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Topic: Integration with variable limits
Replies: 17   Last Post: Aug 20, 2013 10:20 AM

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Sanaa

Posts: 154
Registered: 3/20/12
Re: Integration with variable limits
Posted: Aug 15, 2013 8:04 AM
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"Steven_Lord" <slord@mathworks.com> wrote in message <ktb57u$el8$1@newscl01ah.mathworks.com>...
>
>
> "Torsten " <Torsten.Hennig@umsicht.fraunhofer.de> wrote in message
> news:ktad6k$gtn$1@newscl01ah.mathworks.com...

> > "Torsten" wrote in message <ktacg3$fkj$1@newscl01ah.mathworks.com>...
> >> "Sanaa" wrote in message <kt9bf4$56h$1@newscl01ah.mathworks.com>...
> >> > "Torsten" wrote in message <kst5mn$h75$1@newscl01ah.mathworks.com>...
> >> > > "Sanaa" wrote in message <kss3h5$50n$1@newscl01ah.mathworks.com>...
> >> > > > "Torsten" wrote in message
> >> > > > <ksrcel$2k$1@newscl01ah.mathworks.com>...

> >> > > > > "Sanaa" wrote in message
> >> > > > > <ksrbia$r0l$1@newscl01ah.mathworks.com>...

> >> > > > > > Hi all,
> >> > > > > > How to integrate a function within a loop and the limits of
> >> > > > > > integration are variables?
> >> > > > > > For instance, y(i)*(1-y(i)) is the function I wish to integrate
> >> > > > > > from 0.25*i to t where t \in (0.25*i, 0.25*(i+1)).
> >> > > > > > Any help is really appreciated.
> >> > > > > > Sanaa Moussa

> >> > > > >
> >> > > > > I think y(i)*(1-y(i)) is just a real number, not a function,
> >> > > > > isn't it ?
> >> > > > >
> >> > > > > Best wishes
> >> > > > > Torsten.

> >> > > >
> >> > > > Thanks for replying. You are right, but I mean the function is
> >> > > > y*(1-y) which I want to put it into a loop
> >> > > > My code is
> >> > > > itermax=300;min=itermax-9;% That is we plot from 291:300 i.e. 10
> >> > > > values of x only.
> >> > > > r=0.25;
> >> > > > for rho=0:0.001:4
> >> > > > %x0=0.1;
> >> > > > x(1)=0.1;
> >> > > > for i=1:itermax-1
> >> > > > %t=linspace(i*r,(i+1)*r,itermax)
> >> > > > y(i+1)=x(i);
> >> > > > x(i+1)=x(i)+ rho*int(y(i)*(1-y(i)),i*r,(i+1)*r);
> >> > > > end
> >> > > > %fix(y)
> >> > > > plot(rho*ones(10),x(min:itermax),'b.','linewidth',0.1)
> >> > > > hold on
> >> > > > end
> >> > > > fsize=15;
> >> > > > xlabel('\rho','FontSize',fsize)
> >> > > > ylabel('\itx','FontSize',fsize)
> >> > > > %title('r=0.25, \alpha=1','FontSize',fsize)
> >> > > > hold off
> >> > > > % print(gcf, '-djpeg', '-zbuffer', 'bif.png');
> >> > > >
> >> > > > I get the error
> >> > > > Function 'int' is not defined for values of class 'double'.
> >> > > > what does this mean and how to fix it please?
> >> > > > Many thanks in advance.

> >> > >
> >> > > It means what I said before:
> >> > > y(i)*(1-y(i)) is a scalar value and not a function.
> >> > > So "int" does not know how to handle this because it expects a
> >> > > function, not a scalar as its first input argument.
> >> > >
> >> > > Best wishes
> >> > > Torsten.

> >> >
> >> > Sorry for not posing my question correctly. The function I want to
> >> > integrate is
> >> > y(s)*(1-y(s)) ds from (n*0.25) to t, and t/in(n*0.25, (n+1)*0.25).
> >> > Do you have any idea on how to format my code above to solve my
> >> > problem?
> >> > Thanks a lot in advance

> >>
> >> int_{n*0.25}^{t} (y(s)*(1-y(s)) ds = (1/2*t^2 - 1/3*t^3) -
> >> (1/2*(n*0.25)^2 - 1/3*(n*0.25)^3)
> >> for t/in(n*0.25, (n+1)*0.25).
> >> Is it that what you asked for ?
> >>
> >> Best wishes
> >> Torsten.

> >
> > No sorry, this is wrong - I integrated s*(1-s).
> > For integration, you will have to know the explicit form of y as a
> > function of s,
> > e.g. y(s)=sin(s) or something like that.

>
> If Y is only known as data, you will need to do one of two things:
>
> 1) Use TRAPZ or CUMTRAPZ.
> 2) Interpolate the Y data to obtain the value of the "function" at
> intermediate points. This can be dangerous if your data does not
> sufficiently represent your actual function:
>
> x = 0:10
> y = sin(x*pi).^2
> answer1 = integral(@(t) interp1(x, y, t), 0, 10)
> syms z
> answer2 = int(sin(z*pi)^2, 0, 10)
>
> Your numeric data makes your function "look like" the constant function y =
> 0 and so answer1 is very close to 0. The symbolic integration gives an
> answer2 of 5, which is correct. Evaluating y at more points (x = 0:0.1:10
> for example) would show the true form of the curve and gives an answer very
> close to 5.
>
> --
> Steve Lord
> slord@mathworks.com
> To contact Technical Support use the Contact Us link on
> http://www.mathworks.com


Thank you very much for your discussion and sorry for getting back to you very late.
Actually, I don't know how to interpolate the data that's why I intended to use cumptrapz. However, there is no posibility to put the bounds of my integration in that function!!! I want to integrate from i*0.25 to t, and t from i*0.25 to (i+1)*0.25!!



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