Re: bug -- advice sought
- To: mathgroup at smc.vnet.net
- Subject: [mg84568] Re: bug -- advice sought
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Thu, 3 Jan 2008 20:36:05 -0500 (EST)
- References: <7145136.1199359097519.JavaMail.root@m08>
- Reply-to: drmajorbob at bigfoot.com
Maybe I'm missing something, but consider the following steps:
z = 2*Sqrt[2]*Rvt^5*(Rvt^2 + Rx^2)^(3/2)*
w^2*(Sqrt[2*Dc - I*Rx^2*w]*(2*Dc - I*(Rvt^2 + Rx^2)*w)^3 +
Sqrt[2*Dc + I*Rx^2*w]*(2*Dc + I*(Rvt^2 + Rx^2)*w)^3)
2 Sqrt[2] Rvt^5 (Rvt^2 + Rx^2)^(
3/2) w^2 (Sqrt[
2 Dc - \[ImaginaryI] Rx^2 w] (2 Dc - \[ImaginaryI] (Rvt^2 +
Rx^2) w)^3 +
Sqrt[2 Dc + \[ImaginaryI] Rx^2 w] (2 Dc + \[ImaginaryI] (Rvt^2 +
Rx^2) w)^3)
z2 = z // ExpandAll // Simplify
2 Sqrt[2] Rvt^5 (Rvt^2 + Rx^2)^(
3/2) w^2 (-12 \[ImaginaryI] Dc^2 (Rvt^2 + Rx^2) w (Sqrt[
2 Dc - \[ImaginaryI] Rx^2 w] - Sqrt[
2 Dc + \[ImaginaryI] Rx^2 w]) + \[ImaginaryI] (Rvt^2 +
Rx^2)^3 w^3 (Sqrt[2 Dc - \[ImaginaryI] Rx^2 w] - Sqrt[
2 Dc + \[ImaginaryI] Rx^2 w]) +
8 Dc^3 (Sqrt[2 Dc - \[ImaginaryI] Rx^2 w] + Sqrt[
2 Dc + \[ImaginaryI] Rx^2 w]) -
6 Dc (Rvt^2 + Rx^2)^2 w^2 (Sqrt[2 Dc - \[ImaginaryI] Rx^2 w] +
Sqrt[2 Dc + \[ImaginaryI] Rx^2 w]))
z3 = z2 /. {Sqrt[2 Dc - \[ImaginaryI] Rx^2 w] -> u,
Sqrt[2 Dc + \[ImaginaryI] Rx^2 w] -> v}
2 Sqrt[2] Rvt^5 (Rvt^2 + Rx^2)^(
3/2) w^2 (8 Dc^3 (u + v) -
12 \[ImaginaryI] Dc^2 (Rvt^2 + Rx^2) (u - v) w -
6 Dc (Rvt^2 + Rx^2)^2 (u + v) w^2 + \[ImaginaryI] (Rvt^2 +
Rx^2)^3 (u - v) w^3)
z4 = z3 /. {u + v -> sum, u - v -> diff}
2 Sqrt[2] Rvt^5 (Rvt^2 + Rx^2)^(
3/2) w^2 (8 Dc^3 sum - 12 \[ImaginaryI] Dc^2 diff (Rvt^2 + Rx^2) w -
6 Dc (Rvt^2 + Rx^2)^2 sum w^2 + \[ImaginaryI] diff (Rvt^2 +
Rx^2)^3 w^3)
z5 = Collect[z4 // ExpandAll, {sum, diff}, Simplify]
2 \[ImaginaryI] Sqrt[2] diff Rvt^5 (Rvt^2 + Rx^2)^(5/2)
w^3 (-12 Dc^2 + (Rvt^2 + Rx^2)^2 w^2) -
4 Sqrt[2] Dc Rvt^5 (Rvt^2 + Rx^2)^(3/2)
sum w^2 (-4 Dc^2 + 3 (Rvt^2 + Rx^2)^2 w^2)
Now, it's easy to convince myself that u+v and u-v are real, if one
chooses an appropriate branch cut for Sqrt.
And then z5 is real <==> u == v.
