Addition theorems for elliptic functions: is there a painless way? The Weierstrass $\wp$ function satisfies the addition formula
$$\wp(z+Z)+\wp(z)+\wp(Z) = \left(\frac{\wp'(z)-\wp'(Z)}{\wp(z)-\wp(Z)}\right)^2.$$
Of course, this is just the $x$-coordinate of the sum of the points $(\wp'(z), \wp(z))$ and $(\wp'(Z), \wp(Z))$ on the Weierstrass elliptic curve $y^2=4x^3-g_2x-g_3$. If one has an a priori knowledge of this fact, the computation of the addition formula is absolutely trivial. However, it seems that, in the literature, there is no "illuminating" proof from first principles. Perhaps my standards for "illuminating" are too high, but in comparison with proofs of addition formulas for trigonometric functions, the addition formula for $\wp$ is painstaking to establish.
I would like to know how the formula above, or an equivalent formula (perhaps for the Weierstrass $\sigma$-functions?) may be deduced in the most direct possible manner. Of course, a proof such as one that involves the comparison of power series around the origin is very direct, but it requires a first-hand knowledge of the formula, so it's not quite what I'm looking for.
All the best, and thank you! 
 A: The following uses only easy-to-prove properties like the periodicity and the power series of $\wp(z)$ up to $O(z^4)$, together with the fundamental property that any elliptic function is expressible as $R(\wp)+S(\wp) \cdot \wp'$ for rational $R$ and $S$ (this is also quite easy to prove given certain basic facts about the order of an elliptic function). (Of course once you have the latter, the fact that there is an addition formula is not surprising, since $\wp(z+w)$ is an elliptic function with the same periods as $\wp(z)$.)
We assume that $w$ is in general position; any problematic cases can be disposed of with continuity. Following the algorithm for the decomposition suggests considering the functions
$$ f_+(z) = \wp(z+w)+\wp(z-w), \qquad f_-(z) = \wp(z+w)-\wp(z-w), $$
which are even and odd qua functions of $z$. Both have double poles at $z=\pm w$ as their only singularities, and hence have order $4$.
Looking first at $f_-$, we know that it has a simple zero at $z=0$. The other three zeros are found at $z=\omega_i/2$, where $\omega_i \in \Lambda$ and $\omega_3=-\omega_1-\omega_2$, because $\wp(\omega_i/2+w) = \wp(-\omega_i/2-w) = \wp(\omega_i/2-w)$ using evenness and periodicity. Hence $f_-/\wp'$ has a quadruple zero at $0$, the double poles at $\pm z$ we mentioned before, and no others. Therefore
$$ \frac{f_-(z)}{\wp'(z)} (\wp(z)-\wp(w))^2 $$
has no poles, since the zeros in the bracket cancel with those at $\pm w$, while the bracket's quadruple pole at zero is cancelled by the quadruple zero at zero of the fraction. Hence this function is constant; we can expand at $z=0$ to find that the constant is $-\wp'(w)$, so
$$ f_-(z) = - \frac{\wp'(z)\wp'(w)}{(\wp(z)-\wp(w))^2}. $$
The same idea works for $f_+$, but I can't find a simple way of doing the computation, because the zeros are difficult to find a priori: note that $f_+(z)(\wp(z)-\wp(w))^2$ has only a quadruple pole at $z=0$, so it can be written as a quadratic in $z^2$. Expanding
$$ f_+(z)(\wp(z)-\wp(w))^2 = A\wp(z)^2 + B\wp(z) +C $$
 about $z=0$ up to $O(1)$ and equating coefficients gives expressions for the coefficients in terms of $\wp(w)$, $\wp''(w)$, $\wp^{(4)}(w)$ and $g_2$, and differentiating $\wp'^2=4\wp^3-g_2\wp-g_3$ once and thrice gives expressions for the derivatives, leading to
$$ f_+(z) = \frac{(\wp(z)+\wp(w))(2\wp(z)\wp(w)-g_2/2)-g_3}{(\wp(z)-\wp(w))^2}. $$
Using the differential equation and polynomial division, the right-hand side can be massaged into the form
$$ -2\wp(z)-2\wp(w) + \frac{\wp'(z)^2+\wp'(w)^2}{2(\wp(z)-\wp(w))^2}. $$
Using the formula $\wp(z)=(f_+(z)+f_-(z))/2$ then gives the addition formula.
A: I have found exactly the sort of proof I was looking for in Lang's Elliptic Curves: Diophantine Analysis. It shows precisely how the addition theorem relates to addition on the curve. I'll write it here in case anyone else is interested.
