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For the first assertion, fix $x_0\in [0,1]$ and $\epsilon >0$. By uniform convergence, choose $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x$. By continuity of $f_N$ at $x_0$, there is $\delta >0$ such that $|f_N(x)-f_N(x_0)|<\epsilon /3$ whenever $|x-x_0|<\delta$. The triangle inequality then givesThus a uniform limit of continuous functions is continuous, including at the endpoints with their relative neighborhoods.

The converse implication involving integrals is false. Set $f_1=0$ and, for $n\ge 2$, use the shrinking continuous spikes with vanishing integralTheir supports are $[1/n-1/n^2,1/n+1/n^2]$. For each $x>0$ they eventually vanish at $x$, and they vanish at zero for every $n$. Hence $f_n\to 0$ pointwise, with continuous limit. Each triangular area is $1/n^2$, so ${\int }_0^1{f}_n\to 0={\int }_0^{1}0$. Yet $\sup |f_n|=1$, excluding uniform convergence.

The final assertion is also false. Take $f_n(x)=x^n$. Then the pointwise limit isThis limit is discontinuous at one but is Riemann integrable, with integral zero. Each $f_n$ is continuous and ${\int }_0^1{f}_n=1/(n+1)\to 0$. Thus convergence of the integrals to the integral of an integrable pointwise limit does not enforce its continuity.

Sylvester's law of inertia states that a real quadratic form can be brought by an invertible real change of coordinates towith $r$ remaining zero coordinates, and that the integers $(p,q,r)$ are uniquely determined. To prove existence, represent it by a symmetric matrix. The allowed orthogonal diagonalization yields real eigenvalues, and scaling each nonzero eigen-coordinate by the square root of its absolute eigenvalue changes its coefficient to $+1$ or $-1$.

For uniqueness, $p$ is the largest dimension of a subspace on which the form is positive definite. The positive coordinate subspace attains $p$. Any subspace of dimension greater than $p$ intersects the $(q+r)$-dimensional span of the negative and zero coordinates nontrivially, by the dimension formula. A nonzero vector in that intersection has nonpositive quadratic value, contradicting positive definiteness. Similarly, $q$ is the maximal dimension of a negative-definite subspace. These descriptions are independent of coordinates, and $r=\dim V-p-q$ is then fixed. This proves existence and uniqueness, including degenerate forms.

For the final application, nonsingularity gives $p+q=2m$. In diagonal coordinates write $v=(v_{+},v_{-})$, with $Q(v)=|v_{+}{|}^2-|v_{-}{|}^2$. If $v\in U$ projects to zero in the positive coordinates, $Q(v)=0$ forces $v_{-}=0$, so projection $U\to {\mathbb{R}}^p$ is injective. The same argument applies to the negative coordinates. Hence $m\le p$ and $m\le q$, and their sum is $2m$. ThereforeThe inertia signature is $(m,m)$, or signature difference $p-q=0$ if that convention is used. This is the isotropic dimension bound from real inertia.

For a finite directed flow network with nonnegative finite edge capacities, a source $s$ and a distinct sink $t$, the max-flow min-cut theorem states that the largest feasible flow value equals the smallest capacity of a cut separating $s$ from $t$. A feasible flow has $0\le f_e\le c_e$ on each directed edge and flow conservation at all other vertices. Its value is the net outflow at $s$.

For any cut $S$ containing $s$ but not $t$, sum conservation over vertices of $S$. Internal edges cancel, givingThus every flow value is bounded by every cut capacity.

A maximum flow exists: the feasible set is nonempty, closed and bounded in a finite-dimensional edge-coordinate space, hence compact, and the flow value is continuous. For such a flow, form its residual network: each original edge has forward residual capacity $c_e-f_e$ and backward residual capacity $f_e$. Keep these as edge-labelled arcs if the original network already has edges in both directions. If a directed positive-residual path joined $s$ to $t$, increasing the flow along its forward arcs and decreasing it along its backward arcs by the path's positive bottleneck capacity would preserve feasibility and increase the value. That contradicts maximality.

Let $S$ be the vertices reachable from $s$ by positive residual arcs. Then $t\notin S$. Every original edge from $S$ to $S^c$ is saturated, otherwise its forward residual arc would extend reachability. Every original edge from $S^c$ to $S$ has zero flow, otherwise its backward residual arc would extend reachability. ThereforeTogether with the bound for every flow and cut, this provesThe compactness step makes the proof valid for arbitrary real capacities; it does not assume that a particular augmenting-path algorithm terminates after finitely many augmentations.

