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confusion about sampled points in NIntegrate[...,Method->Oscillatory]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg70085] confusion about sampled points in NIntegrate[...,Method->Oscillatory]
*From*: dimmechan at yahoo.com
*Date*: Mon, 2 Oct 2006 00:34:26 -0400 (EDT)
Dear all,
I posted this message again since of his unreadable format there
was no response. I will concentrate only on my queries ommiting
unecessary details.
-----------------------------------------------------------------------
I am very confused about the points sampled by NIntegrate when the
option Method->Oscillatory is selected.
Remove["Global`*"]
$Version
5.2 for Microsoft Windows (June 20, 2005)
Consider the following rapidly oscillatory function
hh[x_]:=ArcTan[3/x] Sin[-9 x]
Here is its plot
Plot[hh[x],{x,0,10},PlotPoints\[Rule]1000];
Here is the analytic result of the integral over the range
{x,0,Infinity}
Integrate[hh[x],{x,0,Infinity}]//N
-0.1745329251991049
Here is one numerical estimation
NIntegrate[hh[x],{x,0,Infinity},Method\[Rule]Oscillatory]
-0.174532925199098
Previous command is equivalent to
NIntegrate[hh[x],{x,0,3(1/9)Pi}]+NSum[NIntegrate[hh[x],{x,k(1/9)Pi,(k+1)(1/9)Pi}],{k,3,Infinity},Method\[Rule]SequenceLimit,VerifyConvergence\[Rule]False]
NIntegrate::nlim: x = 0.349066 k is not a valid limit of integration.
-0.17453292519909863
Here is the number of sampled points.
Block[{Message},Length[Reap[NIntegrate[hh[x],{x,0,3(1/9)Pi},EvaluationMonitor\[RuleDelayed]Sow[x]]+NSum[NIntegrate[hh[x],{x,k(1/9)Pi,(k+1)(1/9)Pi},EvaluationMonitor\[RuleDelayed]Sow[x]],{k,3,Infinity},Method\[Rule]SequenceLimit,VerifyConvergence\[Rule]False]][[2,1]]]]
374
Below is another implementation
psum[i_?NumberQ]:=NIntegrate[hh[x],{x,i(1/9)Pi,(i+1)(1/9)Pi}]
NSum[psum[i],{i,0,Infinity},Method\[Rule]SequenceLimit,VerifyConvergence\[Rule]False]
-0.1745329251991767
Consider now the following function which keeps track of the points
at which is evaluated, in a list called sampledPoints.
sampledPoints={};
f[x_?InexactNumberQ]:=(AppendTo[sampledPoints,x];-ArcTan[3/x])
Then
NIntegrate[f[x]*Sin[9x],{x,0,Infinity},Method\[Rule]Oscillatory]
-0.17453292519909797
However, executing the following command someone get the idea
that there are almost 4000 sampling points!
Length[sampledPoints]
3894
I do not understand why now exists this big difference in the
sampling points.
Note that executing the next commands show that the two approaches (the
first with Reap...Sow, the second with AppendTo) give the same list of
sample points.
Block[{Message},Length[Reap[NIntegrate[f[x]*Sin[9x],{x,
0,Infinity},EvaluationMonitor\[RuleDelayed]Sow[x]]][[2,1]]]]
286
sampledPoints={};
Block[{Message},NIntegrate[f[x]*Sin[9x],{x,0,Infinity}]]
1.8915013450362048
Length[sampledPoints]
286
Note also that you cannot apply the Reap...Sow approach directly since
Reap[NIntegrate[f[x]Sin[9x],{x,
0,Infinity},Method\[Rule]Oscillatory,EvaluationMonitor\[RuleDelayed]Sow[x]]]
{-0.17453292519909797, {{x}}}
I really appreciate some guidance.
Thanks in advance for any assistance.
Dimitrios Anagnostou
NTUA
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