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This is probably a very standard question in complex analysis, but it doesn't seem to have been asked here yet.

If we have two $\mathbb{R}$-linearly independent non-zero complex numbers $\omega_{1}$ and $\omega_{2}$, and we let $\Omega = \{m\omega_{1} + n\omega_{2}: m, n \in \mathbb{Z}\}$ and $\Omega^{*} = \Omega - \{0\}$, then my definition of the Weierstrass $\wp$-function is $$\wp (z) = \frac{1}{z^2} + \sum_{\omega \in \Omega^{*}} \bigg( \frac{1}{(z-\omega )^2} - \frac{1}{\omega ^2}\bigg).$$

Working from this page: http://mathworld.wolfram.com/WeierstrassEllipticFunction.html, we let $f(z) = \wp (z) - \frac{1}{z^2}$, so that $f(0) = 0$ and $f$ is an even function. Thus all of its odd-order derivatives are $0$ at $z=0$. Expanding $f$ as a Maclaurin series, bearing in mind that all the odd terms are $0$, we get $$f(z) = \frac{z^2}{2!}f''(0) + \frac{z^4}{4!}f^{(4)}(0) + \cdots.$$

That's perfectly OK, but the MathWorld page then goes on to say that we can calculate the derivatives of $f$ from this, and then from there we can get derivatives of $\wp$. This is difficult to work out.

Also, it's easy to show that $\wp ' = \sum_{\omega \in \Omega}\frac{-2}{(z-\omega )^3}$, and getting explicit expressions for higher derivatives of $\wp$ is also easy. However, getting them in terms of just $\wp$ and $\wp '$ isn't so easy, so my question is, are we on the right track to get these by finding $f$ and its Maclaurin series as above? In the post High-order antiderivatives of the Weierstrass P-function, it says that $\wp '' = 6(\wp)^2 - \frac{1}{2}g_{2}$, which might be what I'm after, but what is $g_2$?

Basically, I'm looking for the simplest method of getting $\wp ^{(n)}$ in terms of $\wp$ and $\wp '$. Thanks for any help.

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  • $\begingroup$ $g_2$ and $g_3$ are defined in terms of powers sums of the elements of $\Omega^*$, see equations (24) and (25) of the Wolfram page you are linking. $\endgroup$ – Jack D'Aurizio Feb 10 '18 at 15:40
  • $\begingroup$ That's not such a problem -- more confusing is how they get from equation (16) to equations (17) - (20). Any tips there? EDIT: Especially because those equations (17) - (20) are written as derivatives of $f$, but look like they're derivatives of $\wp$. $\endgroup$ – SPS Feb 10 '18 at 16:58
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Recall that the Weierstraß $\wp$ function satisfies the differential equation $$ (\wp')^2 = 4 \wp^3 - g_2 \wp - g_3$$ where $g_2, g_3$ are the so-called Weierstraß invariants of the corresponding lattice $\Omega$ with $$ g_2 := 60 \sum \limits_{\omega \in \Omega^*} \frac{1}{\omega^4}$$ and $$ g_3 := 140 \sum \limits_{\omega \in \Omega^*} \frac{1}{\omega^6}$$ So if you differentiate the above-mentioned differential equation, you arive at $2 \wp' \wp'' = 12 \wp^2 \wp' - g_2 \wp'$, which simplifies to $$2\wp'' = 12 \wp - g_2$$ If you differentiate this equation once again and keep the first DE in mind, you'll find a formula for any higher-order derivative $\wp^{(k)}$ for $k \in \mathbb{N}$.

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    $\begingroup$ Thank you, and I'll add an answer when I have time deriving that first equation (which I found out about since posting this question) just so anyone with a similar problem has all the steps. $\endgroup$ – SPS Feb 13 '18 at 11:18
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@ComplexFlo answered the question perfectly well, but I thought I'd add to this thread how to derive the expression for $\wp '$ (i.e. the differential equation) we see at the beginning of that answer, in case anyone else wants to work out $\wp^{(k)}$ from scratch.

A Laurent series for $\wp$ around the origin is not difficult to derive and can be found in many places on the internet. (For completeness, I'll write it up if anyone wants it.) We find that $\wp(z) = \frac{1}{z^2} + \sum_{k \in \mathbb{N}} a_{2k} z^{2k}$, where $a_{2k}$ is given by $(2k + 1)\sum_{\omega \in \Omega^*} \frac{1}{\omega^{2k + 2}}.$

Therefore $$\wp '(z) = \frac{-2}{z^3} + 2a_2 z +4a_4 z^3 + \cdots$$ and so $$(\wp '(z))^2 = \frac{4}{z^6} -\frac{8a_2}{z^2} - 16a_4 + h(z),$$ where $h(z)$ is a polynomial with no constant term (i.e. a sum of only positive powers of $z$). Also, $$(\wp (z))^3 = \frac{1}{z^6} + \frac{3a_2}{z^2} + 3a_4 + g(z),$$ where $g(z)$ is also a polynomial with no constant term.

Using these previous two equations, it's clear that $(\wp '(z))^2 - 4(\wp (z))^3$ has no remaining terms in $\frac{1}{z^6}$ and its only term in a negative power of $z$ is $\frac{-20a_2}{z^2}.$ The constant term is $-28a_4.$ Recalling that the Laurent expansion for $\wp$ starts with the term $\frac{1}{z^2}$ and otherwise has only positive powers of $z$, it follows that the following sum has no negative powers of $z$ and no constant: $$(\wp '(z))^2 - 4(\wp (z))^3 + 20a_2 (\wp (z)) + 28a_4$$

Since this function has no non-positive powers of $z$ in its Laurent expansion (by construction), it has no poles. Since it's elliptic, it follows that the function is constant. Since for $z=0$ the function is obviously $0$ (there is a factor of $0$ in each of its terms), the function is constantly $0$. So, letting $g_2 = 20a_2 = 60\sum_{\omega \in \Omega^*} \frac{1}{\omega^{4}}$ and $g_3 = 28a_4 = 140\sum_{\omega \in \Omega^*} \frac{1}{\omega^{6}}$ by the definition of $a_{2k}$, we get the result $$(\wp ')^2 = 4\wp^3 - g_2\wp - g_3.$$

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