Suppose an approximation has an error expansion $A(h)=I+c{h}^p+O(h^q)$ with $q>p$. With refinement factor $r>1$, combining the two approximations eliminates the leading term:If further expansion terms are known, repeat the elimination at their successive powers. For an even-power trapezium expansion and repeated halving, this gives the Romberg recurrence $R_{k,j}=(4^j{R}_{k,j-1}-R_{k-1,j-1})/(4^j-1)$, starting from $R_{k,0}=T(2^{-k})$.

For the stated cubic the one-panel trapezium value is $T(1)=[0^3+1^3]/2=1/2$. The two-panel value isThe leading power is two, so one extrapolation step givesThis equals the exact integral $[x^4/4{]}_0^1=1/4$; for a cubic the trapezium error has only the leading $h^2$ term, which this combination removes exactly.

For a complex matrix $C=(c_{ij})$, Gershgorin's theorem puts every eigenvalue in the union of the row discs $D_i=\{z:|z-c_{ii}|\le R_i\}$, where $R_i={\sum }_{j\ne i}|c_{ij}|$. To prove it, choose an eigenvector $v$ for an eigenvalue $\lambda$ and an index $i$ maximizing $|v_i|>0$. Its $i$th equation givesDivision proves $\lambda \in D_i$.

The usual counting refinement says that a connected component made of $k$ discs contains exactly $k$ eigenvalues, with algebraic multiplicity. For completeness, deform the matrix from its diagonal part to the original by multiplying all off-diagonal entries by $t\in [0,1]$. Its discs have radii $t{R}_i$ and stay inside the original discs. Continuous variation of the roots of the characteristic polynomial keeps the number inside any contour separating one original component from the others constant. At $t=0$, the diagonal entries give exactly $k$ roots there, proving the refinement.

For the application put $s=-\mu >0$. The matrix isIts three row discs have positive real centres and leftmost real coordinatesAll are strictly positive, so zero lies in none of the discs. It cannot be an eigenvalue. Thus $A-\mu B$ is nonsingular for every real $\mu <0$.

A $100(1-\alpha )\%$ confidence interval procedure produces random endpoints $L,U$ such that ${\mathbb{P}}_{\mu }(L\le \mu \le U)=1-\alpha$, under the model. In everyday terms, repeating the same sampling and interval construction many times would give intervals containing the true mean in about $100(1-\alpha )\%$ of repetitions.

For $n\ge 2$, define the sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ and the sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\bar{X}{)}^2$. For independent observations from a normal distribution with ${\sigma }^2>0$, the standard sampling results areThe first follows from the independent normal sample mean and sample variance. Let $q=t_{n-1,1-\alpha /2}$ be the positive Student quantile. Symmetry gives the central confidence intervalIts width is $W=2qS/\sqrt{n}$, so

States $i,j$ communicate when each is accessible from the other: $(P^m{)}_{ij}>0$ and $(P^n{)}_{ji}>0$ for some nonnegative integers $m,n$. The communicating classes areIndeed $1$ and $4$ move only between themselves; $2\to 5\to 3\to 2$ links the second class, but it has exits to $1,4,6$ and no return from the closed classes; state 6 is absorbing.

Let $h_i$ be the probability of ever reaching 6 from state $i$. Then $h_1=h_4=0$, $h_6=1$, and first-step conditioning givesThus $h_3=h_2=3h_5/4$ and $3h_5=h_3+1$, giving $h_5=4/9$. Therefore

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ where finite. For independent $X,Y$, independence factors the expectation, giving $M_{X+Y}(t)=M_X(t)M_Y(t)$ wherever both factors are finite.

For the given probability density function and $t<1$, put $s=(1-t)x$ to obtainHere $\Gamma (n)=(n-1)!$, obtained by repeated integration by parts. For $t\ge 1$ the defining integral diverges.

Each $Y_i$ has moment-generating function $(1-t{)}^{-1}$, so independence givesThe uniqueness theorem for moment-generating functions applies when the functions are finite on a neighborhood of zero. Hence the sum has the same distribution as $X_n$, a Gamma distribution of shape $n$ and rate one, with the displayed probability density function in the source.

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$. Its complex velocity is $u-iv=dw/dz$. Hereso the far-field motion has velocity $(U,0)$ and speed $U$. In polar coordinates the stream function, shifted by a harmless constant, isIt is zero on $r=a$, proving the cylinder is a streamline.

Stagnation points are zeros of $dw/dz$, hence roots of $U{z}^2-i\Gamma z/(2\pi )-U{a}^2=0$. Put $\gamma =\Gamma /(4\pi Ua)\in (0,1)$. The two roots areBoth have modulus $a$ and are distinct. Their angles satisfy $\sin \theta =\gamma$, namely $\arcsin \gamma$ and $\pi -\arcsin \gamma$. The tangential velocity on the cylinder is $-2U\sin \theta +\Gamma /(2\pi a)$, confirming the same zeros. Streamlines are level curves of $\psi$; the positive circulation places both stagnation points on the upper side and increases the flow speed below the cylinder.

