# Laurent series expansion of ℘(z).

## The Attempt at a Solution

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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 $2$ ?

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Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

View attachment 203874

## The Attempt at a Solution

[/B]
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 $2$ ?

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.

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,
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 $n=2$

using work that you do not show.
you know me well...

What does called ?

Ray Vickson
Homework Helper
Dearly Missed
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...
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
$$(1-t)^{-2} = \frac{d}{dt} (1-t)^{-1} = \frac{d}{dt} (1 + t + t^2 + \cdots)$$

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

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 $n=2$
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.

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Dick