Elliptic functions: Weierstrass psi function limit

[binbagsss]

Homework Statement

Show that
##\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2})) =1##

where ##\psi(z)=\frac{1}{z^2}+\sum\limits_{w \in \Omega}' \frac{1}{(z-w)^2}-\frac{1}{w^2}##

where ##\Omega## are the periods of ##\psi(z)##

Homework Equations

The Attempt at a Solution

##\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2}))= 1 + \sum\limits_{w\in \Omega}'\frac{z^2}{(z-w)^2}-\frac{z^2}{w^2}##

The last time clearly vanishes.

For the second term I can write this as ##\frac{1}{(1-\frac{w}{z})^2} \to 1 ## as ##z \to 0 ##

So I get

##\lim_{z \to 0} z^2( \psi(z)-\psi(\frac{w_j}{2})) =1+ \sum_{w \in \Omega}' 1##

This is probably a stupid question but I don't really understand the summation second term here.

##\sum_{n=1}^{n=n} 1 = n## right?

So there's an infintie number of ## w \in \Omega ## so obviously I don't want to do this, are you in affect looking at the limit '##mod \Omega ##', so where the first term corresponds to ##w=0 ## ?

Many thanks

 

[mighty2000]

in advance,
Dear [Forum User],

Thank you for your question. Your solution looks correct to me. The summation in the second term does indeed represent an infinite number of terms, which can appear daunting. However, in this particular case, the terms are all equal to 1, and therefore the summation can be simplified to ##\infty \cdot 1 = \infty##. In this context, the summation is used to represent the infinite number of periods in the function ##\psi(z)##. As you correctly pointed out, the first term corresponds to ##w=0##, and the remaining terms correspond to all the other periods in the function.

I hope this clarifies things for you. Let me know if you have any further questions.