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$, $Q={\sum }_i(X_i-\bar{X}{)}^2$, and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal likelihood is proportional to $({\sigma }^2{)}^{-n/2}\exp [-{\sum }_i(X_i-\mu {)}^2/(2{\sigma }^2)]$. Its unrestricted maximizers are $\hat{\mu }=\bar{X}$, ${\hat{\sigma }}^2=Q/n$. Under the null the variance maximizer is $Q_0/n$. Substitution gives the generalized likelihood-ratio test statisticThe ratio decreases as $|T|$ increases, so the likelihood-ratio rejection region is two-sided in $T$. Under $H_0$, $\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal and $Q/{\sigma }^2$ is independent chi-squared with $n-1$ degrees of freedom. Independence follows by resolving the isotropic Gaussian sample vector into its projection on the constant vector and its orthogonal complement. Hence $T$ has the exact Student t-distribution with $n-1$ degrees of freedom.

For a chosen significance level $\alpha$, compute the sample mean and the sample standard deviation with denominator $n-1$, and reject precisely whenEquivalently, report the two-sided $p$-value $2[1-F_{t_{n-1}}(|T|)]$. This requires $n\ge 2$ and positive population variance; $S>0$ holds almost surely under that model. The nuisance variance is removed by the exact null distribution, so one should not substitute a normal critical value or a large-sample chi-squared approximation. This is the normal-mean likelihood ratio with unknown variance.

The first matrix column is already the unit vector $e_1$, so the first QR elimination needs no reflection. In the remaining three coordinates, the second column is $v=(2,2,1{)}^T$, with norm three. To map it to $(-3,0,0{)}^T$, use the Householder transformationIt is symmetric and orthogonal and satisfies $Hv=(-3,0,0{)}^T$. Extend it to $Q^T=\mathrm{diag}(1,H)$. The QR decomposition and transformed right side areFor the lower right-side block, $w^T(1,4,-1{)}^T=12$, so its transformed value is $(1,4,-1{)}^T-(12/15)w=(-3,12/5,-9/5{)}^T$.

Orthogonality preserves the least-squares residual norm. Thus the first two squared residual terms can and must be set to zero, while the last two are independent of the unknowns:The minimum squared residual is $(12/5{)}^2+(-9/5{)}^2=9$. The solution is unique since the triangular factor has nonzero diagonal entries. No normal equations are used.

Use the position four-vector $X^a=(ct,\mathbf{x})$ and Minkowski signature $(+,-,-,-)$. Along a massive particle's timelike path,Thus its four-velocity is $U^a=d{X}^a/d\tau =\gamma (c,\mathbf{v})$, and multiplying by the invariant rest mass givesIts temporal component is $E/c$, and the invariant relation is $(E/c{)}^2-|\mathbf{p}{|}^2=M^2{c}^2$.

For a massless particle, use the null energy–momentum relation rather than substituting $M=0$ into $M\gamma$: a massless particle has no proper-time parametrization of a timelike rest frame. With positive energy, $p^a{p}_a=0$ implies $E/c=|\mathbf{p}|$. HenceThe original PDF retains the spatial vector in this expression; the TeX conversion incorrectly replaced it by its magnitude.

The partial fraction decomposition is $f(z)=1/(z-i)+1/(z+i)$. Put $w=z-1$ and expand each reciprocal by a geometric series:Adding gives the Taylor seriesEquivalently, since $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$, its coefficient is $(-1{)}^n{2}^{(1-n)/2}\cos ((n+1)\pi /4)$. The first terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$.

The nearest nonremovable poles are $i$ and $-i$, both at distance $\sqrt{2}$ from the center. The function is holomorphic throughout the open disk of that radius and cannot be holomorphic across either pole. Therefore the largest radius of convergence is $r=\sqrt{2}$.

For a geodesic spherical triangle on a sphere of radius $R$, the Gauss-Bonnet theorem giveswhere $\alpha ,\beta ,\gamma$ are its interior angles. Indeed, its geodesic sides have zero geodesic curvature, its Gaussian curvature is $R^{-2}$, and its three exterior turning angles sum to $3\pi -(\alpha +\beta +\gamma )$.

Choose a point strictly inside a full-dimensional convex polyhedron and radially project its boundary onto the unit sphere. Triangulate each projected face by diagonals, using only its existing vertices. Convexity ensures that the resulting spherical triangles cover the sphere once, with disjoint interiors. A face with $k$ edges gives $k-2$ triangles, so their number is $T={\sum }_{\mathrm{faces}}(k-2)=2E-2F$, since every original edge belongs to two faces.

Sum the spherical excess formula over all triangles. The total area is $4\pi$, and the angles incident to each projected vertex sum to $2\pi$. No additional vertices were introduced by the face triangulations. HenceThis is the spherical-triangle proof of the polyhedron Euler formula.

In the given basis, the matrix is $A=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$. Let $\omega =e^{2\pi i/3}$, a primitive cube root of unity. For each $\xi \in \{1,\omega ,{\omega }^2\}$,Thus these are eigenvectors with distinct eigenvalues $1,{\omega }^2,\omega$, respectively, and form an eigenbasis. SetThe columns are mutually orthogonal with squared norm three, so $P^{-1}=P^{\ast }/3$. The diagonalization and characteristic polynomial areScaling $P$ by $1/\sqrt{3}$ gives a unitary diagonalizing matrix.