In this section we examine the relationship between three addition formula for the Weiersrass $\wp$ function.
The results are then extended to arbitrary elliptic functions of order 2.
This link lists various addition formula for the Weierstrass $\wp$ function in particular the classic formula
\begin{equation*}
\wp(u+v) \space = \space \frac 1 4 \left[\frac {\wp'(u) - \wp'(v)} {\wp(u) - \wp(v)} \right]^2 \space - \space \wp(u) \space - \space \wp(v)
\end{equation*}
Level 1 Addition Formula
This is the $n = 3$ case of the Frobenius-Stickelberger addition formula .
\begin{equation*}
E(u,v,w) \space = \space - \tfrac 1 2
\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & \wp(w) & \wp'(w)
\end{vmatrix}
\space = \space \frac {\sigma(u+v+w)\sigma(u-v)\sigma(u-w)\sigma(v-w)} {\sigma(u)^3\sigma(v)^3\sigma(w)^3}
\end{equation*}
We have
\begin{equation*}
E(u,v,w) = 0 \qquad \iff \qquad u + v + w = 0 \quad \text{or} \quad u=v \quad \text{or} \quad u=w \quad \text{or} \quad v=w
\end{equation*}
Level 2 Addition Formula
The LHS of this formula is essentially classic, arranged into a polynomial with $u + v$ replaced by $w$.
\begin{equation*}
F(u,v,w) \space = \space
\left[\wp(u) + \wp(v) + \wp(w)\right] \left[\wp(u) - \wp(v)\right]^2 - \tfrac 1 4 \left[\wp'(u) - \wp'(v)\right]^2 \space = \space \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v)^2}
{\sigma(u)^4\sigma(v)^4\sigma(w)^2}
\end{equation*}
This formula can be derived from fsp by computing
\begin{equation*}
F(u,v,w) = \frac {E(u,v,w) E(u,v,-w)} {P(u,w)P(v,w)}
\end{equation*}
where $P$ is the $n = 2$ case of the Frobenius-Stickelberger addition formula
\begin{equation*}
P(u,v) = \wp(u) - \wp(v) = -\frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)}
\end{equation*}
We have
\begin{equation*}
F(u,v,w) = 0 \qquad \iff \qquad u + v \pm w = 0 \quad \text{or} \quad u=v
\end{equation*}
Level 3 Addition Formula
The LHS of this formula is the algebraic relation implied by Abels addition theorem for Abelian integrals, when applied to the Weierstrass elliptic integral.
\begin{equation*}
G(u,v,w) \space = \space
\left[\wp(u)\wp(v) + \wp(u)\wp(w) + \wp(v)\wp(w) + \tfrac 1 4 g_2\right]^2 - 4\left[\wp(u) + \wp(v) + \wp(w)\right] \left[\wp(u)\wp(v)\wp(w) - \tfrac 1 4 g_3\right]
\space = \space \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v+w)\sigma(u-v-w)} {\sigma(u)^4\sigma(v)^4\sigma(w)^4}
\end{equation*}
It can be derived from wp and p by computing
\begin{equation*}
G(u,v,w) = \frac {F(u,v,w) F(u,-v,w)} {P(u,v)^2}
\end{equation*}
We have
\begin{equation*}
G(u,v,w) = 0 \qquad \iff \qquad u \pm v \pm w = 0
\end{equation*}
Relationship Between The Addition Formulae
Let
\begin{equation*}
S(u,v,w) \space = \space \frac {\sigma(u+v+w)\sigma(u-v+w)\sigma(u+v-w)\sigma(u-v-w)\sigma(u-v)^2\sigma(u+v)^2\sigma(u-w)^2\sigma(u+w)^2\sigma(v-w)^2\sigma(v+w)^2}
