# Subtraction of Power Series

1. Mar 3, 2012

### Poopsilon

Say we have two power series $\sum_{n=0}^{\infty}a_n z^n$ and $\sum_{n=0}^{\infty}b_n z^n$ which both converge in the open unit disk. Is there anything we can say about the radius of convergence of the power series formed by their difference? i.e. $\sum_{n=0}^{\infty}(a_n-b_n) z^n$

What about if we know that their difference is bounded on compact sets? i.e. $|\sum_{n=0}^{\infty}(a_n-b_n) z^n| < M$ for all z in a compact subset of the open unit disk.

Last edited: Mar 3, 2012
2. Mar 3, 2012

### Office_Shredder

Staff Emeritus
Given that you're so focused on subtraction as opposed to addition, does nothing that
$$\sum a_n z^n - \sum b_n z^n = \sum a_n z^n + \sum (-b_n) z^n$$

3. Mar 3, 2012

### Poopsilon

Um, they seem to be equivalent so I can't see any reason why what you've proposed wouldn't be ok.

4. Mar 3, 2012

### Office_Shredder

Staff Emeritus
OK I thought maybe you were confused about how subtraction works as opposed to addition.

If we're adding two power series, as long as both power series converge their sum converges (you should be able to prove this using that limits split into sums). So as long as both of your power series converge, the new one will - hence the radius of convergence is AT LEAST the minimum of the radius of convergence of the orginal two series

For the second question, are you saying it converges on every compact subset of the disk? If so then clearly the radius of convergence is at least 1

5. Mar 3, 2012

### Poopsilon

Yes every compact subset of the disk, although I think without the assumption of boundedness what you've given me should be sufficient for my purposes, thanks =].

Actually I'm curious though, does the boundedness assumption give us anything more in this case? It doesn't seem like it would since if we already know it's convergent on the disk then it must be bounded there as well.

Oh shoot, also while I have your attention, does a power series converge absolutely uniformly on compact sets of its radius of convergence? I'm guessing it does since absolute convergence would create another power series with the same radius of convergence, meaning it converges uniformly on compact sets of that disk.

Last edited: Mar 3, 2012
6. Mar 3, 2012

### Office_Shredder

Staff Emeritus
No, if we know it's convergent on the OPEN disk we don't know it's bounded there. For example $$\sum z^k = 1/(1-z)$$ is a power series which converges on the open disk but is not bounded (for example of z is a positive real number it goes to infinity as z goes to 1). Power series are continuous, so we know on each compact set it's going to be bounded, but if it's bounded by the same number on all the compact subsets of the unit disk then I suspect you can prove that it converges on the boundary of the disk as well (proof not forthcoming). Wikipedia has an example of a power series which is bounded on the unit disk but has radius of convergence 1