Power series expansion

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.

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

squenshl

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

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?

squenshl

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

Susanne217

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

Recall Sir,

[tex]log(1+z) = \sum_{j=1}^\infty} \frac{(-1)^{j+1}}{j}z^{j} = z - \frac{z^2}{2} + \frac{z^3}{3}-\cdots[/tex]

so by very very simply replacing z with -z

you get [tex]-z - \frac{(-z)^2}{2} + \frac{(-z)^3}{3}-\cdots[/tex]

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

Is [tex]P_{n} = -\sum_{j=1}^{n} \frac{z^n}{n}[/tex]

vela

Quote by 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?

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

squenshl

The second term in the series.

squenshl

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

squenshl

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

vela

Yup, that's right.

When you say "residue theorem," you're not referring to the Cauchy residue theorem, right? Because that's what most people mean by that term.

http://mathworld.wolfram.com/ResidueTheorem.html

squenshl

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

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.

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squenshl	Power series expansion Tweet 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.

squenshl	Power series expansion 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?

Susanne217 Blog Entries: 2	Make me think of the power series of log(1+z) Recall Sir, [tex]log(1+z) = \sum_{j=1}^\infty} \frac{(-1)^{j+1}}{j}z^{j} = z - \frac{z^2}{2} + \frac{z^3}{3}-\cdots[/tex] so by very very simply replacing z with -z you get [tex]-z - \frac{(-z)^2}{2} + \frac{(-z)^3}{3}-\cdots[/tex] So the power series expansion of log(1-z) Is [tex]P_{n} = -\sum_{j=1}^{n} \frac{z^n}{n}[/tex]

vela Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus	Yup, that's right. When you say "residue theorem," you're not referring to the Cauchy residue theorem, right? Because that's what most people mean by that term. http://mathworld.wolfram.com/ResidueTheorem.html

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.