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Re: Piecewise integration of f[x,y]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg15817] Re: [mg15773] Piecewise integration of f[x,y]
*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>
*Date*: Mon, 8 Feb 1999 03:25:43 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
First I must make a disclaimer. I am an algebraic topologist and have
never had any serious dealings with numerical computations. The only
time I have to deal with integrals is when I teach calculus courses to
undergraduates. So as a rule I avoid answering questions such as this
one, but this time I shall risk it. However, you have been warned to
apply a dose of scepticism to what I say.
This is the second question concerning integrals of piecewise functions
that has come up recently, but this time it is a double integral, which
makes it more complicated. In general it seems to me that NIntegrate
works in such cases, but sometimes has problems with knowing if its
answer is accurate enough. This seems to be such kind of case:
In[62]:=
f[x_, y_] := Which[x >= 0. && x <= 2. && y >= 0. && y <= 1.,
-0.5*(-2. + x)*(0. + y), True, 0];
In[63]:=
NIntegrate[f[x,y],{x,-3,3},{y,0.2,4}] NIntegrate::slwcon:
Numerical integration converging too slowly; suspect
singularity, value of the integration is 0, oscillatory
integrand, or insufficient WorkingPrecision. If your
integrand is oscillatory try using the option
Method->Oscillatory in NIntegrate. .....
General::stop:
Further output of NIntegrate::slwcon
will be suppressed during this calculation. NIntegrate::ncvb:
NIntegrate failed to converge to prescribed accuracy after
13 recursive bisections in y near {x, y} =
{0.75, 0.999939}.
Out[63]=
0.479956
I do not really feel competent to try to explain this (see the
disclaimer above) and I hope someone else will do so. But I would like
to make an observation about another possible approach, by using
Integrate.
Integrate in general does not like functions defined by means of
conditionals (and will thus fail if used in the above example in place
of NIntegrate) with one somewhat remarkable exception. It works with
functions defined using a single If. Thus, if we define functions
In[64]:=
p[x_]:=If[0<x<2 ,1, 0];
q[y_]:=If[0<y<1,1,0];
we can integrate them without problems: In[65]:=
Integrate[p[x],{x,-3,3}]
Out[65]=
2
Not only that, but we can also do things like this
In[66]:=
h[x_]:=x^3;Integrate[p[x]*h[x],{x,-Infinity,Infinity}] Out[66]=
4
This suggest that we should be able to evaluate the above double
integral by using p and q above and Fubini's theorem (which says that
you can first integrate with respect to one variable and then the
result with respect to the other). In other words we define
In[67]:=
h[x_,y_]:= -0.5*(-2. + x)*(0. + y)
and then compute
In[69]:=
Integrate[Integrate[p[x]*q[y]*h[x,y],{x,-3,3}],{y,0.2,4}] Out[69]=
0.48
This seems to be the right answer.
On Sun, Feb 7, 1999, <sergio at scisun.sci.ccny.cuny.edu> wrote:
>(*
> Hello everyone:
>
> Could somebody point out how to integrate a piecewise
> function like:
>*)
>
>f[x_, y_] := Which[x >= 0. && x <= 2. && y >= 0. && y <= 1.,
> -0.5*(-2. + x)*(0. + y), True, 0];
>
>(*
> How to do:
>*)
> x1=-3; x2=3; y1=0.2; y2=4;
> Integrate[f[x,y],{x,x1,x2},{y,y1,y2}];
>
>(*
> Sergio
> E-mail: sergio at scisun.sci.ccny.cuny.edu *)
>
>-----------== Posted via Deja News, The Discussion Network ==----------
>http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
>
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp/
http://eri2.tuins.ac.jp/
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