Codex @codex  0

For length-$n$ binary linear codes $C_2\subseteq C_1$, their bar product of binary linear codes isThe map $(u,v)\mapsto (u,u+v)$ is injective and linear, soIf $d_i$ is the minimum distance of $C_i$, thenFor $v\ne 0$ this is at least $d_2$; for $v=0$ a nonzero word has weight at least $2d_1$. Words $(0,v)$ and $(u,u)$ attain these bounds, proving

A parity-check matrix $P$ for a code $C$ is a matrix satisfying $C=\ker P$. A pair $(x,y)$ belongs to $C_1|C_2$ exactly whenThus a parity-check matrix isThis is the parity-check matrix of a bar product.

The Reed-Muller code $\mathrm{RM}(d,r)$ is the length-$2^d$ binary code obtained by evaluating every square-free polynomial in $d$ Boolean variables of degree at most $r$ on all points of ${\mathbb{F}}_2^d$. Equivalently,The square-free monomials form a basis, soA word in $\mathrm{RM}(d,1)$ evaluates an affine function $a_0+a\cdot x$. The two constant functions have weights $0$ and $2^d$. Every nonconstant affine function takes each value equally often, so all other codewords have weight $2^{d-1}$.

A monomial of degree $j$ has weight $2^{d-j}$. Hence every generator, and therefore every codeword, has even weight when $r<d$. For $r=d$, the monomial $x_1\cdots x_d$ has weight one. All words have even weight exactly for $0\le r\le d-1$.

Gaussian elimination without pivoting givesThus $\det A=3(\lambda -20)$ and the solution is unique exactly when $\lambda \ne 20$.

For $\lambda =20$, forward substitution in a triangular system in $Ly=b$ gives $y=(1,1,3,\mu -10{)}^T$. Consistency of $Ux=y$ requires $\boxed{\mu =10}$. backward substitution in a triangular system then gives

Taking the divergence of the Ampère-Maxwell equation and using the Gauss's law $\nabla \cdot E=\rho /{\epsilon }_0$ giveswhich is charge conservation from Maxwell equations ${\rho }_t+\nabla \cdot J=0$.

By Ohm's law, $J=\sigma E$, so $\nabla \cdot J=\sigma \rho /{\epsilon }_0$. Thereforewith charge-relaxation rate $\sigma /{\epsilon }_0$ and time ${\epsilon }_0/\sigma$.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrödinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Separation of variables with $u=X(x)Y(y)$ and the three homogeneous boundary conditions gives $X_n=\sin (n\pi x)$ and $Y_n=\sinh (n\pi y)$. The boundary value at $y=1$ contains only the third Fourier sine series mode. ThereforeThis harmonic function has all four prescribed boundary values.

The Bolzano-Weierstrass theorem in $\mathbb{R}$ says that every bounded real sequence has a convergent subsequence. For a bounded sequence in ${\mathbb{R}}^n$, successively extract subsequences on which the first, second, and so on through the $n$th coordinate converge. The final diagonal subsequence converges in Euclidean space, proving the theorem in ${\mathbb{R}}^n$.

Put $K=D\setminus {\bigcup }_{z\in S}{B}_{\epsilon }(z)$. Suppose the asserted $\delta$ did not exist. Then there would be $x_k\in D$ and $y_k\in K$ with $\Vert x_k-y_k\Vert <1/k$ but $|f(x_k)-f(y_k)|\ge \epsilon$. Since $K$ is closed and bounded, Bolzano-Weierstrass gives a subsequence $y_k\to y\in K$, and then $x_k\to y$. The point $y$ is not in $S$, because otherwise $y\in B_{\epsilon }(y)$, so $f$ has continuity at $y$. Both $f(x_k)$ and $f(y_k)$ tend to $f(y)$, a contradiction. Hence the required uniform $\delta >0$ exists.

A cut is a partition $(S,T)$ of the vertices with the source in $S$ and sink in $T$; its capacity is the sum of capacities of directed edges from $S$ to $T$. The max-flow min-cut theorem says that maximum flow value equals minimum cut capacity. With integral capacities, there is an integral max-flow theorem attaining the maximum.

Construct a bipartite graph with one vertex for each row and column and an edge $r_i\to c_j$ whenever $A_{ij}=1$. Add source-to-row and column-to-sink edges of capacity $1$, and give row-to-column edges infinite capacity. An integral flow is precisely a set of independent $1$s, so its maximum value is the largest such set.

A finite-capacity cut places some rows on the sink side and some columns on the source side; these lines cover every $1$, since an uncovered row-to-column edge would cross the cut with infinite capacity. Its capacity is the number of selected lines. Conversely every line cover defines such a cut. Max-flow min-cut therefore proves the maximum number of independent $1$s equals the minimum number of covering lines, the matrix form of Kőnig's theorem for bipartite matching.

A quadratic form on a finite-dimensional real vector space $V$ is a function $q(v)=B(v,v)$ for some symmetric bilinear form $B$. A positive-definite quadratic form satisfies $q(v)>0$ for every $v\ne 0$.

Completing squares givesSet $u=x+y$, $v=y+z$, $w=z$. The corresponding ordered basis in the original coordinates isand the diagonal form is $u^2+v^2+2w^2$. All diagonal coefficients are positive, so the form is positive definite.