# Stationary point of $f(\zeta) =\frac{1}{\log{2}} \left[ \log\left(2 - \zeta \right) - \left( \frac {1-\zeta} {2-\zeta} \right) \zeta \right]$

Let us define the following:

$$\zeta(\beta;\alpha) = \frac{1}{2}\left[ 1+\alpha+\beta - \sqrt{4\beta+(\alpha+\beta-1)^2} \right]$$

$$F(\beta;\alpha) = f(\zeta) = %%% \frac{1}{\log{2}} \left[ \log\left(2 - \zeta \right) - \left( \frac {1-\zeta} {2-\zeta} \right) \zeta \right]$$

and

$$G(\beta;\alpha) = \frac{1}{\beta}F(\beta;\alpha)$$

where $\alpha,\beta \in \mathbb{R}^+$. It can be shown that $\zeta,F,f\in(0,1)$ and $f(0)=1$ and $f(1)=0$.

I want to find for what $\alpha$ $\exists\beta^*>0:$ $$\frac{\partial G}{\partial \beta}(\beta^*;\alpha) = 0$$

I have attempted multiple ways of doing this, I will give my working for the brute force derivative below, but first I would like to ask why the following method is incorrect:

\begin{align} \frac{\partial G}{\partial \beta} %%% &= %%% \frac{\partial}{\partial \beta} \left( \frac{1}{\beta} F\left(\beta;\alpha\right) \right)\\ %%% &= %%% \frac{1}{\beta} \frac{\partial}{\partial \beta} \left( F\left(\beta;\alpha\right) \right) - \frac{1}{\beta^2} F\left(\beta;\alpha\right) \end{align}

Therefore we can write \begin{align} A(\alpha) = F\left(1;\alpha\right) \end{align}

This isn't holding under a numerical check, can't spot a mistake. Also apologies I could not properly define the problem in the title.

Here are my brute force derivative workings:

Consider \begin{align} \log{2} \frac{\partial}{\partial \beta} \left( F\left(\beta;\alpha\right) \right) %%% &= %%% \frac{\partial}{\partial \beta} \left( \log\left(2 - \zeta \right) \right) - \frac{\partial}{\partial \beta} \left[ \left( \frac {1-\zeta} {2-\zeta} \right) \zeta \right]\\ %%% &= %%% - \frac{1}{2 - \zeta} \frac{\partial\zeta}{\partial\beta} - \left( \frac {1-\zeta} {2-\zeta} \right) \frac{\partial\zeta}{\partial\beta} - \zeta \left[ \frac{\partial}{\partial \beta} \left( \frac {1-\zeta} {2-\zeta} \right) \right]\\ %%% &= %%% - \frac{1}{2 - \zeta} \frac{\partial\zeta}{\partial\beta} - \left( \frac {1-\zeta} {2-\zeta} \right) \frac{\partial\zeta}{\partial\beta} - \zeta \left[ - \left( \frac {1} {(2-\zeta)^2} \right) \frac{\partial \zeta}{\partial \beta} \right]\\ %%% &= %%% \left\{ \frac {\zeta} {(2-\zeta)^2} - \frac{1}{2 - \zeta} - \frac {1-\zeta} {2-\zeta} \right\} \frac{\partial \zeta}{\partial \beta}\\ %%% &= %%% \left\{ \frac {\zeta}{(2-\zeta)^2} - \frac{2-\zeta}{2-\zeta} \right\} \frac{\partial \zeta}{\partial \beta}\\ %%% &= %%% \left\{ \frac {\zeta}{(2-\zeta)^2} - 1 \right\} \frac{\partial \zeta}{\partial \beta} \end{align}

Next consider \begin{align} 2\frac{\partial \zeta}{\partial \beta} %%% &= %%% \frac{\partial}{\partial \beta} \left( 1+\alpha+\beta - \sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1} \right)\\ %%% &= %%% 1 - \frac{\partial}{\partial \beta} \sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1}\\ %%% &= %%% 1 - \frac{1}{2} \frac { \frac{\partial}{\partial \beta} \left( \left(\alpha+\beta\right)^2+2(\beta-\alpha)+1 \right) } {\sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1}}\\ %%% &= %%% 1 - \frac{1}{2} \frac { 2\left(\alpha+\beta\right)+2 } {\sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1}}\\ %%% &= %%% 1 - \frac { 1+\alpha+\beta } {\sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1}} \end{align}

Note that \begin{align} \sqrt{\left(\alpha+\beta\right)^2+2(\beta-\alpha)+1} %%% &= %%% 1+\alpha+\beta-2\zeta \end{align}

therefore \begin{align} 2\frac{\partial \zeta}{\partial \beta} %%% &= %%% 1 - \frac { 1+\alpha+\beta } {1+\alpha+\beta-2\zeta}\\ %%% &= %%% \frac{1+\alpha+\beta-2\zeta}{1+\alpha+\beta-2\zeta} - \frac{1+\alpha+\beta}{1+\alpha+\beta-2\zeta}\\ %%% &= %%% -\frac{2\zeta}{1+\alpha+\beta-2\zeta}\\ %%%% \therefore\quad \frac{\partial \zeta}{\partial \beta} &= -\frac{\zeta}{1+\alpha+\beta-2\zeta} \end{align}

Therefore \begin{align} \frac{\partial}{\partial \beta} \left( F\left(\beta;\alpha\right) \right) %%% &= %%% \frac{1}{\log{2}} \left\{ \frac {\zeta}{(2-\zeta)^2} - 1 \right\} \frac{\partial \zeta}{\partial \beta}\\ %%% &= %%% \frac{1}{\log{2}} \left\{ 1-\frac{\zeta}{(2-\zeta)^2} \right\} \frac{\zeta}{1+\alpha+\beta-2\zeta} \end{align}

Therefore \begin{align} \beta \frac{\partial}{\partial \beta} \left( F\left(\beta;\alpha\right) \right) &= F\left(\beta;\alpha\right)\\ %%% \Rightarrow\quad %%% \frac{1}{\log{2}} \left\{ 1-\frac{\zeta}{(2-\zeta)^2} \right\} \frac{\beta\zeta}{1+\alpha+\beta-2\zeta} %%% &= %%% \frac{1}{\log{2}} \left[ \log\left(2 - \zeta \right) - \left( \frac {1-\zeta} {2-\zeta} \right) \zeta \right]\\ %%% \Rightarrow\quad %%% \left\{ 1-\frac{\zeta}{(2-\zeta)^2} \right\} \frac{\beta\zeta}{1+\alpha+\beta-2\zeta} %%% &= %%% \log\left(2 - \zeta \right) - \left( \frac {1-\zeta} {2-\zeta} \right) \zeta\\ %%% \Rightarrow\quad %%% \left\{ \frac{(2-\zeta)^2-\zeta}{(2-\zeta)^2} \right\} \frac{\beta\zeta}{1+\alpha+\beta-2\zeta} %%% &= %%% \log\left(2 - \zeta \right) - \left( \frac {1-\zeta} {2-\zeta} \right) \zeta\\ %%% \Rightarrow\quad %%% \left\{ \frac {\zeta^2-5\zeta+4} {2-\zeta} \right\} \frac{\beta\zeta}{1+\alpha+\beta-2\zeta} %%% &= %%% (2-\zeta)\log\left(2 - \zeta \right) - (1-\zeta)\zeta\\ %%% \Rightarrow\quad %%% \frac {\beta\zeta(\zeta-4)(\zeta-1)} {1+\alpha+\beta-2\zeta} %%% &= %%% (2-\zeta)^2\log\left(2 - \zeta \right) - (1-\zeta)(2-\zeta)\zeta \end{align}