# Frobenius Method

1. Aug 14, 2008

### bishy

I'm a little confused with ODEs. After two weeks of trying to figure out Frobenius I have finally realized that there seems to be two different power set used by all of my three books for the y substitution but I am unsure when to use either one. Here are the two sets that i'm talking about:

Set 1 from section 6.1

$$y = \sum_{n=0}^\infty C_n x^n$$

$$y\prime = \sum_{n=1}^\infty n C_n x^{n-1}$$

$$y\prime\prime = \sum_{n=2}^\infty n(n-1) C_n x^{n-2}$$

set 2 from section 6.2

$$y = \sum_{n=0}^\infty n C_n x^{n-1}$$

$$y\prime = \sum_{n=0}^\infty n C_n x^{n-1}$$

$$y\prime\prime = \sum_{n=0}^\infty n(n-1) C_n x^{n-2}$$

I think I have come to understand that I should use set 1 if and only if all of the singular points are irregular and set 2 when I have at least one regular singular point. Is this correct? If not, when is it appropriate to use set 1 rather than set 2? Or is set 2 only designed to work with Frobenius' method while set 1 only works lacking a Taylor power series expansion?

2. Aug 14, 2008

### NoMoreExams

In your 2nd set, why are you starting the series from 0? for y' you would get 0 for n = 0 and for y'' you would get 0 for n = 0 and n = 1. It just seems like they are shifting indices around.

3. Aug 14, 2008

### bishy

That's what I thought as well when I first realized that little difference within the DE books I am basing the sets on. Considering that they we're using this method to get something backed by primary source papers I stopped trying to pass it off as a typo within my books. Apparently with the solutions manual I have available to me, both sets are being used, when and why this is the case, I really have no idea. Hopefully somebody does.

4. Aug 15, 2008

### HallsofIvy

Neither of those is correct for Frobenius' method.

If you have a regular singular point then Frobenius' method uses
$$\sum_{n=0}^\infty C_n x^{n+ c}$$
where c is determined from the indicial equation, essentially requiring that C0 not be 0. c in not necessarily positive or even an integer.

If you have an irregular singular point, there may not be a series solution at all.

5. Aug 16, 2008

### matematikawan

I agree with HallsofIvy.

If x=0 is an ordinary point then either set 1 or set 2 is a valid expansion. However when using set 2 we must assume that Cn=0 if n is a negative integer (in case we want to replace the dummy index n).