Re: Re: bug --
- To: mathgroup at smc.vnet.net
- Subject: [mg84569] Re: [mg84547] Re: bug --
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Fri, 4 Jan 2008 05:39:00 -0500 (EST)
- References: <200801031035.FAA15853@smc.vnet.net> <200801040125.UAA11973@smc.vnet.net>
Daniel Lichtblau wrote:
> 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.
>>[...]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,
>>[...]
>
> I'm not sure and don't have a lot of time to experiment. I'll sidestep
> by showing how to determine whether z is always positive subject to the
> stated assumptions.
>
> 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);
>
> Timing[inst = FindInstance[{z<0,rx>0,w>0,rvt>0,dc>0},
> {rx,w,rvt,dc}, Complexes]]
>
> Out[49]= {48.663, {{rx -> 84, w -> 67, rvt -> 5, dc -> 54137}}}
> [...]
Oops. Misread the question. You only want to know that z is real, not
necessarily positive. Same idea as above.
In[2]:= Timing[inst = FindInstance[{Im[z]!=0,rx>0,w>0,rvt>0,dc>0},
{rx,w,rvt,dc}, Complexes]]
Out[2]= {8.54853, {}}
This shows there are no allowed values of the parameters that can give z
a nonzero imaginary part, ergo it is real subject to the stated conditions.
Daniel Lichtblau
Wolfram Research
- References:
- Complex elimination and possible Integrate[] bug -- advice sought
- From: UHAP023@alpha1.rhbnc.ac.uk
- Re: bug --
- From: Daniel Lichtblau <danl@wolfram.com>
- Complex elimination and possible Integrate[] bug -- advice sought