And a branch cut might be chosen to make THAT true... but is it the same
branch cut as the one that makes u+v real?
Bobby
On Thu, 03 Jan 2008 04:35:21 -0600, <UHAP023 at alpha1.rhbnc.ac.uk> wrote:
> Dear All,
> A couple of queries;
>
> (1) Regarding the following expression,
>
> z=2*Sqrt[2]*Rvt^5*(Rvt^2 + Rx^2)^(3/2)*w^2*
> (Sqrt[2*Dc - I*Rx^2*w]*(2*Dc - I*(Rvt^2 + Rx^2)*w)^3 +
> Sqrt[2*Dc + I*Rx^2*w]*(2*Dc + I*(Rvt^2 + Rx^2)*w)^3)
>
> all the variables are of Real type and are +ve. I do the following
> FullSimplify[],
>
> z = FullSimplify[
> z, {Rx \[Element] Reals, Rvt \[Element] Reals, Dc \[Element] Reals
,
> w \[Element] Reals, Rx > 0, Rvt > 0, Dc > 0, w > 0}]
>
> and get an unchanged result. Now I believe that this expression
> should *always* produce a real result for real, +ve parameter values.
> A simple but obviously non-rigorous test is to substitute arbitrary
> real,+ve values for the parameters and observe the numeric result,
> eg.,
>
> Rx = Random[]; Rvt = Random[]; w = Random[]; Dc = Random[]; z
>
> The result is indeed always real (ie. Im[z]==0). Given that
> FullSimplify[] tries to return the simplest result rather than
> eliminate complex expressions as I want, I tried doing,
>
> FullSimplify[z, {Rx \[Element] Reals, Rvt \[Element] Reals,
> Dc \[Element] Reals, w \[Element] Reals, Rx > 0, Rvt > 0, Dc > 0,=
> w > 0}, ComplexityFunction -> (Count[{#1}, _Complex , \[Infinity]]=
=
> &)]
>
> which I hoped would do what I want. Unfortunately it returns a more
> complicated result (fair enough) but which has the same number of I's
> as the input expression (not OK). What am I doing wrong here? Is
> there a way to do this?
>
> =========================
=========================
=========================
=====
>
> (2) The above query was precipitated indirectly by the following
> strangeness. I am integrating the following integrand,
>
> igrand=(Dc*Rx^2*Cos[theta]^2)/((4*Dc^2 + Rx^4*w^2*Cos[theta]^4)*
> (Rvt^2 + Rx^2*Sin[theta]^2)^3)
>
> Again the parameters are all of Real type and +ve. Moreover the
> expression only contains a fraction, sums and products. The integral
> below is definite and both Cos[] and Sin[] are +ve over the
> integration limits, so the area under the graph should also be Real
> and +ve.
>
> igral = Integrate[igrand, {theta, 0, Pi/2},
> Assumptions -> {Rx \[Element] Reals, Rvt \[Element] Reals,
> w \[Element] Reals, Dc \[Element] Reals, theta \[Element] Real=
s,
> Rx > 0, Rvt > 0, w > 0, Dc > 0, theta >= 0}]
>
> However the integral contains many complex terms which is
> understandable AIUI because Integrate[] tries many transforms/pattern
> matches to do the integration -- some of which will produce a complex
> result. The thing that puzzles me is that the integration result
> contains lots of terms such as 'Sign[Rvt]^2'. My point is that
> Integrate[] has been explicitly told in its Assumptions argument that
> Rvt in this case is Real and +ve, so why does it do this? Is this a
> bug? Admittedly a,
>
> Simplify[igral, {Rx \[Element] Reals, Rvt \[Element] Reals,
> w \[Element] Reals, Dc \[Element] Reals, Rx > 0, Rvt > 0, w > 0, D=
c =
> > 0}]
>
> will remove terms like 'Sign[Rvt]^2' but I still end up with a complex=
> and unwieldy expression which is much larger that the integrand.
>
> Any suggestions?
>
> Many thanks
> Tom Crane
>
> Ps. I'm using Mathematica 4.0 and the From: field in the message heade=
r =
> is
> invalid. It is just a spam-trap.
>
-- =
DrMajorBob at bigfoot.com