{\sigma(u)^{12}\sigma(v)^{12}\sigma(w)^{12}}
\end{equation*}
then using fsp, wp, p and ap we get
\begin{equation*}
S(u,v,w) \space = \space E(u,v,w) E(u,v,-w) E(u,-v,w) E(u,-v,-w) \space = \space F(u,v,w) F(u,-v,w) P(u,w)^2 P(v,w)^2 \space = \space \thinspace G(u,v,w) P(u,v)^2 P(u,w)^2 P(v,w)^2
\end{equation*}
Alternate Level 2 Addition Formula
Using p with four different pairs of arguments we get
\begin{equation*}
P(u+v,w) P(u,w) P(v,w) P(u,v)^2 \space = \space \tfrac 1 4 E(u,v,w) E(u,v,-w)
\end{equation*}
which rearranges to
\begin{equation*}
\wp(u+v) \space - \space \wp(w) \space = \space \frac
{\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & \wp(w) & \wp'(w) \\
\end{vmatrix} \space
\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & \wp(w) & -\wp'(w) \\
\end{vmatrix}}
{4 \big[\wp(u) - \wp(w)\big] \big[\wp(v) - \wp(w)\big] \big[\wp(u) - \wp(v)\big]^2}
\end{equation*}
Choosing $w$ so that $\wp(w)=0$ in wp2 gives
\begin{equation*}
\wp(u+v) \space = \space \frac
{\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & 0 & \sqrt{g_3} \\
\end{vmatrix}
\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & 0 & -\sqrt{g_3} \\
\end{vmatrix}}
{4 \wp(u) \wp(v) \big[\wp(u) - \wp(v)\big]^2}
\end{equation*}
Taking the product of wp2 over the half periods, $w \in \{\omega_1,\omega_2,\omega_3\}$ followed by a square root gives
\begin{equation*}
\wp'(u+v) \space = \space - \frac
{\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & e_1 & 0 \\
\end{vmatrix}
\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & e_2 & 0 \\
\end{vmatrix}
\begin{vmatrix}
1 & \wp(u) & \wp'(u) \\
1 & \wp(v) & \wp'(v) \\
1 & e_3 & 0 \\
\end{vmatrix}}
{\wp'(u) \wp'(v) \big[\wp(u) - \wp(v)\big]^3}
\end{equation*}
The correct sign for the square root can be determined by replacing $v$ by $-v$ and letting $v \rightarrow u$.
These two addition formulae are nothing other than these formulae.
Taking the product of dwpadd over $v \in \{\omega_1,\omega_2,\omega_3\}$ gives the slightly surprising formula
\begin{equation*}
\wp'(u) \thinspace \wp'(u + \omega_1) \thinspace \wp'(u + \omega_2) \thinspace \wp'(u + \omega_3) \space = \space 16 (e_1-e_2)^2(e_1-e_3)^2(e_2-e_3)^2
\end{equation*}
Only slightly surprising, because it's quite easy to see that all the zeroes and poles of the LHS cancel out.
Putting $v = \omega_1$ in dwpadd gives
\begin{equation*}
\wp'(u + \omega_1) \space = \space - \frac {\left[\wp(u)-e_1\right]\wp'(u)(e_2-e_1)\wp'(u)(e_3-e_1)} {\wp'(u) \left[\wp(u) - e_1\right]^3} \space = \space
- \frac {\wp'(u)(e_2-e_1)(e_3-e_1)} {\left[\wp(u) - e_1\right]^2}
\end{equation*}
and similar formulae for $\omega_2$ and $\omega_3$.
The result follows by taking the product of these three formulae and simplifying.
