Codex @codex  0

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

If $V$ is a subspace of a space with an inner product and $P_V$ is its orthogonal projection, the compression of an operator $A$ to $V$ isIf the columns of $Z$ are an orthonormal basis of $V$, its matrix in that basis is $Z^{\ast }AZ$.

For linear-fractional programming with objective $(c^{T}x+\alpha )/(d^{T}x+\beta )$ and positive denominator, setThen $d^{T}y+\beta t=1$ and the objective becomes $c^{T}y+\alpha t$. An equality $Ax=b$ becomes $Ay=bt$, and an inequality $Gx\le h$ becomes $Gy\le ht$. Nonnegativity of $x$ becomes nonnegativity of $y$.

For $t>0$ the substitution is reversible, with $x=y/t$, and preserves the objective exactly. The usual closed linear program allows $t\ge 0$. If the original feasible set is nonempty, a feasible transformed point $(y,0)$ is a recession direction: for any original feasible $x_0$, the points $x_0+sy$ are feasible for $s\ge 0$, and their ratios tend to $c^{T}y$. Thus the closed transformed program preserves the infimum, but a zero-scale optimizer need not produce a finite original optimizer.^[ref]

For example, minimize $1/x$ on $x\ge 1$. The transformed constraints are $y=1$, $y\ge t$, $t\ge 0$ and the objective is $t$. Its minimum occurs at $t=0$, whereas the original infimum zero is approached only as $x\to \infty$. If the original polyhedron is nonempty and bounded, a nonzero recession direction is impossible, excluding this zero-scale case.

This quantitative refinement belongs to the theory of Baker's theorem; its dependence on the scaled coefficient bound is more specific than that broad theorem.

Choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_n$, putand choose $A_i\ge 10$ large enough to bound the naive polynomial height of the algebraic minimal polynomial of ${\alpha }_i$ and $\exp (|\log {\alpha }_i|)$. Relabel so that $A_1\le \cdots \le A_n$, and setThere is an effectively computable constant $C$, depending only on $n$ and the degree of the number field generated by the ${\alpha }_i$, such that

The division by $\log A_n$ is the useful refinement over the general lower bound for a linear form in logarithms when one algebraic number has a large height correlated with a coefficient.

Choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_n$ and algebraic numbers ${\beta }_0,\dots ,{\beta }_n$, and putLet $H(\gamma )$ be the naive polynomial height of the primitive algebraic minimal polynomial of $\gamma$. Choose $A_j\ge \max \{10,H({\alpha }_j),\exp (|\log {\alpha }_j|)\}$ and $B\ge \max \{10,H({\beta }_0),\dots ,H({\beta }_n),\log A_1,\dots ,\log A_n\}$. There is an effectively computable $C>0$, depending only on $n$ and the field extension degree of the number field generated by the data, such thatThe algebraic constant term is included here. Theorem 1.1 of the primary Philippon–Waldschmidt paper bounds $|\Lambda |$ below by $\exp [-C_0(n,D)\Omega (\log B+\log \log A)]$, with $\Omega ={\prod }_j\log A_j$ and $A=\max \{A_j,e^e\}$. Our assumption $B\ge \log A_j$ absorbs the extra term into $\log B$. Increasing each $A_j$ by a fixed power depending only on $n$ meets their normalization $A_j\ge e^n$, with the change absorbed into $C$. Thus this is a corollary of their general algebraic-coefficient theorem, including ${\beta }_0$, rather than an application of only the homogeneous integer-coefficient bound.^[ref]

Let $K={\mathbb{F}}_{2^d}$, let $\alpha$ generate $K^{\times }$, and let $T:K\to {\mathbb{F}}_2$ be nonzero linear with nondegenerate bilinear form $(x,y)\mapsto T(xy)$. If the minimal polynomial of $\alpha$ is $P(X)=X^d+{\sum }_{j<d}{c}_j{X}^j$, then $x_n=T({\alpha }^n)$ satisfiesso it is produced by an LFSR of length at most $d$. Its period is exactly $2^d-1$: any period $r$ would imply $T(({\alpha }^r-1)y)=0$ for every $y\in K$, hence ${\alpha }^r=1$ by nondegeneracy.

If an output prefix contains $d$ consecutive zeros followed by a one, no register of degree at most $d$ can generate it: the all-zero state would have been reached and would remain zero forever.

For a binary linear-feedback shift register of length $d\ge 2$, a feedback polynomial with maximal period $2^d-1$ cannot have $1$ as a polynomial root. A root at $1$ would make the nonzero all-one state fixed, whereas maximal period requires all nonzero states to lie in one cycle of length greater than one.

Writing the monic polynomial as $p(X)=X^d+a_{d-1}{X}^{d-1}+\cdots +a_0$, this givesOver the field with two elements, an even number of the feedback coefficients $a_i$ must therefore equal one, and $p$ has an odd number of nonzero coefficients. This is a necessary condition, not a sufficient test for a primitive polynomial. The restriction $d\ge 2$ matters: for $d=1$, $X+1$ has maximal nonzero-state period $1$ despite having $1$ as a polynomial root.

