Re: Integrate this how?

*To*: mathgroup at smc.vnet.net*Subject*: [mg116329] Re: Integrate this how?*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Fri, 11 Feb 2011 04:15:00 -0500 (EST)

I get this in version 8 (with error messages for one or more methods): r[t_] = (1 + Sin[t]) (1 + 9/10 Cos[8 t]) (1 + 1/10 Cos[24 t]) (9/10 + 5/100 Cos[200 t]); arcLength[t_] = Sqrt[r[t]^2 + D[r[t], t]^2] // Simplify; NIntegrate[arcLength@t, {t, -Pi, Pi}, Method -> #] & /@ {"GlobalAdaptive", "LocalAdaptive", "MonteCarlo", "AdaptiveMonteCarlo", "QuasiMonteCarlo", "AdaptiveQuasiMonteCarlo"} {48.37, 48.0666, 48.0852, 48.3858, 47.985, 48.0663} but Integrate simply fails. The following plot suggests how messy this is, numerically: Plot[arcLength@t, {t, -Pi, Pi}] I think you're trying to approximate a fractal, and that fractal has infinite arc length... so it's not surprising you run into difficulties. Bobby On Thu, 10 Feb 2011 04:21:36 -0600, BoLe <bole79 at gmail.com> wrote: > Curve is closed (t from -Pi to Pi) and in polar form > > r[t_]:=(1+Sin[t]) (1+9/10 Cos[8 t]) (1+1/10 Cos[24 t]) (9/10+5/100 > Cos[200 t]) > > Would like to find out its arc length, so the integral of > > Sqrt[r[t]^2+D[r[t],t]^2] > > Version 7 complains but returns a somewhat reasonable number > (something around 50; drew together with a circle of the same length), > version 8 says nothing but gives just too low a number (five and > something). How can this be? New Mathematica should have no problem > with such high oscillatory integrands, shouldn't she? > -- DrMajorBob at yahoo.com