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Let $f$ be an elliptic function of order $m$. Show that its derivative is an elliptic function of order $n$ with $m+1\leq n\leq 2m$.

It's somewhat intrinsically clear to me that the derivative needs to be an elliptic function as well, but how would I actually go about showing that?

As for the bounds on the order:

The lower bound is easy to "prove" since I know that the Weierstrass elliptic function has order 2 and its derivative (which is elliptic as well) has order 3. To get the upper bound I imagine we have to work with the fact that taking the derivative of function ups the order of a Pole by 1.

Assuming we count the different poles of $f$ without counting multiplicity modulo the lattice used to define $f$ and calling that number $j$ we should have that $n=m+j$ which, if all poles are different, works out to $2m$. With the above example on the other hand it works out to $m+1$ since $\displaystyle {\wp }$ has one such pole.

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The order of an elliptic function is the sum of the orders of the poles in a fundamental region. Differentiating adds one to the order of each pole. So the order of the elliptic function increases by the number of distinct poles in a fundamental region. An elliptic function of order $m$ has between $1$ and $m$ distinct poles, so the derivative has order between $m+1$ and $2m$.

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  • $\begingroup$ Sorry, that is what I meant with "different" poles. English isn't my native language. $\endgroup$
    – jfsjio
    Commented Dec 16, 2017 at 21:37
  • $\begingroup$ Any idea on the derivative part? $\endgroup$
    – jfsjio
    Commented Dec 17, 2017 at 0:45
  • $\begingroup$ If $f(z+a)=f(z)$ for all $z$, then $f'(z+a)=f'(z)$ for all $z$. @jfsjio $\endgroup$ Commented Dec 17, 2017 at 11:06

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