Dual Level 1 Addition Formula
Due to algebraic symmetry in ap we have
\begin{equation*}
H(u,v,w) \space = \space
\begin{vmatrix}
1 & e_1 & \sqrt{(\wp(u) - e_1)(\wp(v) - e_1)(\wp(w) - e_1)} \\
1 & e_2 & \sqrt{(\wp(u) - e_2)(\wp(v) - e_2)(\wp(w) - e_2)} \\
1 & e_3 & \sqrt{(\wp(u) - e_3)(\wp(v) - e_3)(\wp(w) - e_3)} \\
\end{vmatrix} \space = \space 2
\begin{vmatrix}
1 & e_1 & e_1^2 \\
1 & e_2 & e_2^2 \\
1 & e_3 & e_3^2 \\
\end{vmatrix}
\frac {\sigma(\frac u 2 + \frac v 2 + \frac w 2)\sigma(\frac u 2 + \frac v 2 - \frac w 2)\sigma(\frac u 2 - \frac v 2 + \frac w 2)\sigma(- \frac u 2 + \frac v 2 + \frac w 2)}
{\sigma(u)\sigma(v)\sigma(w)}
\end{equation*}
where the sign of the square roots is chosen so that residue at $u,v,w=0$ is $+1$.
$H$ is meromorphic and elliptic on the doubled period lattice and we have
\begin{equation*}
H(u,v,w) = 0 \qquad \iff \qquad u \pm v \pm w = 0
\end{equation*}
See Part 2 for more details.
Addition Formulae For Elliptic Function With Double Pole
Formulae fsp, wp and ap with minor adjustments they are valid for any elliptic function $f(z)$ with a double pole at zero.
For reference all the above formulae, along with the 4 variable level 1 addition formula are reproduced here.
If $f$ is an elliptic function with a double pole, $f'^2 = K(f - e_1)(f - e_2)(f - e_3)$ with $f(0)=\infty$ then
\begin{align*}
-\tfrac 1 4 K f'(u) &= \frac {\sigma(2u)} {\sigma^4(u)} \\\\
-\tfrac 1 4 K \big[f(u) - f(v)\big] &= \frac {\sigma(u-v)\sigma(u+v)} {\sigma^2(u)\sigma^2(v)} \\\\
-\tfrac 1 {32} K^2
\begin{vmatrix}
1 & f(u) & f'(u) \\
1 & f(v) & f'(v) \\
1 & f(w) & f'(w)
\end{vmatrix}
\space &= \space \frac {\sigma(u+v+w)\sigma(u-v)\sigma(u-w)\sigma(v-w)} {\sigma(u)^3\sigma(v)^3\sigma(w)^3} \\\\
\tfrac 1 {64} K^3 \left[f(u) + f(v) + f(w) - (e_1 + e_2 + e_3)\right] \left[f(u) - f(v)\right]^2 - \tfrac 1 {64} K^2 \left[f'(u) - f'(v)\right]^2 \space
&= \space \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v)^2} {\sigma(u)^4\sigma(v)^4\sigma(w)^2} \\\\
\tfrac 1 {256} K^4 \Big[\big[f(u)f(v) + f(u)f(w) + f(v)f(w) - (e_1e_2 + e_1e_3 + e_2e_3)\big]^2 - 4\big[f(u) + f(v) + f(w) - &(e_1 + e_2 + e_3)\big] \big[f(u)f(v)f(w) - e_1e_2e_3\big]\Big] \\\\
&= \space \frac {\sigma(u+v+w)\sigma(u+v-w)\sigma(u-v+w)\sigma(u-v-w)} {\sigma(u)^4\sigma(v)^4\sigma(w)^4} \\\\
\tfrac 1 {512} K^4 \begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
1 & f(x) & f(x)^2 & f'(x)
\end{vmatrix}
\space &= \frac {\sigma(u+v+w+x)\sigma(u-v)\sigma(u-w)\sigma(u-x)\sigma(v-w)\sigma(v-x)\sigma(w-x)} {\sigma(u)^4\sigma(v)^4\sigma(w)^4\sigma(x)^4}
\end{align*}
Addition Formulae For Elliptic Function With Two Simple Poles
If $f$ is an elliptic function with two simple poles, $f'^2=K(f-e_1)(f-e_2)(f-e_3)(f-e_4)$ with $f'(0)=0$ and $f(\pm \rho)=\infty$ and $\residue(\rho) = 1/\sqrt{K}$ then
\begin{equation*}
\tfrac 1 2 K^2
\begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
1 & f(x) & f(x)^2 & f'(x)
\end{vmatrix}
\space = \space \frac {\sigma(u+v+w+x)\sigma(u-v)\sigma(u-w)\sigma(u-x)\sigma(v-w)\sigma(v-x)\sigma(w-x)\sigma(2\rho)^4}
{\sigma(u-\rho)^2\sigma(u+\rho)^2\sigma(v-\rho)^2\sigma(v+\rho)^2\sigma(w-\rho)^2\sigma(w+\rho)^2\sigma(x-\rho)^2\sigma(x+\rho)^2}
\end{equation*}
This follows from the Extended Frobenius-Stickelberger formula
because $1,f,f^2,f'$ is a basis for the vector space of elliptic functions with a pair of poles of order at most 2 at $\pm \rho$.
