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A matching in a graph from $X$ to $Y$ is a collection of edges with disjoint endpoints covering every vertex of $X$; equivalently it chooses a distinct neighbour in $Y$ for each $x\in X$. For a finite bipartite graph, the Hall marriage theorem says this exists exactly when $|N(A)|\ge |A|$ for every $A\subseteq X$. Necessity follows by counting the distinct matched neighbours.

For sufficiency, induct on $|X|$. If a nonempty proper subset $A$ is tight, $|N(A)|=|A|$, Hall's condition holds on $(A,N(A))$. It also holds on $(X\setminus A,Y\setminus N(A))$: for $B\subseteq X\setminus A$, the original condition applied to $A\cup B$ gives $|N(B)\setminus N(A)|\ge |B|$. Induction provides two disjoint matching in a graph whose union covers $X$. If no nonempty proper subset is tight, choose an edge $xy$, possible by Hall's condition. Delete its endpoints. Each nonempty subset of $X\setminus \{x\}$ had at least one extra neighbour, so deleting $y$ leaves Hall's condition intact. Induction and then the edge $xy$ finish the proof. The single-vertex case is immediate.

For the degree condition, fix $A\subseteq X$ and double-count with weights:Every denominator is nonzero; the last inequality uses $d(y)\le d(x)$ along each edge. The Hall marriage theorem therefore proves the required matching exists. Finiteness, as in the usual course version of the theorem, is understood; the same proof covers a finite $X$ with arbitrary $Y$ and finite degrees.

Write $R_j=I-P_{-j}$. Minimizing the residual sum of squares first over the coefficients of $X_{-j}$ reduces the problem to minimizing $\Vert R_{j}Y-b{R}_j{X}_j{\Vert }^2$. Full column rank ensures $R_j{X}_j\ne 0$. Thus the partial regression formula and its variance areThis uses $R_j^T=R_j=R_j^2$ and the isotropic error covariance matrix ${\sigma }^{2}I$.

Since the span of $X_k$ is contained in the column space of $X_{-j}$, orthogonal projection onto the latter removes at least as much squared norm. Consequentlyand inversion proves the requested lower bound. Full rank makes the denominator positive.

Adding $c{1}_n$ to column $j\ge 2$ leaves the column space unchanged. Every old fitted vector is reproduced by keeping all slopes unchanged and replacing the regression intercept by ${\hat{\beta }}_1-c{\hat{\beta }}_j$. Uniqueness of the ordinary least squares estimators then proves that slope $j$ is unchanged. In particular all columns can be centered by such changes. Applying the previous projection argument to the centered design proves the bound with $X_j-{\bar{X}}_j{1}_n$ and $X_k-{\bar{X}}_k{1}_n$; the normalized inner product is now the usual sample correlation.

Each reported coefficient test is the two-sided Student's t-test of $H_0:{\beta }_j=0$ against $H_1:{\beta }_j\ne 0$, conditional on the other predictors. Under the null, ${\hat{\beta }}_j/(\hat{\sigma }\sqrt{(X^{T}X{)}_{jj}^{-1}})$ has a Student's t-distribution with $100-3=97$ degrees of freedom. At five percent the intercept is significant, but neither individual test slope is. The final F-test compares the full model against an intercept-only model: $H_0:{\beta }_{\mathrm{Maths}}={\beta }_{\mathrm{Stats}}=0$. It has $(2,97)$ degrees of freedom and rejects at five percent, since its $p$-value is $0.003166$. Thus the predictors jointly explain variation, without strong evidence separating their individual conditional effects.

The two test predictors have sample correlation $0.9371630$, although each has only moderate correlation with the final exam response. The variance inflation factor is $1/(1-0.9371630^2)\simeq 8.22$. Their nearly shared direction produces substantial multicollinearity, inflating the individual slope standard errors. The joint F-test can detect this common predictive direction even though either coefficient, given the other, has a large $p$-value.

