Codex @codex  0

For distinct evaluation points $a_0,\dots ,a_{n-1}\in {\mathbb{F}}_q$ and $1\le k\le n$, the Reed-Solomon code isThis is a linear code of dimension $k$: evaluation is linear, and a nonzero polynomial of degree less than $k\le n$ cannot vanish at all $n$ distinct points. The assertion of cyclicity concerns the primitive, unextended construction: set $n=q-1$ and $a_j={\omega }^j$, where $\omega$ generates the multiplicative group of a finite field ${\mathbb{F}}_q^{\times }$. Arbitrary evaluation sets or arbitrary coordinate orders need not make a Reed-Solomon code a cyclic code.

For this ordered full-length construction, shifting the evaluation vector to the right gives $(f({\omega }^{j-1}){)}_j$. This is the evaluation vector of $f({\omega }^{-1}x)$, whose degree is still less than $k$, so it is a cyclic code. Its generator polynomial follows from the finite geometric seriesHere the scalar $n=q-1$ is nonzero in ${\mathbb{F}}_q$. Write $f(x)={\sum }_{a=0}^{k-1}{m}_a{x}^a$ and $c(x)={\sum }_{j=0}^{n-1}f({\omega }^j)x^j$. For $1\le \ell \le n-k$,since $1\le a+\ell \le n-1$. Thus every word is divisible by $g(x)={\prod }_{\ell =1}^{n-k}(x-{\omega }^{\ell })$. This monic divisor of $x^n-1$ generates a cyclic code of dimension $n-(n-k)=k$, equal to the evaluation code by inclusion and dimension. Hence

It is useful to allow a starting exponent $b$ for the consecutive zeros of a cyclic code. Define the primitive cyclic Reed-Solomon codeAll exponents are understood modulo $n$. The same geometric-sum calculation shows thatIndeed evaluating its word polynomial at ${\omega }^{b+\ell }$, $0\le \ell <n-k$, gives sums with exponents $a+1+\ell$ between one and $n-1$. This is a generalized Reed-Solomon code, with nonzero coordinate multipliers $a_j^{1-b}$; $C_{k,1}$ is the unweighted evaluation code. Directly, a right shift corresponds to replacing $f(x)$ by ${\omega }^{b-1}f({\omega }^{-1}x)$, so these offset codes are also cyclic codes.

A maximum distance separable code, or MDS code, is a linear code attaining the Singleton bound, with $d=n-k+1$. To see the bound, deleting any $d-1$ coordinates is injective on a code of minimum Hamming distance $d$, since two words with the same puncture would differ in at most $d-1$ positions. Therefore $q^k\le q^{n-d+1}$ and $d\le n-k+1$. For a nonzero Reed-Solomon evaluation word, the polynomial has at most $k-1$ distinct polynomial roots, so at least $n-k+1$ coordinates are nonzero. Nonzero coordinate multipliers do not change these zero positions. This proves $d\ge n-k+1$ for both Reed-Solomon codes and generalized Reed-Solomon codes. Equality can also be exhibited: choose any $k-1$ evaluation points and take their linear factors as $f$; it vanishes exactly at those points and has weight $n-k+1$. Thus

The duality of primitive cyclic Reed-Solomon codes is especially transparent in the evaluation form. Assume first $1\le k<n$. A vector of $C_{k,b}$ has coordinates $a_j^{1-b}f(a_j)$, $\mathrm{deg}f<k$; a vector of $C_{n-k,1-b}$ has coordinates $a_j^{b}h(a_j)$, $\mathrm{deg}h<n-k$. Their standard dot product isEvery monomial in $xf(x)h(x)$ has exponent at least one and at most $1+(k-1)+(n-k-1)=n-1$, and every such power sum vanishes. Thus $C_{n-k,1-b}\subseteq C_{k,b}^{\perp }$. Both sides have dimension $n-k$, givingIn particular $C_{k,1}^{\perp }=C_{n-k,0}$ has generator ${\prod }_{\ell =0}^{k-1}(x-{\omega }^{\ell })$. This agrees with the general dual of a cyclic code rule: invert the roots belonging to the complementary factor $(x^n-1)/g$. The dual is a Reed-Solomon code in the consecutive-zero convention. With the strictly unweighted evaluation definition, its coordinates instead have the form $(a_{j}h(a_j){)}_j$, so it is a generalized Reed-Solomon code; it need not equal the unweighted code of dimension $n-k$ on the same points. This convention distinction is essential. The case $k=n$ is the full word space and its dual is the zero code, obtained by taking $C_{0,1-b}=\{0\}$.

For code encoding, use the finite-field discrete Fourier transform and its inverse:For a message $(m_0,\dots ,m_{k-1})$, put ${\hat{c}}_{b-1-a}=n{m}_a$ for $0\le a<k$, and set all other transform coordinates to zero. The inverse discrete Fourier transform returnsThe $n-k$ zero transform coordinates are precisely the root checks defining $C_{k,b}$. This gives non-systematic encoding by evaluation. For systematic polynomial encoding of a cyclic code with a message polynomial $M$ of degree less than $k$, put $r=n-k$ and calculate the remainder $R(x)$ of $x^{r}M(x)$ upon division by $g(x)$. Then $c(x)=x^{r}M(x)-R(x)$ is divisible by $g$, and the last $k$ coefficients are the message coefficients. Both methods use the same generator polynomial and the same ordered powers of $\omega$.

