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For a real central potential and real $k$, radial-current conservation gives $S_l(k{)}^{\ast }{S}_l(k)=1$. Invariance of the radial equation under $k\mapsto -k$ gives $S_l(k)S_l(-k)=1$, so a continuous phase convention has ${\delta }_l(-k)=-{\delta }_l(k)$.

For a program code $e$, input $x$, and time bound $t$, the truncated computation simulates program $e$ on $x$ for at most $t$ steps and reports either its output or that no halt occurred within the bound. This is a total computable function of $(e,x,t)$.

A total computable function is a computable function whose domain is the entire input space.

Applying the Parseval identity to the repeated exponential convolution gives

On the unit sphere with Riemannian metricfix a latitude $0<\theta <\pi$ and follow increasing $\varphi$ through one circuit. In the orthonormal frame $e_{\theta }={\partial }_{\theta }$, $e_{\varphi }=(\sin \theta {)}^{-1}{\partial }_{\varphi }$, the components $u,v$ of a parallel tangent vector satisfyFor the initial tangent vector with coordinate components $(V^{\theta },V^{\varphi })=(1,0)$, solving this constant-coefficient ordinary differential equation and converting $v$ back to a coordinate component givesThe parallel transport is a rotation through $-2\pi \cos \theta$ in the orthonormal frame. Modulo $2\pi$, this equals the area $2\pi (1-\cos \theta )$ of the polar cap, as predicted by the Gauss-Bonnet theorem. The coordinate formula excludes the poles because ${\partial }_{\varphi }$ degenerates there.

On a homogeneous background with $X=\bar{X}$, a $P(X,\varphi )$ perturbation has sound speedThe denominator is its time-kinetic coefficient and $P_X$ is its spatial-gradient coefficient.

A connection 1-form records how a chosen local frame changes under a covariant derivative. If $e_b$ is the local frame and ${\theta }^a$ its dual coframe, thenThus evaluating ${\omega }^a{}_b$ on a tangent vector gives the mixing of vectors of the local frame under differentiation in that direction. The coefficients depend on the chosen local frame; their curvature 2-form describes the geometrical obstruction to locally consistent parallel transport around a loop.

For the Levi-Civita connection, the torsion tensor vanishes, giving Cartan's first structure equationIn an orthonormal frame, metric compatibility also gives ${\omega }_{ab}=-{\omega }_{ba}$ after lowering the first index with the metric tensor in that local frame. In a general local frame the metric components need not be constant, so this simple antisymmetry is not an unrestricted statement about every connection matrix.

A Jacobi matrix is a real symmetric tridiagonal matrix with positive off-diagonal entries. Its diagonal entries $b_j$ and off-diagonal entries $a_j>0$ encode the three-term recurrence of orthonormal polynomials:where $p_{-1}=0$ and the $a_0{p}_{-1}$ term is omitted. In the orthonormal polynomial basis this is the matrix of multiplication by $x$, truncated to finitely many basis vectors. Positivity of $a_j$ excludes a reducible split into independent blocks.^[ref]

The determinant $D_n(x)=\det (xI-J_n)$ of the leading $n$-by-$n$ block satisfies $D_{n+1}=(x-b_n)D_n-a_n^2{D}_{n-1}$, with $D_0=1$. Thus its eigenvalues are the zeros of orthogonal polynomials of degree $n$. For unit weight on $[-1,1]$, the first two normalized polynomials give $J_2=\left(\begin{array}{cc}0& 1/\sqrt{3}\\ 1/\sqrt{3}& 0\end{array}\right)$, whose eigenvalues $\pm 1/\sqrt{3}$ are the two-node Gaussian quadrature nodes.

A real degree-$n$ orthogonal polynomial for a positive weight on an interval, with the required finite moments, has exactly $n$ distinct simple zeros in the interval's interior. This places the quadrature nodes of Gaussian quadrature inside its integration domain.^[ref]

To prove it, list the distinct interior roots where the polynomial $p_n$ changes sign, and let $q$ be the product of their linear factors. Then $p_{n}q$ has one constant sign on the interval and is not identically zero. If there were fewer than $n$ sign-changing roots, $q$ would have lower polynomial degree, so orthogonality would give $\int p_{n}qwdx=0$, contradicting positivity of the integral after a choice of sign. There must therefore be $n$ such roots. They exhaust the polynomial degree, proving both simplicity and the absence of additional roots.

The physicists' Hermite polynomials areThey are orthogonal on the real line with weight $e^{-x^2}$. The first examples are $1$, $2x$ and $4x^2-2$; they solve the Hermite differential equation $H_n^{\prime \prime }-2x{H}_n^{\prime }+2n{H}_n=0$.

Multiplication by a Gaussian factor converts them into quantum harmonic-oscillator eigenfunctions. The probabilists' normalization instead has weight $e^{-x^2/2}$ and a rescaled variable. Stating the convention prevents factors of two from being hidden in the name. The polynomials are named after Charles Hermite.^[ref]

A monic orthogonal polynomial has leading coefficient one and is orthogonal to every lower-degree polynomial in the chosen positive weighted inner product. Suppose the measure has finite moments of the required orders and its inner product is positive definite on polynomials. Subtracting the orthogonal projection onto a finite-dimensional subspace of $x^n$ onto the polynomials of degree below $n$ constructs the degree-$n$ monic orthogonal polynomial.

It is unique: two candidates differ by a lower-degree polynomial $r$ orthogonal to itself, forcing $r=0$ by positive definiteness. With unit weight on $[-1,1]$, the first three are $1,x,x^2-1/3$, since ${\int }_{-1}^1{x}^{2}dx/{\int }_{-1}^{1}1dx=1/3$. A measure on too few support points can make the inner product degenerate and prevent this construction at higher degrees.^[ref]

A matrix is skew-symmetric when $A^T=-A$.

The special orthogonal group isEvery element of $SO(2)$ is a rotation in the plane. Every element of $SO(3)$ fixes an axis and restricts to a planar rotation on its perpendicular plane.

A real square matrix is orthogonal when $R^{T}R=I$, equivalently when its columns form an orthonormal basis.

An orientation of a surface induces the boundary direction used in Stokes theorem. Walking in the positive boundary direction keeps an upward-oriented surface on the left.

A multiplicative set $S$ in a ring $R$, with $1\in S$ and $0\notin S$, is left Ore when, for every $s\in S$ and $r\in R$, there are $s^{\prime }\in S$ and $r^{\prime }\in R$ such that $s^{\prime }r=r^{\prime }s$.

The ordinary generating function of the sequence $a_n=\lfloor n/2\rfloor$ isThe floor function takes the same value $k$ at $n=2k$ and $n=2k+1$. Grouping those adjacent terms giveswhich simplifies to the displayed expression. The middle identity follows by differentiating the geometric series. The calculation is valid as an identity of formal power series, or analytically for $|x|<1$.

Expanding begins $x^2+x^3+2x^4+2x^5+3x^6+\cdots$, making the paired values of the sequence visible in the coefficients.

A first integral is a function of time, state, and derivatives that remains constant along every solution.

A linear ordinary differential equation is linear in its unknown function and all its derivatives. Its order-$n$ scalar form iswhere the coefficients and forcing are prescribed functions of the independent variable. They need not be constant: $y^{\prime }+ty=\sin t$ is linear, whereas $y{y}^{\prime }+y=0$ is nonlinear. On an interval where $a_n\ne 0$ and the normalized coefficients and forcing are continuous, $n$ initial values determine a unique solution throughout that interval.^[ref]

If $L$ denotes the linear differential expression, $L(\alpha u+\beta v)=\alpha Lu+\beta Lv$. Thus the homogeneous solutions form a vector space, while the solutions for fixed nonzero forcing form an affine translate $y_p+\ker L$ of that space. This is why a fundamental system of solutions and one particular solution suffice. For $y^{\prime \prime }+y=1$, every solution is $1+A\cos t+B\sin t$; substitution verifies all terms. Initial conditions choose $A,B$, whereas boundary conditions can produce compatibility restrictions or nonuniqueness.

A differentiable real function satisfyingon an interval has $f(0)=0$ and $f^{\prime }=C(1+f^2)$, hence $f(x)=\tan (Cx)$ wherever the tangent remains finite.