Re: Real integrand->complex result.
- To: mathgroup at smc.vnet.net
- Subject: [mg20308] Re: [mg20293] Real integrand->complex result.
- From: David Withoff <withoff at wolfram.com>
- Date: Fri, 15 Oct 1999 20:20:40 -0400
- Sender: owner-wri-mathgroup at wolfram.com
> I get a complex result of the form f(t)=(a+bi)g(t) when evaluating a
> real integrand F(x,t)*phi(x) over the interval {x:0<x<L). All of the
> variables have been properly declared as real and the functions have
> been declared as real for real arguments (using the package "ReIm").
>
> To see the problem more clearly, the code can be copied into Mathematica
> (3 or 4) and executed. (The Traditional or Standard Forms will not paste
> into a text document such as this)
>
> When evaluating the integral
>
> \!\(\[Integral]\_0\%\[ScriptL]\( simpF\[CurlyPhi]\_n\) \[DifferentialD]x\)
>
> over the interval (0,ScriptL) where ScriptL=1. The integrand
> \!\(simpF\[CurlyPhi]\_n\ is, for n=1, given by
>
> \!\(TraditionalForm
> \`\(-\(1\/6\)\)\ \[ExponentialE]\^\(\(\(-3\)\ \[Tau]\)/2\)\
> \((\((3 + 2\ \[Pi]\^2)\)\ \(cos(\[Pi]\ x)\) -
> 6\ \[ExponentialE]\^\(x/2\) + 3)\)\ \((cos(3.66558239083868908`\ x) +
> sin(3.66558239083868908`\ x))\)\)
>
> The evaluation of the integral is also quite slow, even for n=2, even
> when the expression is simplified as much as possible, and the final
> program will need to evaluate a larger number of terms.
>
> Thanks in advance, Bill
The imaginary 0.0 I parts are an artifact of numerical error. You can
discard those parts using Chop. For example:
In[1]:= Integrate[(-(1/6) (3 - 6 E^(x/2) + (3 + 2 Pi^2) Cos[Pi x])*
(Cos[3.665582390838689 x] + Sin[3.665582390838689 x]))/
E^((3 tau)/2), {x, 0, L}] //Chop
Out[1]= -(22.7991 + 1.38775 (-0.589747 Cos[3.66558 L] +
0.5 L
> 1. E Cos[3.66558 L] - 16.839 Cos[3.14159 L] Cos[3.66558 L] -
> 14.4319 Cos[3.66558 L] Sin[3.14159 L] + 0.589747 Sin[3.66558 L] -
0.5 L
> 1.3159 E Sin[3.66558 L] +
> 16.839 Cos[3.14159 L] Sin[3.66558 L] -
(3 tau)/2
> 14.4319 Sin[3.14159 L] Sin[3.66558 L])) / (6 E )
That calculation finished in a fraction of a second when I tried it.
The slowness in your calculation is almost certainly due to the ReIm
package, which will consume a lot of time and will not be helpful,
because there isn't anything that this or any other package can do to
avoid numerical error.
Dave Withoff
Wolfram Research