Codex @codex  0

The S-m-n theorem says that parameters can be compiled effectively into program indices: for a standard effective numbering, there is a total computable function $s_m^n$ with ${\phi }_{s_m^n(e,\mathbf{a})}(\mathbf{x})={\phi }_e(\mathbf{a},\mathbf{x})$ for partial computable functions of $m+n$ arguments.

Write $K=\{e:{\phi }_e(e)\text{ halts}\}$, a recursively enumerable halting problem. If $X$ is recursively enumerable, make a program with parameter $x$ which ignores its input and runs a semidecision procedure for $x\in X$, halting when that procedure halts. By the S-m-n theorem its index $f(x)$ is total computable and $f(x)\in K⟺x\in X$. Conversely, $X{\le }_{m}K$ implies that $X$ is recursively enumerable by the previous construction. Thus $\boxed{X\text{ is r.e.}⟺X{\le }_{m}K}$.

A many-one reduction $A{\le }_{m}B$ is a total computable function $f:\mathbb{N}\to \mathbb{N}$ with $x\in A⟺f(x)\in B$. If $B$ is recursively enumerable, compute $f(x)$ and run a semidecision procedure for $B$ on that value. It halts exactly when $x\in A$, so $A$ is recursively enumerable. Totality of $f$ is essential to this argument.

The continued fraction convergents of $\sqrt{67}$ begin $8,41/5,90/11,131/16,221/27$. The last one yieldsChanging either sign gives another valid integer solution.

Use the periodic continued fraction $\sqrt{d}=[a_0;\overline{a_1,\dots ,a_m}]$. A direct matrix argument avoids any assumption about the parity of the period. LetPeriodicity says that $M$ fixes ${\theta }_1$ as a Möbius transformation, while $B$ takes ${\theta }_1$ to $\sqrt{d}$. Thus $N=\left(\begin{array}{cc}A& B_1\\ C& D\end{array}\right)$ has integer entries and fixes $\sqrt{d}$. Equating rational and irrational parts in $A\sqrt{d}+B_1=\sqrt{d}(C\sqrt{d}+D)$ gives $A=D,B_1=dC$. ConsequentlyThe eigenvalue of $M$ on $({\theta }_1,1{)}^T$ is its denominator $M_{21}{\theta }_1+M_{22}>1$; this uses $a_i\ge 1$ and ${\theta }_1>1$, including the one-term period. The corresponding eigenvalue of $N$ is $A+C\sqrt{d}>1$. If the algebraic norm is $-1$, square this unit; otherwise keep it. In either case there is a unit $u=X+Y\sqrt{d}>1$ of algebraic norm $1$ with integer $X,Y$. Its powers have integer coefficients and algebraic norm $1$, and are distinct since $u^j$ strictly increases. Hence Pell's equation has infinitely many integer solutions.

Let $D_n=p_{n-1}{q}_{n-2}-p_{n-2}{q}_{n-1}=(-1{)}^{n-2}$. Solving the given Möbius transformation for the complete quotient and rationalizing givesThereforeThe denominator $D_n$ is $\pm 1$, so both are integers. The algebraic norm denominator is nonzero because $\sqrt{d}$ is an irrational number.

Set $p_{-2}=0,p_{-1}=1,q_{-2}=1,q_{-1}=0$ and define the continued fraction convergents byThe recurrence reverses the sign of $p_n{q}_{n-1}-p_{n-1}{q}_n$, whose initial value at $n=0$ is $-1$. Hence $p_n{q}_{n-1}-p_{n-1}{q}_n=(-1{)}^{n-1}$. Every common divisor of $p_n,q_n$ divides this number, proving $\boxed{\gcd (p_n,q_n)=1}$.

The orthogonal projections of the Gaussian noise onto the column space of $X$ and its orthogonal complement are independent. Consequently ${\hat{\beta }}_0$ and the residual sum of squares are independent, andThus under the alternative,This is the noncentral t-distribution. The rejection region is fixed by the design, $\nu$ and the chosen level, while this probability distribution depends on $\beta$ only through ${\beta }_0$. Therefore the statistical power does not depend on the other regression coefficients, with the design and $\sigma$ fixed.

