*Sr*Sin[a*k]);
d1 = Flatten[Table[{Re[#1], Im[#1]} &[
ee], {k, Pi, 2 Pi, 0.01*Pi}, {a, 0, 1, 0.01}, {Sr, 1, 1, 1}], 2];
ListPlot[d1];
Regards,
On Sun, 27 Feb 2005 01:29:07 -0500 (EST), Paul Abbott
<paul at physics.uwa.edu.au> wrote:
> In article <cvk3hc$d6p$1 at smc.vnet.net>,
> "Hugh" <h.g.d.goyder at cranfield.ac.uk> wrote:
>
> > I wish to find bounds for the complex trig expression ee below. The
> > expression depends on the real variables Sr, a and k which lie in the
> > range
> > 0 < Sr <1
> > 0 < a < 1
> > 0 < k
> > A blind numerical evaluation of many values, plotted below, suggests
> > that the real part is bounded by (-32, 16) while the imaginary part is
> > bounded by approximately (-26, 26). I am happy bounding in a
> > rectangular region on the complex plane although the numerical plot
> > suggests an elliptical region.
> >
> > Is there a non numerical approach to finding the bounds? Possibly by
> > replacing Cos and Sin by all permutation of + and - 1?
>
> A semi-numerical approach is to use Interval.
>
> > I have more expressions like this to tackle so I would like an approach
> > that can be generalized.
> >
> > ee = (-1 + Cos[a*k] + I*Sin[a*k])*(6 - 3*Sr + 3*(2 + Sr)*Cos[(-2 +
> > a)*k] - (2 + Sr)*Cos[2*(-1 + a)*k] -
> > 2*Cos[a*k] + Sr*Cos[a*k] - 6*I*Sin[(-2 + a)*k] - 3*I*Sr*Sin[(-2 +
> > a)*k] + 2*I*Sin[2*(-1 + a)*k] +
> > I*Sr*Sin[2*(-1 + a)*k] - 2*I*Sin[a*k] + I*Sr*Sin[a*k]);
>
> Entering
>
> FullSimplify[ee /.
> {a -> Interval[{0, 1}],
> Sr -> Interval[{0, 1}],
> k -> Interval[{0, 2 Pi}]}]
>
> gives a cruder bound. This bound is not tight because the intervals are
> treated as independent. Also, consider expressions like
>
> FullSimplify[Exp[I Interval[{0, 2 Pi}]]]
>
> The resulting rectangular bound encloses the unit circle.
>
> Finally, I note that ee can be written as
>
> (z^a - 1) ((Sr + 2) (3 - z^(-a)) z^(2 - a) + (Sr - 2) (z^a - 3))
>
> where z = Exp[I k]. Perhaps this form can be used to find a tighter
> bound?
>
> Cheers,
> Paul
>
> --
> Paul Abbott Phone: +61 8 6488 2734
> School of Physics, M013 Fax: +61 8 6488 1014
> The University of Western Australia (CRICOS Provider No 00126G)
> 35 Stirling Highway
> Crawley WA 6009 mailto:paul at physics.uwa.edu.au
> AUSTRALIA http://physics.uwa.edu.au/~paul
>
>
--
Chris Chiasson
Kettering University
Mechanical Engineering
Graduate Student
1 810 265 3161
- References:
- Re: Bounds for Trig expression
- From: Paul Abbott <paul@physics.uwa.edu.au>
- Re: Bounds for Trig expression