And we can compute the multiplicative constant $\kappa$ from the Laurent expansions at the poles $f(z) = \alpha(z-\rho)^{-1} + \dots$ etc. where $\alpha=1/\sqrt{K}$
\begin{equation*}
\kappa \space = \space \begin{vmatrix}
0 & 0 & \alpha^2 & -\alpha \\
0 & \alpha & \cdot & \cdot \\
0 & 0 & \alpha^2 & \alpha \\
1 & \cdot & \cdot & \cdot
\end{vmatrix}
\space \sigma(2\rho)^4 \space = \space \frac 2 {K^2} \space \sigma(2\rho)^4
\end{equation*}
Taking the limit as $x \rightarrow -\rho$ gives an off centre 3 variable addition formula for $f$
\begin{equation*}
-\tfrac 1 2 K
\begin{vmatrix}
1 & f(u) & f'(u) - \sqrt{K}f(u)^2 \\
1 & f(v) & f'(v) - \sqrt{K}f(v)^2 \\
1 & f(w) & f'(w) - \sqrt{K}f(w)^2
\end{vmatrix}
\space = \space \frac {\sigma(u+v+w-\rho)\sigma(u-v)\sigma(u-w)\sigma(v-w)\sigma(2\rho)^2}
{\sigma(u-\rho)^2\sigma(v-\rho)^2\sigma(w-\rho)^2\sigma(u+\rho)\sigma(v+\rho)\sigma(w+\rho)}
\end{equation*}
Taking the limit again, as $w \rightarrow \rho$ gives the analogue of p for $f$
\begin{equation*}
-\sqrt{K}\big[f(u) - f(v)\big]\space = \space \frac {\sigma(u+v)\sigma(u-v)\sigma(2\rho)} {\sigma(u-\rho)\sigma(u+\rho)\sigma(v-\rho)\sigma(v+\rho)}
\end{equation*}
Taking the limit as $v \rightarrow u$ gives a formula for the derivative
\begin{equation*}
-\sqrt{K}f'(u) \space = \space \frac {\sigma(2u)\sigma(2\rho)} {\sigma(u-\rho)^2\sigma(u+\rho)^2}
\end{equation*}
Substituting fsx11 and fsx1 into the differential equation and rearranging slightly gives a formula closely related to the
$\sigma$ duplication formula
\begin{equation*}
\frac {\sigma(2u)^2} {\sigma(2\rho)^2} \space = \space
\frac {\sigma(u)\sigma(u-\omega_1)\sigma(u-\omega_2)\sigma(u-\omega_3)} {\sigma(\rho)\sigma(\rho-\omega_1)\sigma(\rho-\omega_2)\sigma(\rho-\omega_3)} \space
\frac {\sigma(u)\sigma(u+\omega_1)\sigma(u+\omega_2)\sigma(u+\omega_3)} {\sigma(\rho)\sigma(\rho+\omega_1)\sigma(\rho+\omega_2)\sigma(\rho+\omega_3)}
\end{equation*}
Substituting fsx11 into the cross-ratio gives a formula closely related to the
$\sigma$ addition formula
\begin{equation*}
\frac {\big[f(u) - f(v)\big]\big[f(w) - f(x)\big]} {\big[f(u) - f(w)\big]\big[f(v) - f(x)\big]} \space = \space
\frac {\sigma(u-v)\sigma(u+v)\sigma(w-x)\sigma(w+x)} {\sigma(u-w)\sigma(u+w)\sigma(v-x)\sigma(v+x)}
