Seek the solutions that are power series to the differential equation:

I assume we first need to assume that so that can be differentiated.

I have that

Printable View

- March 26th 2011, 02:08 AMMondreusPower series - Differential equation
Seek the solutions that are power series to the differential equation:

I assume we first need to assume that so that can be differentiated.

I have that - March 26th 2011, 02:45 AMProve It
Do you really need to use power series?

.

Go from here... - March 26th 2011, 02:55 AMMondreus
Yes, the question is in the chapter on power series and it explicitly says that the answer should be a power series. The book hasn't covered complex Taylor and Mclaurin series yet, so even if I know that the answer is some branch of -2log(1-z) I can't use that to derive the power series representation of it.

- March 26th 2011, 04:47 AMHallsofIvy
I have that [/QUOTE]

Yes, that derivative, and the change in index, is correct. Now put them into your differential equation:

Now you have to change the index in that second sum. let i= n+1. Then

or, changing the dummy index back to n, [tex]\sum_{n= 1}^\infty n c_n z^n[/itex].

Now, we can write the differential equation as

Since that second sum does not begin until n= 1, we need to do n= 0 separately. If n= 0 then

For n> 0, equating coefficients of the same powers of z,

or

That is, [tex]c_1= 2c_0, , , , etc. - March 26th 2011, 05:43 AMMondreus
You made some small errors, particulary in the last part. It should be: and finally