Laurent series expansion of ℘(z).

1. May 20, 2017

binbagsss

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Hi,

I am trying to understand the 2nd equality .

I thought perhaps it is an expansion of $(1-\frac{z}{w})^{-2}$ (and then the $1$ cancels with the $1$ in $( (1-\frac{z}{w})^{-2}) -1 )$) in the form $(1-x)^{-2}$, however this doesn't give me the right answer, can someone please help explain how they got from line $1$ to $2$ ?

Many thanks in advance.

2. May 20, 2017

Ray Vickson

I get the right answer when I expand $(1 - (z/w))^{-2}$ in a Maclauren series about $z = 0$, or use the binomial expansion involving the "negative binomial" $C^{-2}_n$. It sounds like that is what you did too, but got the wrong answer for reasons you do not explain, using work that you do not show.

3. May 20, 2017

binbagsss

oh okay, thank you for your reply.
So for $(1+x)^{-2}$ where $x= \frac{z}{w}$ is of the form $1+nx+\frac{n(n+1)x^2}{1.2}+\frac{n(n+1)(n+2)x^3}{1.2.3}+...$I don't know how to express this as a summation over $n$ unless it simplifies somehow for the case here $n=2$

you know me well...

4. May 21, 2017

Buffu

What does called ?

5. May 21, 2017

Ray Vickson

Your expansion is wrong. It should be
$$(1-t)^{-2} = 1 + 2 t + 3t^2 + 4 t^3 + \cdots = \sum_{n=0}^{\infty} (n+1)t^n.$$
Note that $(1-t)^{-2} = \sum_n (-1)^n C(-2,n) t^n$, and $C(-2,n) = (-1)^n (n+1).$

Of course, you can also get it by looking at
$$(1-t)^{-2} = \frac{d}{dt} (1-t)^{-1} = \frac{d}{dt} (1 + t + t^2 + \cdots)$$

Last edited: May 21, 2017
6. May 21, 2017

binbagsss

Apologies it should have read $(1-x)^{-2}$ , wrong sign, but plugging in $n=2$ gives cancellations to give the same as your expansion there

7. Jun 22, 2017

binbagsss

Can I just ask, for general $n$ there is no simplified general expression as a summation ?

For example considering $n=3$ the most I can simplify this to is:

$1+3x+\frac{(3)(4)x^2}{(2)(1)}+\frac{(3)(4)(5)x^3}{(3)(2)(1)}+\frac{(3)(4)(5)(6)x^4}{(4)(3)(2)(1)}+\frac{(3)(4)(5)(6)(7)x^5}{(5)(4)(3)(2)(1)}+...$
$=1+3x+6x^2+\frac{(4)(5)x^3}{(2)(1)}+\frac{(5)(6)x^4}{(2)(1)}+\frac{(6)(7)x^5}{(2)(1)}$
$=1+3x+6x^2+\sum\limits_{k=3}^{k=\infty} \frac{(k+1)(k+1)x^k}{2}$

so for a general expression the $x^{n-1}$ terms will not be included in some general summation? (for $n$ integer)

Is there a more simplified way to write this or not?

Many thanks.

Last edited: Jun 22, 2017
8. Jun 22, 2017