\end{equation*}
Similar to wp2 we have
\begin{equation*}
f(u + v + w) \space - \space f(x) \enspace = \enspace \frac 1 {\big[f(u)-f(x)\big]\big[f(v)-f(x)\big]\big[f(w)-f(x)\big]}
\frac {\begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
1 & f(x) & f(x)^2 & f'(x) \\
\end{vmatrix} \space
\begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
1 & f(x) & f(x)^2 & -f'(x) \\
\end{vmatrix}}
{\begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
0 & 0 & 1 & \sqrt{K} \\
\end{vmatrix} \space
\begin{vmatrix}
1 & f(u) & f(u)^2 & f'(u) \\
1 & f(v) & f(v)^2 & f'(v) \\
1 & f(w) & f(w)^2 & f'(w) \\
0 & 0 & 1 & -\sqrt{K} \\
\end{vmatrix}}
\end{equation*}
Also similar to wp2 we can by choosing $x$ such that $f(x)=0$ to get a formula for $f(u+v+w)$.
And by taking the product of fadd over $x \in \{0, \omega_1, \omega_2, \omega_3\}$ to get a formula for $f'(u+v+w)$.
Four Variable Symmetric Addition Formulae For Elliptic Function With Two Simple Poles
Taking the definition of $\sigma, \omega_i, \eta_i$ from NIST
we have $\omega_1+\omega_2+\omega_3 = 0$ and $\eta_1+\eta_2+\eta_3 = 0$ and $\sigma(z + \omega_i) = -e^{2\eta_i z} \sigma(z - \omega_i)$ for $i = 1,2,3$
Adding $\omega_4=\eta_4=0$ for notational convenience, we have $e_i = f(\omega_i)$ for $i=1 \ldots 4$ and
\begin{equation*}
f'(z)^2 \space = \space K \prod\limits_{i=1}^4 \left(f(z) - f(\omega_i)\right)
\end{equation*}
Defining $\rho$ to be the pole of $f$ with residue $K^{-1/2}$ we have
\begin{equation*}
f(z_1) - f(z_2) \space = \space - \frac {\sigma(2\rho)} {K^{1/2}} \cdot \frac {\sigma(z_1 - z_2)\sigma(z_1 + z_2)} {\sigma(z_1 - \rho)\sigma(z_1 + \rho)\sigma(z_2 - \rho)\sigma(z_2 +
\rho)}\qquad\qquad\qquad
f'(z) \space = \space - \frac {\sigma(2\rho)} {K^{1/2}} \cdot \frac {\sigma(2z)} {\sigma^2(z - \rho)\sigma^2(z + \rho)}
\end{equation*}
and therefore
\begin{equation*}
\frac {\sigma(2z)} {\sigma(2\rho)} \space = \space
\prod_{i=1}^4\frac {\sigma(z - \omega_i)} {\sigma(\rho - \omega_i)}
\end{equation*}
Vandermonde determinants are
\begin{equation*}
\begin{vmatrix}
1 & f(z_1) & f(z_1)^2 \\
1 & f(z_2) & f(z_2)^2 \\
1 & f(z_3) & f(z_3)^2 \\
\end{vmatrix} \enspace = \enspace
