- #1
opticaltempest
- 135
- 0
I uploaded a scanned page from Schaum's Outline of Complex Variables. I have some questions on how they found the Laurent series in Example 27.
http://img19.imageshack.us/img19/7172/i0001.jpg
If the image doesn't load, go to - http://img19.imageshack.us/img19/7172/i0001.jpg"
In Example 27 part (b), they used the Laurent series they obtained from part (a) and subtracted this series from the Laurent series they obtained for |z|>3 to get a Laurent series that is valid for |z|>3.
It appears that they are making substitutions into the geometric series to find the power series for 1/(2*(z+1)) (that is convergent?) for |z|>1 and the power series for 1/(2*(z+1)) (that is convergent?) for |z|>3. However, I thought the geometric series did not converge for |z|>1. How can we do these substitutions and say that we have a convergent series for |z|>1 and |z|>3? In other words, how do we know that the series for |z|>1 and |z|>3 actually converge so that we can subtract them?
http://img19.imageshack.us/img19/7172/i0001.jpg
If the image doesn't load, go to - http://img19.imageshack.us/img19/7172/i0001.jpg"
In Example 27 part (b), they used the Laurent series they obtained from part (a) and subtracted this series from the Laurent series they obtained for |z|>3 to get a Laurent series that is valid for |z|>3.
It appears that they are making substitutions into the geometric series to find the power series for 1/(2*(z+1)) (that is convergent?) for |z|>1 and the power series for 1/(2*(z+1)) (that is convergent?) for |z|>3. However, I thought the geometric series did not converge for |z|>1. How can we do these substitutions and say that we have a convergent series for |z|>1 and |z|>3? In other words, how do we know that the series for |z|>1 and |z|>3 actually converge so that we can subtract them?
Last edited by a moderator: