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In the limit $d\to\infty$ the op's difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2})\,. \end{align} Now we rewrite the $x_t$-recursion in terms of $\tau$, too, and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau) &=\frac{x(\tau-\delta\tau)+\sqrt{x^2(\tau-\delta\tau)-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau-\delta\tau)+\sqrt{x^2(\tau-\delta\tau)-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}\,, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau)-x(\tau-\delta\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}\,. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}\,. \end{align} The convexity for $0\leq\tau\leq1$ follows.

In the limit $d\to\infty$ the op's difference equation can be transformed into an ordinary differential equation, which can be solved exactly.

We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2})\,. \end{align} Next, we rewrite the $x_t$-recursion in terms of $\tau$, too, and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau) &=\frac{x(\tau{-}\delta\tau)+\sqrt{x^2(\tau{-}\delta\tau)-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau{-}\delta\tau)+\sqrt{x^2(\tau{-}\delta\tau)-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}\,, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau)-x(\tau{-}\delta\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}\,. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}\,. \end{align} The convexity for $0\leq\tau\leq1$ follows.

In the limit $d\to\infty$ the difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2}). \end{align} Now we rewrite the $x_t$-recursion also in terms of $\tau$ and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau+\delta\tau) &=\frac{x(\tau)+\sqrt{x(\tau)^2-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau)+\sqrt{x(\tau)^2-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau+\delta\tau)-x(\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}. \end{align} The convexity for $0\leq\tau\leq1$ follows.

In the limit $d\to\infty$ the op's difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2})\,. \end{align} Now we rewrite the $x_t$-recursion in terms of $\tau$, too, and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau) &=\frac{x(\tau-\delta\tau)+\sqrt{x^2(\tau-\delta\tau)-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau-\delta\tau)+\sqrt{x^2(\tau-\delta\tau)-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}\,, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau)-x(\tau-\delta\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}\,. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}\,. \end{align} The convexity for $0\leq\tau\leq1$ follows.

In the limit $d\to\infty$ the difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2}). \end{align} Now we rewrite the $x_t$-recursion also in terms of $\tau$ and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau+\delta\tau) &=\frac{x(\tau)+\sqrt{x(\tau)^2-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau)+\sqrt{x(\tau)^2-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau+\delta\tau)-x(\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}. \end{align} The convexity follows.

In the limit $d\to\infty$ the difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand $\beta(\tau) = \beta_{\tau d}$ around $d=\infty$, \begin{align}\tag{1}\label{eq:1} \beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2}). \end{align} Now we rewrite the $x_t$-recursion also in terms of $\tau$ and get (with $\delta\tau=1/d$) \begin{align}\tag{2a}\label{eq:2a} x(\tau+\delta\tau) &=\frac{x(\tau)+\sqrt{x(\tau)^2-4\beta(\tau)}}{2}\\ \tag{2b}\label{eq:2b} &=\frac{x(\tau)+\sqrt{x(\tau)^2-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2}, \end{align} such that \begin{align} \tag{3a}\label{eq:3a} x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau+\delta\tau)-x(\tau)}{\delta\tau}\\ \tag{3b}\label{eq:3b} &=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}. \end{align} The solution of this ODE with initial condition $x(0)=1$ reads \begin{align} \tag{4}\label{eq:4} x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}. \end{align} The convexity for $0\leq\tau\leq1$ follows.