Laurent series

Yes I think it is correct.

But the series is powers of $(1/z)$. The symbol is not epsilon. It's called sigma ($\Sigma$)

It does agree with the book. Why do you think it doesn't?

Vasco

I have a few more questions, including the exercise at the end of this subsection 1:
1. Starting from something basic, for the series $\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$ looking at the left hand side sum, it is easy to see that it only converges to some value when $|z| < 1$ but can you see it using the right hand formula as well? I know that there are various methods of calculating radius of convergence i.e a ratio test which here returns $\frac{a_{n+1}}{a_{n}} = \frac{1}{1} = 1$ and we manually test endpoints to discover $-1 < R < 1$
2. For $\sum_{n=0}^{\infty} (\frac{z}{2})^n$ we say it is valid for $|z| < 2$ why? UPDATE. Ok it makes sense even intuitively, we can now have two times bigger $z$ as we divide by two. As for the ratio test it's value shall be now divided by $2$ and hence we have $ L = \frac{1}{2}$ so $R = \frac{1}{L} = 2$
3. For the exercise for $|z| > 2$ how was the new expansion obtained? We know that for $|z| > 2$ there are no more singularities but in the same time both series are divergent. So two questions,
a. Why do we care about a Laurent series anymore if we know that there are no more singularities? Shouldn't Taylor series be enough here?
b. Both series are divergent at $|z| > 2$

UPDATE:

for 3.a, I think we still want Laurent series to also cover the singularities before but strictly speaking, if we only care about arguments $|z| > 2$ we don't need a Laurent series anymore right?
b. Surprisingly, the two examples that are already in the book show that by slightly modifying these partial fraction we can get wide range of convergence. This observation allowed me to tackle this exercise. 

Here is my attempt. We need $|z| > 2$ and for the first partial fraction we already have it, namely $- \sum_{n=0}^{\infty} \frac{1}{z}^{n+1}$ it converges for $|z| > 1$ so will also do for $|z| > 2$. Next we nee to adjust $\frac{1}{(2-z)}$ so it converges in the same circle. It is not hard to come up with $\frac{1}{z} \sum_{n=0}^{\infty}(1-\frac{2}{z})$ 
Next expanding just first few terms shows: first term cancels out for both sums. The second term, one gives $\frac{1}{z^2}$ while the other $\frac{2}{z^2}$ which sums up to $-\frac{1}{z^2}$. In the same way next terms sum up to $-\frac{3}{z^3}$. Which complies with what's in the book (ignoring the first term $\frac{1}{z}$ which is already marked as an error and removed in a newer editions of the book)

Thank you.

Mondo

1. You can't say we can see that the series converges. How can you know? You obviously haven't read the last few paragraphs on page 433. Forget all those ratio tests we don't need them and they don't give us any insight as to what is happening.

2.Same as 1.

3. You need to distinguish between Taylor series and Laurent series. The two Taylor series are divergent but what about the Laurent series. Again this shows that you haven't read this chapter and understood it.

3a We want to find a series which converges for $|z|>2$ that's why we care about a Laurent series. Taylor series are divergent here!!!!

3b Both Taylor series are divergent, but what about the Laurent series?

Comments regarding your UPDATE

3a No not right. We do need a Laurent series here as The two Taylor series are divergent here!!

b What you have written about modifying the partial fractions to get convergence is totally wrong. Those manipulations are to make it easier to write the series as a summation that's all.

As a result your attempt at 3 cannot be right. Also the $1/z$ is still in the latest printing.

Vasco

Mondo


Mondo

I also agree with this, quoted from your reply #3 above. Apologies again.

Vasco

Vasco,
I got caught up by work, will respond as soon as I can.
Thank you!

Vasco,

but $P(w)=a_0+a_1w+a_2w^2+...$ where $w=1/(z-a)$ means we have a radius of convergence equal the distance to the nearest singularity, which in this case is $w=1$. Hence we can do $|w| < 1 => |\frac{1}{(z-a)}| < 1 = z > 1 + a$ But this is not quite what we wanted.

Vasco,
what is the last statement "So the power series Q.." based on? How is that true?