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The Hermitian conjugate is $A^{†}={\overline{A}}^{,T}$. A matrix is unitary when $A^{†}A=A{A}^{†}=I$, and Hermitian when $A^{†}=A$.

The intermediate value theorem says that a continuous $f:[a,b]\to \mathbb{R}$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

Set $\varphi (a)=0$ and $\varphi (x)=\sin (1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\varphi$ take every value between $0=\varphi (a)$ and $\varphi (b)$, yet $\varphi$ is discontinuous at $a$.

A monotone function can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right limits would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda \in (l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original sequences.

The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2{a}^2)$.

The cosmological perfect-fluid continuity equation isDuring expansion, $\dot{a}>0$, and thereforeIf $C=\rho (t_0)a(t_0{)}^3$, then throughout the later expanding phaseMultiplying the Friedmann equation by $a^2$ givesThe right-hand side is negative oncewhich is impossible because ${\dot{a}}^2\ge 0$. Thus the scale factor cannot grow without bound.

In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the Friedmann acceleration equationThe curvature term in the first Friedmann equation also impliesWhile $a$ remains below the finite bound just found, $\ddot{a}$ is bounded above by a strictly negative constant. Hence $\dot{a}$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is recollapse of a closed Friedmann universe with nonnegative pressure.

Taking downward distance as positive, the Lagrangian is

At equilibrium the lower spring supports only the lower mass, while the upper spring supports both masses:Hence

Let $q_i=z_i-z_i^{(0)}$. The linear terms cancel by the equilibrium equations, and the quadratic Lagrangian iswith the small-oscillation mass and stiffness matrices

The generalized eigenvalue problem for small oscillations isIts characteristic equation issoCorresponding displacement eigenvectors are

They are not orthogonal in the ordinary Euclidean inner product. They are orthogonal in the kinetic-energy, or mass-matrix, inner product:This is mass-matrix orthogonality of normal modes.

The Papperitz symboldescribes the two-dimensional solution space of a second-order homogeneous Fuchsian differential equation on the Riemann sphere with exactly three regular singular points $z_1,z_2,z_3$. Near $z_i$, its two independent Frobenius behaviours are locallysubject to logarithmic modifications in resonant cases. The six characteristic exponents obey the Fuchs relation

For the Gauss hypergeometric equation, the exponents at infinity are $a$ and $b$. Put $w=z^{-1}$. Transforming the Papperitz symbol from $z$ to $w$ shows thatandare local solutions near $z=\infty$, with respective leading behaviours $z^{-a}$ and $z^{-b}$. When $a-b\notin \mathbb{Z}$, these behaviours are distinct and the two solutions are linearly independent. They therefore form a basis of the solution space on any simply connected common domain with compatible branch choices.

The solution $F(a,b;c;z)$ normalized at zero can be analytically continued into that domain. Since it solves the same second-order equation, it must be a constant linear combination of the basis at infinity:The constants depend on $a,b,c$ and the branch convention, but not on $z$. This is the hypergeometric connection formula at infinity.