Recently, I found this property of the elementary symmetric polynomial (ESP), further read on ESP: https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial, I tried expressing the ESP in terms of standing-alone elements in one recursive fraction.

Let's define the following:

$$e_k\big(x_1,...,x_n\big) = I^{k-1}_{n-k-1}(e^{(2)}_k)\quad\text{ for }0<k<n, n>1$$

With the starting point $e^{(2)}_k$ being:

$$e^{(2)}_k = \frac{x^2_1-x^2_2}{x^{2-k}_1-x^{2-k}_2}$$

Define the recursive step $I$ as:

$$ I^d_c(x^a_b)= \begin{cases} \frac{I^{d-1}_c\big(x^{a+1}_1\big) - I^{d-1}_c\big(x^{a+1}_{b+1}\big)}{I^{d-1}_c\big(x^{2}_1\big) - I^{d-1}_c\big(x^{2}_{b+1}\big)} &\quad d > 0\\ \frac{I^d_{c-1}\big(x^{a+1}_1\big) - I^d_{c-1}\big(x^{a+1}_{b+1}\big)}{I^d_{c-1}\big(x_1\big) - I^d_{c-1}\big(x_{b+1}\big)} &\quad c > 0 \land d = 0\\ x^a_b &\quad c=0\land d=0 \\ \end{cases} $$

Some cases, for your understanding:

\begin{align*} e_1\big(a,b,c\big) &= \frac{\frac{a^3-b^3}{a-b}-\frac{a^3-c^3}{a-c}}{\frac{a^2-b^2}{a-b}-\frac{a^2-c^2}{a-c}} = a + b + c\\ e_2\big(a,b,c,d\big) &= \frac{\frac{\frac{a^4-b^4}{a-b}-\frac{a^4-c^4}{a-c}}{\frac{a^3-b^3}{a-b}-\frac{a^3-c^3}{a-c}} - \frac{\frac{a^4-b^4}{a-b}-\frac{a^4-d^4}{a-d}}{\frac{a^3-b^3}{a-b}-\frac{a^3-d^3}{a-d}}}{\frac{\frac{a^2-b^2}{a-b}-\frac{a^2-c^2}{a-c}}{\frac{a^3-b^3}{a-b}-\frac{a^3-c^3}{a-c}} - \frac{\frac{a^2-b^2}{a-b}-\frac{a^2-d^2}{a-d}}{\frac{a^3-b^3}{a-b}-\frac{a^3-d^3}{a-d}}} = ab + ac + ad + bc + ... \end{align*}

I've made a Javaprogram to check cases, and it seems to hold. I tried using some form of induction to prove this identity, but in the end it failed. So my question is: how would one prove that this identity holds?


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