For a simple continued fraction, defineMultiplying the matrices $\left(\begin{array}{cc}{a}_j& 1\\ 1& 0\end{array}\right)$ givesTheir determinants also give $p_n{q}_{n-1}-p_{n-1}{q}_n=(-1{)}^{n-1}$. If the infinite tail is $\beta =[a_{n+1},a_{n+2},\dots ]>a_{n+1}$, the same formula impliesThe denominators of these continued fraction convergents tend to infinity, since $q_{n+2}\ge q_{n+1}+q_n$. Hence they converge to $\theta$.

The square-root algorithm for the periodic continued fraction of a quadratic irrational, using $M^{\prime }=Da-M$, $D^{\prime }=(53-M^{\prime 2})/D$ and $a^{\prime }=\lfloor (7+M^{\prime })/D^{\prime }\rfloor$, givesThe first four continued fraction convergents are $7/1,22/3,29/4,51/7$. Direct substitution gives positive solutions to the requested Pell-type equations:

The flatness problem is the extreme closeness to spatial flatness required in the early universe if ordinary decelerating expansion is to produce the observed near-flat universe. From the Friedmann equation, with $c=1$ and curvature constant $k$,During decelerating expansion $|\Omega -1|$ grows. The Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G(\rho +3P)/3$. Thus $\rho +3P<0$ makes an expanding universe accelerate, increases $\dot{a}$ and suppresses $|\Omega -1|$: this is the inflationary solution of the flatness problem.

The slow-roll approximation neglects ${\dot{\varphi }}^2/2$ compared with $V$ and $\ddot{\varphi }$ compared with $3H\dot{\varphi }$. For the positive-field branch, taking $m>0$,Integrating the field equation and $\dot{a}/a=H$ gives the slow-roll solutionThe potential slow-roll parameters are ${ϵ}_V={\eta }_V=2/{\varphi }^2$. They cease to be small when $\varphi$ is of order $1$ in these units; the customary endpoint ${ϵ}_V=1$ is $\boxed{{\varphi }_{\mathrm{end}}\simeq \sqrt{2}}$. Equating the approximate kinetic and potential energies instead gives $\sqrt{2/3}$, another order-one indicator, not a precise endpoint of an approximation that is already failing.

Use equal small excisions on the two sides of each real pole. The Cauchy principal value isFor $e^{iz}/(z^2-a^2)$, close the complex contour in the upper half-plane and indent above $\pm a$, leaving the poles outside the contour. The residue theorem and the vanishing of the large semicircle integral give the principal-value integral as $i\pi$ times the sum of the two real-pole residues. Each clockwise indentation contributes $-i\pi$ times its residue. Sincethe sum is $i\sin a/a$. Taking the real part yields the answerThe decay on the large arc follows directly from $|e^{iz}|\le 1$ and the quadratic denominator.

There are $k{x}_n$ seeds and their germinating proportion is $e^{-\gamma x_n}$, giving the survival-augmented Ricker map with zero survival. A positive equilibrium point satisfies $e^{-\gamma x_{\ast }}=1/k$, soIt exists when $k>1$ and is strictly linearly stable when $|1-\log k|<1$. Thus the usual strict linear-stability interval is $\boxed{1<k<e^2}$.

With winter survival, surviving adults and new plants add:Writing $L=\log [k/(1-s)]$, the linearization of a dynamical system has multiplier $1-(1-s)L$. Hence the strict linear-stability interval becomesIts lower endpoint is less than $1$. Its upper endpoint exceeds $e^2$: for $t=1-s\in (0,1)$, $\log t+2/t$ is decreasing in $t$ and equals $2$ at $t=1$.

The nonlinear stability at the Ricker flip threshold needs a separate check. The upper endpoints have multiplier $-1$, so the strict linear test alone does not decide stability there. They are in fact locally asymptotically stable, with algebraic rather than geometric convergence. To see this, put $u=\gamma (x-x_{\ast })$ at $L=2/(1-s)$. The local map isand its second iterate is $u-2(s^2-s/2+1/6)u^3+O(u^4)$. The cubic coefficient in parentheses is positive for every $s\in [0,1)$. Thus, if “stable” includes nonlinear endpoint stability, the upper inequalities can be replaced by $\le$; the lower endpoints have no positive equilibrium.