Use the usual normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design $X\in {\mathbb{R}}^{n\times p}$, $n>p$, and $\epsilon \sim N(0,{\sigma }^2{I}_n)$. These Gaussian distribution and constant-variance hypotheses are necessary for an exact finite-sample Student's t-test. Put $V=(X^{T}X{)}^{-1}$ and $\nu =n-p$. The vector of ordinary least squares estimators satisfiesWith $s^2=\Vert Y-X\hat{\beta }{\Vert }^2/\nu$, the test statistic isUnder the null hypothesis, $T$ has Student's t-distribution with $\nu$ degrees of freedom. A two-sided level-$5\%$ Student's t-test rejects when $|T|>t_{\nu ,0.975}$. If the intended test is one-sided against ${\beta }_0>0$, it rejects when $T>t_{\nu ,0.95}$; the statistical power conclusion below holds for either convention.

Let $A$ be the interior-grid matrix representing ${\Gamma }_9$, with zero boundary error. It is real symmetric and has the stated sine eigenbasis. Put $s={\sin }^2(k\pi h/2)$ and $t={\sin }^2(\ell \pi h/2)$; the negative of its eigenvalue isThis increases in each variable on $[0,1{]}^2$. With $s_1={\sin }^2(\pi h/2)\ge h^2$, the smallest absolute eigenvalue is at least $8s_1-(8/3)s_1^2\ge (16/3)h^2$. Hence its Euclidean norm operator norm satisfiesThe corrected scheme has an unscaled residual $r$ whose individual entries are uniformly $O(h^6)$. There are $m^2$ entries and $m<h^{-1}$, so $\Vert r{\Vert }_2=O(h^5)$. Subtracting the exact nodal equations from the numerical equations gives $Ae=-r$ (with sign depending on the residual convention), and thereforeChoose a constant slightly larger than the resulting bound to obtain the requested strict $\Vert e\Vert <c{h}^3$ for all sufficiently small $h$. Uniform smoothness is needed for an $h$-independent constant; unspecified smoothness only in the interior would not by itself guarantee it.

The five-point Laplacian applied to $f$ has expansionThe corrected right side $h^{2}f+(h^2/12){\Gamma }_{5}f$ cancels the $h^4\Delta f/12$ term in ${\Gamma }_{9}u$. ConsequentlyMore precisely, the unscaled residual's $h^6$ coefficient is $-(u_{xxxxxx}+u_{yyyyyy})/240+(u_{xxxxyy}+u_{xxyyyy})/144$. This fourth-order correction of the nine-point Poisson stencil does not require $f$ to be harmonic.

If $\Delta f=0$, then ${\Delta }^{2}u=0$, so the $h^4$ term in the previous expansion vanishes. Therefore the normalized local truncation error is $\boxed{O(h^4)}$, and the unscaled residual is $O(h^6)$. This is generally fourth order under the given harmonic-source condition; special solutions can have additional cancellation, but that is not guaranteed by this argument.

Assume sufficient smoothness up to the boundary for uniform Taylor remainders, for example $u\in C^8(\overline{\Omega })$ and $f=\Delta u\in C^6(\overline{\Omega })$. Direct Taylor expansion of the nine-point finite-difference stencil givesThus the normalized local truncation error $h^{-2}{\Gamma }_{9}u-f$ is $\boxed{O(h^2)}$. The unscaled residual ${\Gamma }_{9}u-h^{2}f$ is $O(h^4)$. Stating the normalization avoids ambiguity in the meaning of truncation order.

For the Einstein static universe with $P=0$ and ${\rho }_0>0$, the acceleration equation gives $\Lambda =4\pi {\rho }_0$. The first Friedmann equation then gives $k/a_0^2=4\pi {\rho }_0$, soKeep $\Lambda ,k$ fixed and use the specified perturbation closure $\delta P=\delta \rho /3$. Linearization of the constraint gives $\delta \rho =-3{\rho }_0\delta a/a_0$. In the acceleration equation the background terms cancel, leavingThusThe growing mode establishes linear instability of the static universe; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda =0$, which is not the positive-density static universe considered here.

For $k=1$, put $r=\sin \chi$. The spatial metric tensor becomesThe global closed spatial slice is a 3-sphere with $0\le \chi \le \pi$; the original $r$ chart covers only a hemisphere, so two charts are needed. A circle at fixed $\chi$ in an equatorial angular plane has circumference $\boxed{2\pi a\sin \chi }$. The full spatial volume isIntegrating a single $r\in [0,1]$ chart without its second copy would give only half this volume.