If $f$ is an elliptic function with respect to the lattice $\Omega$, integrating $\frac{1}{2\pi i}z\frac{f'(z)}{f(z)}$ around the boundary of a fundamental parallelogram shows that the sum of the zeroes of $f$ minus the sum of the poles of $f$ inside the parallelogram is $\equiv 0 \mod \Omega$. If $f$ has its only pole at the origin, it shows that the sum of the zeroes of $f$ is $\equiv 0 \mod \Omega$.
Now let $z_1$, $z_2$ be any two points in general position. Find constants $a$ and $b$ such that
$$ \wp(z_1) + a\wp'(z_1) = b$$
$$ \wp(z_2) + a\wp'(z_2) = b.$$
The function $\wp(z)+a\wp'(z)-b$ is of order $3$ and has a triple pole at the origin. It has zeroes at $z_1$ and $z_2$. Thus its third zero must be at $-z_1-z_2$. Thus the point $(\wp(-z_1-z_2), \wp'(-z_1-z_2))=(\wp(z_1+z_2), -\wp'(z_1+z_2))$ is collinear with the points $(\wp(z_1), \wp'(z_1)), (\wp(z_2), \wp'(z_2))$. Since it also lies on the cubic $y^2=4x^3-g_2x-g_3$, it suffices to find the third point of intersection of the line with the cubic!
A: I'm not quite sure if this is the sort of proof that'll satisfy, but I'll write out (a sketch of) one proof in Akhiezer's book.
The prerequisite is the knowledge that an arbitrary elliptic function $f(u)$ can be represented in terms of Weierstrass $\sigma$:
$$f(u)=k\frac{\prod\limits_{j=1}^n\sigma(u-p_i;g_2,g_3)}{\prod\limits_{j=1}^n\sigma(u-q_i;g_2,g_3)}$$
where the $p_i$ are zeroes and $q_i$ are poles within the period parallelogram (multiple poles/zeroes being indicated according to their order), and these satisfy the equation $\sum_i p_i=\sum_i q_i$.
Consider the function $\wp(u)-\wp(v)$ where $v$ is a constant and $\wp(v)$ is finite. (I do not indicate the invariants $g_2,g_3$ here and in the sequel due to my laziness. ;)) The function has a double pole at $u=0$ and two zeroes at $u=v$ and $u=-v$ (hint: $\wp$ is even), so we have the representation
$$\wp(u)-\wp(v)=k\frac{\sigma(u-v)\sigma(u+v)}{\sigma^2(u)}$$
$k$ can be determined by multiplying both sides of the equation by $u^2$ and letting $u\to0$, which gives $k=-\frac1{\sigma^2(v)}$; thus ending up with 
$$\wp(u)-\wp(v)=-\frac{\sigma(u-v)\sigma(u+v)}{\sigma^2(u)\sigma^2(v)}$$
If we logarithmically differentiate both sides, we have
$$\frac{\wp^\prime(u)}{\wp(u)-\wp(v)}=\zeta(u+v)-2\zeta(u)+\zeta(u-v)$$
where $\zeta(u)$ is Weierstrass $\zeta$. If $u$ and $v$ are swapped and we remember that $\zeta$ is odd, we have
$$-\frac{\wp^\prime(v)}{\wp(u)-\wp(v)}=\zeta(u+v)-2\zeta(v)-\zeta(u-v)$$
Adding up these two and rearranging yields the (pseudo-)addition formula for $\zeta$:
$$\zeta(u+v)=\zeta(u)+\zeta(v)+\frac12\frac{\wp^\prime(u)-\wp^\prime(v)}{\wp(u)-\wp(v)}$$
Differentiating the (pseudo-)addition formula with respect to $u$ (and remembering that $\zeta^\prime=-\wp$) gives
$$-\wp(u+v)=-\wp(u)+\frac12\frac{\wp^{\prime\prime}(u)(\wp(u)-\wp(v))-\wp^\prime(u)(\wp^\prime(u)-\wp^\prime(v))}{(\wp(u)-\wp(v))^2}$$
Adding this up with the derivative with respect to $v$, and remembering that $\wp^{\prime\prime}=6\wp^2-\frac{g_2}{2}$, we can simplify the mess and thus obtain the addition formula for $\wp$.
$$\wp(u+v)+\wp(u)+\wp(v)=\frac14\left(\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)}\right)^2$$
(i.e. there's a typo in the OP... ;))