- \prod_{i \lt j \le 3} \left(f(z_i) - f(z_j)\right) \enspace = \enspace
\frac {\sigma^3(2\rho)} {K^{3/2}} \cdot \frac {\prod\limits_{i \lt j \le 3} \sigma(z_i+z_j) \sigma(z_i-z_j)} {\prod\limits_{i=1}^3 \sigma^2(z_i-\rho) \sigma^2(z_i+\rho)}
\end{equation*}
and
\begin{equation*}
\begin{vmatrix}
1 & f(z_1) & f(z_1)^2 & f(z_1)^3 \\
1 & f(z_2) & f(z_2)^2 & f(z_2)^3 \\
1 & f(z_3) & f(z_3)^2 & f(z_3)^3 \\
1 & f(z_4) & f(z_4)^2 & f(z_4)^3 \\
\end{vmatrix} \enspace = \enspace
\prod_{i \lt j \le 4} \left(f(z_i) - f(z_j)\right) \enspace = \enspace
\frac {\sigma^6(2\rho)} {K^3} \cdot \frac {\prod\limits_{i \lt j \le 4} \sigma(z_i+z_j) \sigma(z_i-z_j)} {\prod\limits_{i=1}^4 \sigma^3(z_i-\rho) \sigma^3(z_i+\rho)}
\end{equation*}
We have
\begin{equation*}
\begin{vmatrix}
1 & e_1 & e_1^2 \\
1 & e_2 & e_2^2 \\
1 & e_3 & e_3^2 \\
\end{vmatrix}^2 \enspace = \enspace
\frac {\sigma^6(2\rho)\sigma^4(\omega_1)\sigma^4(\omega_2)\sigma^4(\omega_3)} {K^3\prod\limits_{i=1}^3 \sigma^4(\omega_i-\rho) \sigma^4(\omega_i+\rho)}
\qquad\qquad\textsf{and}\qquad\qquad
\begin{vmatrix}
1 & e_1 & e_1^2 & e_1^3 \\
1 & e_2 & e_2^2 & e_2^3 \\
1 & e_3 & e_3^2 & e_3^3 \\
1 & e_4 & e_4^2 & e_4^3 \\
\end{vmatrix} \enspace = \enspace
\frac {\sigma^6(2\rho)\sigma^4(\omega_1)\sigma^4(\omega_2)\sigma^4(\omega_3)} {K^3\prod\limits_{i=1}^4 \sigma^3(\omega_i-\rho) \sigma^3(\omega_i+\rho)}
\end{equation*}
and
\begin{equation*}
\prod_{i=1}^4 \left(e_4 - f(z_i)\right) \enspace = \enspace
\frac{\sigma^4(2\rho)} {K^2 \sigma^4(\omega_4 - \rho) \sigma^4(\omega_4 + \rho)} \cdot \prod_{i=1}^4 \frac {\sigma(z_i - \omega_4) \sigma(z_i + \omega_4)} {\sigma(z_i - \rho) \sigma(z_i + \rho)}
\end{equation*}
so
\begin{equation*}
\begin{vmatrix}
1 & e_1 & e_1^2 & e_1^3 \\
1 & e_2 & e_2^2 & e_2^3 \\
1 & e_3 & e_3^2 & e_3^3 \\
1 & e_4 & e_4^2 & e_4^3 \\
\end{vmatrix}^{-1}
\begin{vmatrix}
1 & e_1 & e_1^2 \\
1 & e_2 & e_2^2 \\
1 & e_3 & e_3^2 \\
\end{vmatrix}^2
\prod_{i=1}^4 \left(e_4 - f(z_i)\right) \enspace = \enspace
\frac {\sigma^4(2\rho)} {K^2 \prod\limits_{i=1}^4\sigma(\omega_i - \rho) \sigma(\omega_i + \rho) \sigma(z_i - \rho) \sigma(z_i + \rho)}
\cdot \prod\limits_{i=1}^4 \sigma(z_i - \omega_4)\sigma(z_i + \omega_4)
\end{equation*}
so
\begin{equation*}
