Power Series Expansion and Residue Calculation for log(1-z)

In summary, the person is asking for help with the power series expansion of the η(τ)/η(3τ) function. They are stuck on how to do it.
  • #1
squenshl
479
4

Homework Statement


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.


Homework Equations





The Attempt at a Solution


Is this power series the same as the case for real numbers.
 
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  • #2
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
When I manipulate do I use the power series for log(1-z)
 
  • #4
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
How do I do it by the residue theorem and L'Hopitals rule.
 
  • #6
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]
 
Last edited:
  • #7
squenshl said:
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?
 
  • #8
The second term in the series.
 
  • #9
I think I got it now. It is 1 beacuse this is the constant for the z-1 term (the term 1/z)
 
  • #10
Using the formula for the residue at a simple pole (Residue theorem) I also get 1 as my residue.
res0 = 1
 
  • #11
  • #12
True.
Also got 1 using L'Hopitals rule.
Didn't realize it was so easy.
Cheers.
 
  • #13

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.
 

1. What is a power series expansion?

A power series expansion is a mathematical technique used to represent a function as an infinite sum of powers of a variable. It is a useful tool for approximating functions and solving differential equations.

2. How is a power series expansion calculated?

A power series expansion is calculated by taking the derivative of a function and evaluating it at a specific point, then taking the second derivative and evaluating it at the same point, and so on. These values are then plugged into the general formula for a power series, which is a sum of terms involving the variable raised to different powers.

3. What is the purpose of a power series expansion?

A power series expansion is used to approximate functions that are difficult to evaluate directly. It can also be used to solve differential equations and find the behavior of a function at certain points or over a certain interval.

4. Can a power series expansion be used for any function?

No, a power series expansion can only be used for functions that are infinitely differentiable. This means that the function must have derivatives of all orders at any point in its domain.

5. How accurate is a power series expansion?

The accuracy of a power series expansion depends on the number of terms included in the series. The more terms that are included, the more accurate the approximation will be. However, power series expansions can only approximate functions within a certain radius of convergence, so their accuracy is limited in that sense.

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