B. Expressing the Equations (4.14)

In order to write (4.19) as a function of $y_{n}$ the following set of rules is needed:

$$\begin{array}{ccc} (\bar{y}_{n}+y_{n})&=&X_{1}\quad, \\ (\bar{y}_{n}-y_{n})&=&X_{2}\quad, \end{array}$$ (7.10)

or

$$\begin{array}{ccc} \frac{X_{1}+X_{2}}{2}&=&y_{n}\quad, \\ \frac{X_{1}-X_{2}}{2}&=&\bar{y}_{n}\quad. \end{array}$$ (7.11)

The following identities (7.8):

$$\begin{array}(\bar{y}_{n}\pm y_{n})^{2}+\zeta^{2}&=&\left(1\p... ...{n}\, y_{n} &=& \frac{X_{1}^{2}-X_{2}^{2}}{4}\quad, \end{array}$$ (7.12)

allow to write $\bar{y}_{n}$ as function of $y_{n}$,

$$\begin{array}{ccc} \bar{y}_{n}^{2}+y_{n}^{2}+\zeta^{2}&=&1+\b... ...}&=&\frac{1-y_{n}^{2}-\zeta^{2}}{1-y_{n}^{2}}\quad, \end{array}$$ (7.13)

and $y_{n}$ as function of $\bar{y}_{n}$,

$$\begin{array}{ccc} \bar{y}_{n}^{2}+y_{n}^{2}+\zeta^{2}&=&1+\b... ...bar{y}_{n}^{2}-\zeta^{2}}{1-\bar{y}_{n}^{2}} \quad. \end{array}$$ (7.14)

Starting with the fraction of (4.20) in dimensionless form $I\equiv\frac{c(X_{2})-c(X_{1})}{c(X_{2})+c(X_{1})}$, and putting in the definition of $c(X)$, leads to the term below:

$$I\equiv\frac{X_{2}\, \left(\zeta \pm \sqrt{\zeta^2 + X_{1}^{2}}\r... ...1}^{2}}\right)+X_{1}\, \left(\zeta \pm \sqrt{\zeta^2 + X_{2}^{2}}\right)}\quad.$$ (7.15)

Proceeding by substituting $X_{1}$ with (7.10) and $X_{2}$ with (7.10), and using the identity:

$$\begin{array}{ccccc} \sqrt{X_{1,2}^{2}+\zeta^{2}}&=&\vert 1\p... ...2}}{4}\vert&=&\vert 1\pm\bar{y}_{n}y_{n}\vert\quad, \end{array}$$ (7.16)

the term (7.15) can be written as a function of $\bar{y}_{n}$ and $y_{n}$,

$$I\equiv\frac{\left(\bar{y}_{n}-y_{n}\right)\, \left( \zeta \pm \l... ..._{n}\right)\, \left( \zeta \pm \left(1 -\bar{y}_{n}y_{n}\right) \right) }\quad,$$ (7.17)

Expanding the brakets and reorganizing the expression results in:

$$I\equiv -\frac{y_{n} \left(\zeta\pm 1 \mp \bar{y}_{n}^{2}\right)}{\bar{y}_{n} \left(\zeta \pm1 \mp y_{n}^{2}\right)}\quad.$$ (7.18)

After substituting $\sqrt{\frac{1-y_{n}^{2}-\zeta^{2}}{1-y_{n}^{2}}}$ for $\bar{y}_{n}$ the expression simplifies to:

$$I\equiv\frac{-y_{n}\zeta\left(1\pm\frac{\zeta}{1-y_{n}^{2}}\right... ...\pm1\mp y_{n}^{2}\right)\sqrt{\frac{1-y_{n}^{2}-\zeta^{2}}{1-y_{n}^{2}}}}\quad,$$ (7.19)

and can be reformulated to the term used in (4.22):

$$I\equiv \mp \frac{y_{n}\, \zeta}{\sqrt{(1-y_{n}^{2})(1-y_{n}^{2}-\zeta^{2})}}\quad.$$ (7.20)