**Appendix C. Computations of Some Equations**

*Appendix C.1. Computation*

Let *f* be a function of *λ* defined as follows:

$$f(\lambda) = \sum\_{n=1}^{\infty} \left(\frac{1}{2}\right)\_n \frac{1}{n} \frac{(1-\lambda)^n}{n!}.\tag{A21}$$

The multiplication of the derivative of *f* with respect to *λ* by (1 − *λ*) is given as follows

$$(1 - \lambda) \frac{\partial}{\partial \lambda} f(\lambda) = -\sum\_{n=1}^{\infty} \left(\frac{1}{2}\right)\_n \frac{(1 - \lambda)^n}{n!} = 1 - \lambda^{-1/2}.\tag{A22}$$

As a consequence,

$$\frac{\partial}{\partial \lambda} f(\lambda) = \frac{1 - \lambda^{-1/2}}{1 - \lambda} = \frac{-\lambda^{-1/2}}{1 + \lambda^{1/2}}.\tag{A23}$$

Finally,

$$f(\lambda) = -2\ln\frac{1+\lambda^{1/2}}{2}.\tag{A24}$$

*Appendix C.2. Computation*

$$\frac{\partial}{\partial a} \left\{ \,\_2F\_1 \left( a, \frac{1}{2}; a + \frac{3}{2}; 1 - \lambda\_i \right) \right\} \Big|\_{a=0} = \sum\_{n=1}^{\infty} \frac{(\frac{1}{2})\_n (1)\_n}{(\frac{3}{2})\_n n} \frac{(1 - \lambda\_i)^n}{n!} $$

$$= f(\lambda\_i) \tag{A25}$$

where *f* is a function of *λi*. The multiplication of the derivative of *f* with respect to *λ<sup>i</sup>* by (1 − *λi*) is given as follows

$$f(1 - \lambda\_i) \frac{\partial}{\partial \lambda\_i} f(\lambda\_i) = -\sum\_{n=1}^{\infty} \frac{(\frac{1}{2})\_n (1)\_n}{(\frac{3}{2})\_n} \frac{(1 - \lambda\_i)^n}{n!} \tag{A26}$$

$$=-\,\_2F\_1\left(\frac{1}{2}, 1; \frac{3}{2}; 1-\lambda\_{\bar{i}}\right) + 1.\tag{A27}$$

Knowing that

$$\,\_2F\_1\left(\frac{1}{2}, 1; \frac{3}{2}; 1 - \lambda\_i\right) = \frac{\arctan\left(\sqrt{\lambda\_i - 1}\right)}{\sqrt{\lambda\_i - 1}}\tag{A28}$$

$$\eta = \frac{1}{2\sqrt{1-\lambda\_i}} \ln \left( \frac{1+\sqrt{1-\lambda\_i}}{1-\sqrt{1-\lambda\_i}} \right) \tag{A29}$$

we can deduce an expression of

$$\frac{\partial}{\partial \lambda\_i} f(\lambda\_i) = \frac{\arctan(\sqrt{\lambda\_i - 1})}{(\lambda\_i - 1)^{3/2}} + \frac{1}{1 - \lambda\_i}. \tag{A30}$$

Accordingly,

$$f(\lambda\_i) = -\ln \lambda\_i - 2 \frac{\arctan(\sqrt{\lambda\_i - 1})}{\sqrt{\lambda\_i - 1}} + 2 \tag{A31}$$

$$\lambda = -\ln \lambda\_i + \frac{1}{\sqrt{1 - \lambda\_i}} \ln \left( \frac{1 - \sqrt{1 - \lambda\_i}}{1 + \sqrt{1 - \lambda\_i}} \right) + 2. \tag{A32}$$

*Appendix C.3. Computation*

$$\begin{split} \frac{\partial}{\partial a} \left\{ \,\_2F\_1 \left( a, 1; a + \frac{3}{2}; 1 - \lambda \right) \right\} \Big|\_{a=0} &= \sum\_{n=1}^{\infty} \frac{(1)\_n (1)\_n}{(\frac{3}{2})\_n n!} \frac{(1 - \lambda)^n}{n!} \\ &= f(\lambda) \end{split} \tag{A33}$$

where *f* is a function of *λ*. The multiplication of the derivative of *f* with respect to *λ* by (1 − *λ*) is given as follows

$$f(1 - \lambda) \frac{\partial}{\partial \lambda} f(\lambda) = -\sum\_{n=1}^{\infty} \frac{(1)\_n (1)\_n}{(\frac{3}{2})\_n} \frac{(1 - \lambda)^n}{n!} \tag{A34}$$

$$=-\,\_2F\_1\left(1,1;\frac{3}{2};1-\lambda\right)+1.\tag{A35}$$

Knowing that

$$\,\_2F\_1\left(1,1;\frac{3}{2};1-\lambda\right) = \frac{1}{\sqrt{\lambda}}\frac{\arcsin(\sqrt{1-\lambda})}{\sqrt{1-\lambda}}\tag{A36}$$

we can state that

$$\frac{\partial}{\partial \lambda} f(\lambda) = -\frac{1}{\sqrt{\lambda}} \frac{\arcsin(\sqrt{1-\lambda})}{(1-\lambda)^{3/2}} + \frac{1}{1-\lambda}.\tag{A37}$$

As a consequence,

$$f(\lambda) = -\frac{2\sqrt{\lambda}\arcsin(\sqrt{1-\lambda})}{\sqrt{1-\lambda}} + 2\tag{A38}$$

$$\hat{\lambda} = -\frac{2}{\sqrt{1-\lambda^{-1}}} \ln(\sqrt{\lambda} + \sqrt{\lambda - 1}) + 2. \tag{A39}$$
