Re: Need some help with monitoring evaluation

*To*: mathgroup at smc.vnet.net*Subject*: [mg96998] Re: Need some help with monitoring evaluation*From*: Dan629 <dangerrity at gmail.com>*Date*: Sun, 1 Mar 2009 04:57:42 -0500 (EST)*References*: <go0j6f$n2j$1@smc.vnet.net> <go63rc$q1a$1@smc.vnet.net>

On Feb 27, 3:09 am, Dan629 <dangerr... at gmail.com> wrote: > On Feb 26, 5:01 am, congruentialumina... at yahoo.com wrote: > > > > > This has caused me some consternation in the past. By this I mean that > > I make some coding error that causes Mathematica to loop. > > > I would check out: > > > ref/TimeConstrained > > ref/Reap > > ref/StepMonitor > > > HTH. > > > Roger Williams For the curious, TracePrint provided the most useful following information using _Integrate as the pattern. After about 24 hours the InverseFourierTransform could not be done and Mathematica returned the original expression. I was wistfully optimistic, expecting the multiplication in the Fourier domain to give a result when the corresponding integral in the normal domain won't. I don't know if this is a forum where math questions are posted (as opposed to strictly Mathematica questions), but I'll ask anyway. The PDF of the circular distribution has two discontinuities. If I do a numeric convolution with ListConvolve I can easily get a result. Alternatively I can use a summation and get a symbolic result consisting of hundreds or thousands of summands. Neither solution can be readily integrated nor differentiated (the variables shown as constants below are more complicated that that, though they are independent). I'd really like to get a closed form symbolic formula for the convolution. Am I dreaming to think that Mathematica can do something like this? Or it is simply a fact of math that a discontinuous function cannot be convolved with a Gaussian Distribution and no trick or computational genius will make it work? Finally, I've tried to curve fit to the summation solution without much luck. I've used the Taylor and polynomial expansions along with FFTs. The convolution is a beautiful curve based on a sum of Gaussians and so intuitively seems easy to fit, but I can't seem to get it. All variables are real, and both ll and ss are always positive. So, for the illustration, try this: vars = { ll -> .27, mm -> 0, ss -> .06 }; pdf1[ xx_ ] := PDF[ NormalDistribution[ mm, ss ], xx ]; pdf2[ xx_ ] := Piecewise[ { { 1/(Pi * Sqrt[ ll^2 - xx^2 ]), Abs[ xx ] < ll } }, 0 ]; Plot[ { pdf1[ xx ], pdf2[ xx ] } /. vars, { xx, -.3, .3 } ] conv = Sum[ (pdf1[ xx - zz ] pdf2[ zz ]), { zz, -.6, .6, 0.02 } ] / 200; Plot[ Evaluate[ conv /. vars ], { xx, -.5, .5 } ] I'm seeking a symbolic representation of the second curve. Please be kind Jens -- I'm admittedly an amateur!

**Re: wrong solution for double integral of piecewise**

**Question about MathLink and .NET/Link with Mono on Mac OS X and Linux**

**Re: wrong solution for double integral of piecewise**

**Question about MathLink and .NET/Link with Mono on Mac OS X and Linux**