I do not know how to determine if is a regular or irregular singular point for . How do I determine if and are analytic?
For , we wish to know if has a pole of order greater than one or has a pole of order greater than two. If so, then the singular point is irregular. We can find out by simply taking limits that would cancel a simple pole in the former case or a double pole in the later case. So in the case above with the singular point at x=0, we evaluate the following limits:
If both limits exists, then cannot have a pole of order greater than one, and cannot have a pole greater than two. So just take the limits and see what happens.
You have:
and is analytic at zero. It has a removable singularity there. The term is also analytic at zero. Your DE has a regular singular point at the origin. We can see a dramatic difference between ordinary and irregular singular points by considering the two DEs:
The first has an irregular singular point and the second has a regular one. I've plotted the real components of the complex solutions of both close to the origin. Note the qualitative differences: that folding pattern in the first continues infinitely often in any neighborhood of the origin and is the classic morphology of an essential singularity. However, I'm not sure if all DEs with irregular singular points give rise to solutions with essential singularities. I suspect so though.
Well, I would recommend using the definition of "analytic" which is that a (real) function, f, is analytic at x= a if and only if the Taylor's series for f(x) about x= a exists and converges to f(x) in some neighborhod of a. You should know that both and cos(x) have Taylor's series that converge to them for all x and so their product does also.
As for , it is clearly analytic everywhere except, possibly, at x= 0. Now, strictly speaking it is not analytic at x= 0 because it is is not defined there.
But what you really mean is the function that is for x not equal to 0 and is equal to 0 when x= 0. The function has Taylor's series, around x= 0, . The function , then, has Taylor series, around x= 0, . Finally, then, .