\begin{vmatrix}
1 & e_1 & e_1^2 & e_1^3 \\
1 & e_2 & e_2^2 & e_2^3 \\
1 & e_3 & e_3^2 & e_3^3 \\
1 & e_4 & e_4^2 & e_4^3 \\
\end{vmatrix}^{-\tfrac 1 2}
\begin{vmatrix}
1 & e_1 & e_1^2 & \pm\sqrt{(e_1-x_1)(e_1-x_2))(e_1-x_3)(e_1-x_4)} \\
1 & e_2 & e_2^2 & \pm\sqrt{(e_2-x_1)(e_2-x_2))(e_2-x_3)(e_2-x_4)} \\
1 & e_3 & e_3^2 & \pm\sqrt{(e_3-x_1)(e_3-x_2))(e_3-x_3)(e_3-x_4)} \\
1 & e_4 & e_4^2 & \pm\sqrt{(e_4-x_1)(e_4-x_2))(e_4-x_3)(e_4-x_4)} \\
\end{vmatrix} \enspace = \enspace
\frac {\sigma^2(2\rho)} {a \sqrt{\prod\limits_i\sigma(\omega_i - \rho) \sigma(\omega_i + \rho) \sigma(z_i - \rho) \sigma(z_i + \rho)} }
\cdot \sum\limits_j \pm \sqrt{\prod\limits_i \sigma(z_i - \omega_j) \sigma(z_i + \omega_j)}
\end{equation*}
The $k$-th cofactor of the last column is
\begin{equation*}
(-1)^{k} \thinspace f'(z_k) \prod_{\substack{i \lt j \\ i,j \ne k}} \big[f(z_i) - f(z_j)\big] \enspace = \enspace
(-1)^{k} \cdot \frac {\sigma^4(2\rho) \sigma(2z_k)} {K^2} \cdot \frac {\prod\limits_{\substack{i \lt j \\ i,j\ne k}} \sigma(z_i+z_j) \sigma(z_i-z_j)} {\prod\limits_{i=1}^4 \sigma^2(z_i-\rho)
\sigma^2(z_i+\rho)} \enspace = \enspace
\frac {\sigma^4(2\rho) \prod\limits_{i \lt j \le 4} \sigma(z_i-z_j)} {K^2 \prod\limits_{i=1}^4 \sigma^2(z_i-\rho) \sigma^2(z_i+\rho)}
\cdot \frac {\sigma(2z_k) \prod\limits_{\substack{i \lt j \\ i,j\ne k}} \sigma(z_i+z_j)} {\prod\limits_{\substack{i \le 4 \\ i \ne k}} \sigma(z_k-z_i)}
\end{equation*}
where
\begin{equation*}
\prod\limits_{\substack{i \lt j \le 4 \\ i,j\ne k}} \sigma(z_i-z_j) \enspace = \enspace
(-1)^{k} \cdot \frac {\prod\limits_{i \lt j \le 4} \sigma(z_i-z_j)} {\prod\limits_{\substack{i \le 4 \\ i \ne k}} \sigma(z_k-z_i)}
\end{equation*}
so
\begin{equation*}
\begin{vmatrix}
1 & f(z_1) & f(z_1)^2 & f'(z_1) \\
1 & f(z_2) & f(z_2)^2 & f'(z_2) \\
1 & f(z_3) & f(z_3)^2 & f'(z_3) \\
1 & f(z_4) & f(z_4)^2 & f'(z_4) \\
\end{vmatrix} \enspace = \enspace
\frac {\sigma^4(2\rho) \prod\limits_{i \lt j \le 4} \sigma(z_i-z_j)} {K^2 \prod\limits_{i=1}^4 \sigma^2(z_i-\rho) \sigma^2(z_i+\rho)}
\cdot \sum_{k=1}^4 \frac {\prod\limits_{\substack{i \lt j \enspace i,j\ne k}} \sigma(z_i+z_j)} {\prod\limits_{i \le 4 \enspace i \ne k} \sigma(z_k-z_i)} \sigma(2z_k)
\end{equation*}
References