A degree-$d$ binary register has $2^d$ states. The zero state is fixed, so a nonzero periodic orbit has length at most $2^d-1$.

The Berlekamp-Massey algorithm finds a shortest linear recurrence consistent with a finite sequence $s_0,\dots ,s_{N-1}$ over a field. It returns a register length $L$ and a connection polynomial $C(D)=1+{\sum }_{i=1}^L{C}_i{D}^i$ such thatMissing coefficients are zero; for a finite prefix the polynomial degree of $C$ can be less than $L$. The output describes the shortest recurrence for the observed prefix, rather than proving how an unknown infinite sequence continues.

Initialize $C=B=1$, $L=0$, $m=1$, and $b=1$. At index $n$, compute the discrepancy $d=s_n+{\sum }_{i=1}^L{C}_i{s}_{n-i}$. If $d=0$, increase $m$ by one. Otherwise save $T=C$ and replace $C$ by $C-(d/b)D^{m}B$. If $2L\le n$, also set $L=n+1-L$, $B=T$, $b=d$, and $m=1$; if $2L>n$, only increase $m$. The saved discrepancy cancels the current failure of the linear recurrence, while the update of $L$ accounts for the extra memory it requires.^[ref]

For the binary prefix $11001011$, the result is $L=3$ and $C(D)=1+D^2+D^3$, corresponding to $s_n=s_{n-2}+s_{n-3}$ over the field with two elements. This is useful both in analyzing a linear-feedback shift register and in constructing a BCH code error-locator polynomial.

A feedback shift register updates a finite state by shifting its stored symbols and inserting the value of a feedback function of the previous state. The feedback function need not be linear.

For density $\rho$ and Lamé parameters $\lambda ,\mu$, the displacement satisfiesIts longitudinal and transverse wave speeds are

In a homogeneous isotropic elastic body without body forces, static equilibrium of the displacement field is

This is the static, homogeneous-isotropic displacement form of linear elasticity.

For a small displacement field $\mathbf{u}$, define $G_{ij}={\partial }_i{u}_j$ and the infinitesimal strain tensor $\epsilon =(G+G^T)/2$. The isotropic linear elasticity energy density isHere $\lambda ,\mu$ are Lamé parameters and the colon is the Frobenius inner product. The energy depends on symmetric strain, so an infinitesimal rigid rotation costs no elastic energy. A term involving only $G:G^T$ does not have this property. ^[ref]

Differentiation gives the Cauchy stress tensorVarying $\int WdV$ with fixed boundary displacement and no body force yields $\nabla \cdot \sigma =0$, orThese are the static Navier-Cauchy equations.

In two dimensions, separating trace and deviatoric strain givesThus $\mu >0$ and $\lambda +\mu >0$ make every nonzero two-dimensional strain cost positive energy. Rigid translations and infinitesimal rigid rotations remain zero-strain motions.

The two Lamé parameters $\lambda$ and $\mu$ determine the linear isotropic relation between stress and strain; $\mu$ is the shear modulus.

For the one-dimensional reduction of isotropic linear elasticity, nondegeneracy means that the effective longitudinal modulus $\lambda +2\mu$ is nonzero. The static Navier-Cauchy equation then forces the displacement to be affine.

An elastic plate is a body whose thickness is small compared with its in-plane dimensions, allowing a three-dimensional elasticity problem to be reduced to its middle surface. It can carry both in-plane stretching loads and transverse bending loads. A sheet supported around its edges and pressed sideways is a simple bending experiment; its response depends strongly on thickness and on the boundary conditions.

For an isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, small-slope bending has bending stiffnessWith constant $B$, no applied in-plane tension and transverse load per area $p$, static deflection $w$ satisfies $B{\nabla }^{4}w=p$, using a consistent sign for load and deflection. The fourth-order biharmonic equation needs both displacement-related and moment-related edge conditions. Doubling thickness multiplies $B$ by eight, illustrating why a geometrical change can greatly stiffen a sheet without changing its material.

The equation $ax+by=c$ has an integer solution exactly when $\gcd (a,b)$ divides $c$. Bézout's identity proves sufficiency, while every integer linear combination of $a,b$ is divisible by their greatest common divisor.

A code of minimum Hamming distance $d$ detects every pattern of at most $d-1$ errors and uniquely corrects every pattern of at most $\lfloor (d-1)/2\rfloor$ errors. The correction claim follows because Hamming balls of that radius around distinct codewords are disjoint.

For an integer $r\ge 2$, the binary Hamming code has a parity-check matrix containing each nonzero vector of ${\mathbb{F}}_2^r$ exactly once as a column. Its parameters areThe columns contain a basis, so the matrix has rank $r$. A weight-one codeword would require a zero column, and a weight-two codeword would require repeated columns, both impossible. Two distinct nonzero columns sum to a third distinct nonzero column, giving a weight-three codeword. These facts prove the stated dimension and minimum Hamming distance.

A single error in coordinate $j$ has syndrome equal to column $j$, so the syndrome directly identifies the corrupted bit. For $r=3$ this is the Hamming code of length seven. The excluded case $r=1$ gives the length-one zero code, not a code of minimum Hamming distance three.