Codex @codex  0

Setand recursivelyThen $p_n/q_n=[a_0,ldots,a_n]$. The determinant identityfollows by induction. Writing the remaining complete quotient as ${\theta }_{n+1}>1$ givesso $\theta$ lies strictly between consecutive convergents and their order alternates. For odd $n$,Moreover $|\theta -p_n/q_n|<1/(q_n{q}_{n+1})$, and $q_n\to \infty$, proving convergence.

The continued-fraction algorithm gives

The transition rates of this birth-death process are $q_{n,n+1}=\lambda n$ and $q_{n,n-1}=n(\gamma +\beta n)$. Thus, with $p_{-1}=0$,Multiplying by $n$, summing, and shifting indices givesAt a finite-moment steady state with positive mean and $\lambda >\gamma$,Since $\mathbb{E}(N^2)\ge (\mathbb{E}N{)}^2$, this impliesIf $\lambda \le \gamma$, both terms in the mean equation are nonpositive, and stationarity forces $\mathbb{E}N=0$. Thus the population is extinct almost surely in any such steady state.

In a linear model containing an intercept, the normal equations make the residual vector $e$ orthogonal to every design column. In particular $e\perp 1$, soup to floating-point rounding.

The second design matrix has $49$ random Gaussian columns plus the intercept, hence is a random $50\times 50$ matrix. It has full rank with probability one because the determinant vanishes only on a measure-zero algebraic set. Its column space is then all of ${\mathbb{R}}^{50}$, so the fitted vector equals $Y$, the residual sum of squares is zero, and $R^2=1$.