Tanh and elementary symmetric polynomials

Yesterday I wrote a post that looked at the hyperbolic tangent sum

$x \oplus y = \tanh\left(\tanh^{-1}x + \tanh^{-1} y\right) = \frac{x + y}{ 1 + xy}$

for x and y strictly between −1 and 1. This sum arises when adding velocities in special relativity. The post ended with a description of the expression for

$x_1 \oplus x_2 \oplus x_3 \oplus \cdots \oplus x_n$

in terms of elementary symmetric polynomials but did not offer a proof. This post will give a proof and show why elementary symmetric polynomials arise naturally.

We start by noting

$\tanh x = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(-x)} = \frac{\exp(2x) -1}{\exp(2x) + 1}$

and

$\tanh^{-1} x = \frac{1}{2} \log \frac{1+x}{1-x}$

Then

$\begin{align*} \bigoplus_{i=1}^n x_i &= \tanh\left( \sum_{i=1}^n \tanh^{-1}(x_i) \right) \\ &= \tanh\left( \frac{1}{2} \sum_{i=1}^n \log \frac{1+x_i}{1 - x_i} \right) \\ & = \tanh\left( \frac{1}{2} \sum_{i=1}^n \log(1 + x_i) - \frac{1}{2} \sum_{i=1}^n \log(1 - x_i)\right) \\ &= \frac{\prod_{i=1}^n (1 + x_i) - \prod_{i=1}^n (1 - x_i)}{\prod_{i=1}^n (1 + x_i)+\prod_{i=1}^n (1 - x_i)} \end{align*}$

Now

$\prod_{i=1}^n (1 + x_i) = e_0 + e_1 + e_2 + \cdots + e_n$

where

$\begin{align*} e_0 &= 1 \\ e_1 &= \sum_{i=1}^n x_i \\ e_2 &= \sum_{i=1}^n \sum_{j = i+1}^n x_ix_j \\ \phantom{e_1} &\phantom{=}\cdots \\ e_n &= x_1x_2x_3 \cdots x_n \end{align*}$

are the elementary symmetric polynomials.

We get e₀ by choosing the 1 term from each binomial in the product. We get e₁ by choosing the 1 term from all but one of the binomials and choosing an x as the remaining term. We get e₂ by choosing the 1 term from n − 2 of the binomials and choosing x‘s from the two remaining terms, and so on. Finally, we get e_n by choosing an x from each binomial.

Similarly

$\prod_{i=1}^n (1 - x_i) = e_0 - e_1 + e_2 - \cdots +(-1)^n e_n$

Therefore

$\begin{align*} \frac{\prod_{i=1}^n (1 + x_i) - \prod_{i=1}^n (1 - x_i)}{\prod_{i=1}^n (1 + x_i)+\prod_{i=1}^n (1 - x_i)} &= \left. \sum_{i \text{ odd}} e_i\,\,\middle/\sum_{i \text{ even}} e_i \right. \end{align*}$

because the even terms cancel out in the numerator and the odd terms cancel out in the denominator.

In words, the hyperbolic tangent sum of multiple arguments is the ratio of the sums of the odd and even elementary symmetric polynomials in the arguments.

Elementary symmetric polynomials enter the derivation because they are what you get when you expand products of (1 + x_i).