For a material fluid volume, mass conservation yields ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$. Constant mass density reduces this to $\nabla \cdot \mathbf{u}=0$. For inviscid flow, the Cauchy stress tensor is $-pI$; its net surface force is the volume integral of $-\nabla p$ by the divergence theorem. Newton's second law for each material volume therefore gives $\rho D\mathbf{u}/Dt=-\nabla p$. Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$, these are exactly the two stated equations.

In two dimensions they readDifferentiate the second momentum equation in $x$ and subtract the $y$ derivative of the first. The pressure terms cancel. For $\omega =v_x-u_y$, collecting terms givesIncompressibility removes the last term, givingThus each fluid parcel retains its scalar vorticity as it moves. In particular an initially irrotational flow remains irrotational in this smooth two-dimensional inviscid evolution. There is no vortex stretching in this two-dimensional setting.

The time-independent Schrödinger equation isPut $\alpha =m\omega /\hslash$. For $\psi =e^{-\alpha x^2/2}$, differentiation gives ${\psi }^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\psi$. Substitution cancels the terms in $x^2$ and leaves $E_0={\hslash }^2\alpha /(2m)=\hslash \omega /2$.

The Gaussian integral gives $\int e^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$. Hence a normalized wave function isMultiplication by any constant of modulus one gives the same quantum state. The position probability density is $|{\psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$. Its first moment vanishes by oddness. Integration by parts gives $\int x^2{e}^{-\alpha x^2}dx=(2\alpha {)}^{-1}\int e^{-\alpha x^2}dx$, with the boundary term zero. Therefore

Polar coordinates give speed squared ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so unit mass has kinetic energy $T=\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$. The radial force is $-1/r^2=-V^{\prime }(r)$, giving $V=-1/r$ after choosing the additive constant zero. Therefore $L=T-V$ is the stated expression.

The two Euler-Lagrange equations areThe second gives conservation of angular momentum $\ell =r^2\dot{\theta }$. The mechanical energy isDirect differentiation expresses its derivative asThus energy is conserved for the motion wherever $r>0$.

For fixed endpoint values, the Euler-Lagrange equation isIt follows by differentiating the functional along $y+\epsilon \eta$ with $\eta (0)=\eta (\pi )=0$, integrating the $F_{y^{\prime }}{\eta }^{\prime }$ term by parts and applying the fundamental lemma of the calculus of variations.

For this quadratic integrand it becomes $y^{\prime \prime }+ky=0$. If $k>0$, write $\nu =\sqrt{k}$; the condition at zero gives $y=B\sin (\nu x)$, and that at $\pi$ requires $B\sin (\nu \pi )=0$. Consequently $y=0$ is the only extremal when $\sqrt{k}$ is not an integer. If $\sqrt{k}=n$ is a positive integer, all extremals areThe integer case $n=0$ must be handled separately: $y^{\prime \prime }=0$ with both endpoints zero has only $y=0$. If negative real $k$ is allowed, the hyperbolic-function solutions also yield only zero under these endpoints. These statements concern stationary functions; no minimum is assumed.

Continuity means that the inverse image of every open subset of $Y$ is open in $X$. If $x\in \overline{A}$ and $V$ is an open neighborhood of $f(x)$, then $f^{-1}(V)$ is an open neighborhood of $x$, so it meets $A$. Thus $V$ meets $f(A)$. The closure characterization givesIf $B$ is empty the assertion is immediate. Otherwise, if $B$ is bounded, choose a centre $p$ and $R<\infty$ with $d(p,b)\le R$ for all $b\in B$. The closed ball $\{y:d(p,y)\le R\}$ is closed because $|d(p,y)-d(p,z)|\le d(y,z)$. It therefore contains $\overline{B}$. Hence the closure of a bounded set is bounded. Empty sets cause no exception. If $f(A)$ is bounded, its closure is bounded by this result, and the displayed inclusion makes $f(\overline{A})$ bounded too.

A topology on $X$ contains $\varnothing ,X$, is closed under arbitrary unions and under finite intersections. The cofinite family has these properties: a nonempty union contains one cofinite member and therefore has finite complement; the complement of a finite intersection is a finite union of finite sets. The empty union and the sets $X,\varnothing$ are included explicitly.

Any two nonempty open sets intersect, because their intersection has finite complement in the infinite set $X$. Thus no two disjoint nonempty open sets can separate $X$, proving connectedness. For compactness, take any open cover and choose a nonempty member $U$. Only finitely many points lie outside $U$. Choose one covering member for each of those points. Together with $U$ these give a finite subcover. Thus $X$ is compact.