# Homework Help: Power series expansion

1. May 30, 2010

### squenshl

1. The problem statement, all variables and given/known data
Find a power series expansion for log(1-z) about z = 0. Find the residue at 0 of 1/-log(1-z) by manipulation of series, residue theorem and L'Hopitals rule.

2. Relevant equations

3. The attempt at a solution
Is this power series the same as the case for real numbers.

2. May 30, 2010

### squenshl

I have the power series expansion about z = 0 for log(1-z).
-z - z2/2 - z3/3 - ...
But how do I find the residues with the methods mentioned

3. May 30, 2010

### squenshl

When I manipulate do I use the power series for log(1-z)

4. May 31, 2010

### squenshl

My series for -1/log(1-z) is:
1/z - 1/2 - z/12 - z2/24 - ...
So my residue is a-1 = -1/2.
Is that right?

5. May 31, 2010

### squenshl

How do I do it by the residue theorem and L'Hopitals rule.

6. Jun 1, 2010

### Susanne217

Make me think of the power series of log(1+z)

Recall Sir,

$$log(1+z) = \sum_{j=1}^\infty} \frac{(-1)^{j+1}}{j}z^{j} = z - \frac{z^2}{2} + \frac{z^3}{3}-\cdots$$

so by very very simply replacing z with -z

you get $$-z - \frac{(-z)^2}{2} + \frac{(-z)^3}{3}-\cdots$$

So the power series expansion of log(1-z)

Is $$P_{n} = -\sum_{j=1}^{n} \frac{z^n}{n}$$

Last edited: Jun 1, 2010
7. Jun 1, 2010

### vela

Staff Emeritus
No. Where'd you get -1/2 from?

8. Jun 1, 2010

### squenshl

The second term in the series.

9. Jun 1, 2010

### squenshl

I think I got it now. It is 1 beacuse this is the constant for the z-1 term (the term 1/z)

10. Jun 1, 2010

### squenshl

Using the formula for the residue at a simple pole (Residue theorem) I also get 1 as my residue.
res0 = 1

11. Jun 1, 2010

### vela

Staff Emeritus
12. Jun 1, 2010

### squenshl

True.
Also got 1 using L'Hopitals rule.
Didn't realise it was so easy.
Cheers.

13. Oct 25, 2011

### twoforone

Hi Every body!

I wan to compute the power series expansion of dedekind eta function. Specifically, I want to know the power series expansion of η(τ)/η(3τ)? How could I expand this function? I would be happy if you could help me as I am stuck at this state when I am computing the modular polynomial of prime number 3.