Re: Numerical Integration

• To: mathgroup at smc.vnet.net
• Subject: [mg71511] Re: Numerical Integration
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Tue, 21 Nov 2006 07:05:08 -0500 (EST)
• References: <ejrn8h\$9a4\$1@smc.vnet.net><ejtdna\$j3k\$1@smc.vnet.net>

Peter I apologize for the situation.
My thanks was for you but somehow I confused the names!

Best Regards
Dimitris

dimitris wrote:
> Dear David,
>
> Thanks a lot for your nice solution.
>
> Here is another along the same lines.
>
> h[x_] := Tan[BesselJ[0, x]]
>
> Needs["NumericalMath`BesselZeros`"]
>
> lst = BesselJZeros[0, 10];
> lst[[0]] = 0;
>
> f[i_] := NIntegrate[h[x], {x, lst[[i]], lst[[i + 1]]}]
>
> SequenceLimit[FoldList[Plus, 0, Table[f[i], {i, 0, 9}]]]
> 1.45451
>
>
> Best Regards
> Dimitris
>
> Peter Pein wrote:
> > dimitris schrieb:
> > > Dear All,
> > >
> > > I have one question about the numerical integration of one function.
> > >
> > > \$VersionNumber
> > > 5.2
> > >
> > ...
> > > h[x_] := Tan[BesselJ[0, x]]
> > >
> > > Plot[h[x], {x, 0, 40}, PlotPoints -> 100, Axes -> None, Frame -> {True,
> > > True, False, False}, PlotStyle -> AbsoluteThickness[2]]
> > >
> > > Limit[h[x], x -> Infinity]
> > > 0
> > >
> > > I try hard to find any proper settings for getting a numerical
> > > estimation of its integral
> > > over {0,Infinity} but I can't succeed.
> > >
> > > Any help will be greatly appreciate.
> > >
> > > Dimitris
> > >
> >
> > Hi Dimitris,
> >
> > I tried it this way:
> >
> > In[1]:=
> > Needs["NumericalMath`BesselZeros`"];
> > h[x_] := Tan[BesselJ[0, x]];
> > t0 = SessionTime[];
> > bzlist = NestList[BesselJZerosInterval[0, {1, 2}*Last[#1] + {-1/10, 1/10}] & ,
> >     Flatten[{0, BesselJZeros[0, 2]}], 9];
> > v0 = (NIntegrate[h[x], Evaluate[Flatten[{x, #1}]]] & ) /@ bzlist;
> > SequenceLimit[Rest[FoldList[Plus, 0, v0]]]
> > (SessionTime[] - t0)*seconds
> > Out[6]=
> > 1.4545133229307878
> > Out[7]=
> > 1.75*seconds
> >
> > The displayed result (1.45451) does not change any more when increasing the
> > number of intervals from 9 to 10 or more.
> >
> > Peter

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