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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: 162
Registered: 3/20/12
Re: Integration with variable limits
Posted: Aug 15, 2013 9:46 AM
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"Torsten" wrote in message <kuih1o$c4g$1@newscl01ah.mathworks.com>...
> "Sanaa" wrote in message <kuig3n$9d$1@newscl01ah.mathworks.com>...
> > "Steven_Lord" <slord@mathworks.com> wrote in message <ktb57u$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!!

>
> Just for clarification:
> How is your function given you want to integrate ?
> 1. By data points (x1,y1),(x2,y2),...,(xn,yn) you already calculated beforehand,
> 2. By a functional equation, e.g. f(x)=x^2,
> 3. By a differential equation dy/dx=y(x)*(1-y(x)),
> 4. ??
>
> Best wishes
> Torsten.

Tanks a lot for replying. before I make a discretization process, I have the system:
y(t) = x(t-r),
dx/dt = f(y(t)), where f = rho*y(1-y).
After discretization and integration I have the system
y_(n+1)=x_n;
x_(n+1)= x_n + integration (f(y_(n+1)(s))) ds.



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