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Numerical Methods difficulties

Hello everyone :smile:

I'm basically stuck with a report I have to write for a resit unit; my main issue is that I don't quite understand how to do part of it, so here begins one of my questions:

I'm using matlab to implement my own numerical integration method, and I've chosen to use the composite simpsons rule. I've got a list of criteria for my function to complete, and one of these criteria is that my function should also be able to deal with the case that the approximation does not converge (for example, if the integral is unbounded).

Now, I understand what an unbounded integral is, and I understand what it means when an integral doesn't converge, but my main question is how to get a numerical approximation to an integral that doesn't converge? I've done some research, but most of what I have found seems to be too confusing for me to understand.

My second question:

How could I get this integration method to deal with non-finite end points?

(edited 9 years ago)

Reply 1

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Reply 2

atsruser

Original post by BENOO

Hello everyone :smile:

I've got a list of criteria for my function to complete, and one of these criteria is that my function should also be able to deal with the case that the approximation does not converge (for example, if the integral is unbounded).

I guess this just means that your function should be able to identify such a case and report it; you clearly can't produce an approximation to an unbounded integral.

My second question:

How could I get this integration method to deal with non-finite end points?

You mean for something like

\int_1^\infty \frac{1}{x^2} dx

Reply 3

BENOO

Original post by atsruser

I guess this just means that your function should be able to identify such a case and report it; you clearly can't produce an approximation to an unbounded integral.

I guess you're right, I've emailed the guy who set it but I'm not quite sure what the answer is. I can do that though, even if it's not right, and change it dependant on the reply :smile:

Thanks for your help, helpful to know I'm not going crazy worrying about something that I can't do only to not have to do it!

Original post by atsruser

You mean for something like

\int_1^\infty \frac{1}{x^2} dx

And yeah, just like that; really don't understand how I could get a numerical approximation to that.

Reply 4

around

Original post by BENOO

And yeah, just like that; really don't understand how I could get a numerical approximation to that.

You can just integrate between 1 and x for larger and larger x and observe that it converges to some value.

Reply 5

BENOO

Original post by around

You can just integrate between 1 and x for larger and larger x and observe that it converges to some value.

I suppose that's the only way to do it really :P

As for my first question also, am I right in thinking that if the approximated error increases rather than decreases, then the iteration does not converge?

Reply 6

atsruser

Original post by BENOO

I suppose that's the only way to do it really :P

It's one way, but I suspect numerical analysts will have dreamed up others.

As for my first question also, am I right in thinking that if the approximated error increases rather than decreases, then the iteration does not converge?

I don't think so. Consider

\int_1^\infty \frac{\sin x}{x^2} dx

. This converges, but if you truncate the domain of integration at (in)appropriate points, then the values given by Simpson's rule may not be monotonically decreasing, giving the impression that the integral diverges.

Reply 7

BENOO

Original post by atsruser

It's one way, but I suspect numerical analysts will have dreamed up others.

I don't think so. Consider

\int_1^\infty \frac{\sin x}{x^2} dx

So what do you think is the best way to go about stopping my function if it doesn't converge? Is a limit on the number of iterations the best way, or can you think of some other way? :/