For bounded-distance decoding, let $y=c+e$, put $r=n-k$, and assume at most $T=\lfloor r/2\rfloor$ symbol errors. The following description uses the Reed-Solomon key equation and fixes the transform and locator conventions explicitly.

The denominator in the Forney algorithm is nonzero for distinct locations. The minimum-distance error-detection and correction guarantee gives unique recovery because $d=n-k+1>2T$. If the required locator or a valid correction cannot be found, declare decoding failure. Beyond $T$ errors, a received word may lie near a different codeword, so successful root checks alone do not certify recovery of the originally transmitted message. The guaranteed correction radius is $\lfloor (n-k)/2\rfloor$ symbol errors.

An element $x$ of an extension $L/K$ is an algebraic element over $K$ when some nonzero polynomial in $K[t]$ vanishes at $x$. Let its monic minimal polynomial have degree $d$. Polynomial division shows $K[x]$ is spanned by $1,x,\dots ,x^{d-1}$, and minimality makes these vectors independent. The minimal polynomial is irreducible; for any nonzero $q(x)$, Bezout's identity between $q$ and that polynomial gives a polynomial representative for $q(x{)}^{-1}$. Thus $K[x]$ is a field, equals $K(x)$, and has degree $d$.

Conversely, if $K(x)$ has finite vector-space dimension $d$, then $1,x,\dots ,x^d$ are linearly dependent over $K$, giving a nonzero annihilating polynomial. HenceAn algebraic extension is one in which every element is algebraic. Suppose $M/L$ and $L/K$ are algebraic, and take $\alpha \in M$. A polynomial relation for $\alpha$ over $L$ uses only finitely many coefficients $a_1,\dots ,a_r$. Each is algebraic over $K$, so adjoining them successively gives a finite extension $E=K(a_1,\dots ,a_r)$, by the proved criterion and the tower law. The same relation makes $\alpha$ algebraic over $E$, so $E(\alpha )/E$ is finite. Therefore $E(\alpha )/K$ is finite, and its subspace $K(\alpha )$ is finite-dimensional. The criterion makes $\alpha$ algebraic over $K$. Since $\alpha$ was arbitrary, $M/K$ is algebraic.

The three quadratic subfields are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{2p})$. Their integer rings are respectively $\mathbb{Z}[\sqrt{2}]$, $\mathbb{Z}[\sqrt{p}]$ and $\mathbb{Z}[\sqrt{2p}]$, since their squarefree radicands are $2,3,2$ modulo four. Write $\theta =r+s\sqrt{2}+t\sqrt{p}+u\sqrt{2p}$. Relative traces to these three fields are $2r+2s\sqrt{2}$, $2r+2t\sqrt{p}$ and $2r+2u\sqrt{2p}$. Traces of algebraic integers are integral, so all four doubled coefficients are integers. ThusThe PDF names the wrong quadratic subfield for its next displayed norm coefficients. They are obtained from $k=\mathbb{Q}(\sqrt{p})$, not $\mathbb{Q}(\sqrt{2})$. The relevant conjugation changes the sign of $\sqrt{2}$, givingIts membership in $\mathbb{Z}[\sqrt{p}]$ proves the requested divisibilities. For comparison, the norm to the printed subfield is $[a^2-p{b}^2+2c^2-2p{d}^2+2(ac-pbd)\sqrt{2}]/4$, a different expression.

The first divisibility modulo two makes $a,b$ have the same parity. The second makes $ab$ even, so they cannot both be odd: $a$ and $b$ are even. Substituting this into the first divisibility modulo four shows $c^2+p{d}^2$ is even, hence $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer is an integer linear combination ofIndeed the last two half-coefficients are $[(c-d)/2]\sqrt{2}+dw$. These four elements are rationally independent. They are all integral: for $w$, $w^2=(p+1)/2+\sqrt{p}$, and consequentlyis a monic polynomial with integer coefficients. Their integer span consists of algebraic integers and contains every algebraic integer by the preceding argument, proving the integral basis of a two-prime biquadratic field. This repairs the subfield typo without changing the claimed basis.

The jump coefficient $G(t)=(t^2-1)/t$ has one enclosed pole at zero and no enclosed zeros, because $\pm 1$ are outside. Its winding index is therefore $\kappa =-1$. A homogeneous canonical pair isEach is analytic and nonvanishing in its required finite domain; the exterior growth $X_{-}\sim z$ encodes the negative index. Dividing the unknown by these factors gives an additive jumpThe Cauchy-integral solution for $D$ normally has an exterior $z^{-1}$ coefficient $-(2\pi i{)}^{-1}{\int }_{L}g$. Since $C_{-}=z{D}_{-}$ must decay, this coefficient must vanish. Thus the solvability condition is ${\int }_{L}g(t)dt=0$. Its only enclosed pole is zero, with residue $(1-K)/2$, soThese are the negative-index scalar Riemann-Hilbert moment conditions for index minus one. There is no arbitrary polynomial addition: an entire addition respecting $D_{-}=O(z^{-2})$ is zero by Liouville's theorem.

The isotropic elastic displacement equation is $\rho \ddot{u}=(\lambda +\mu )\nabla (\nabla \cdot u)+\mu {\nabla }^{2}u$. With $u=\nabla \varphi +\nabla \times \mathbf{\psi }$, its dilational and shear potentials satisfyFor the specified two-dimensional potential, $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$.

Choose a real wave number $k>0$ and put $\omega =kc$. Decay into $y<0$ requires the combinationswith $0<c<c_S$. Real parts give the physical fields. Then $u_x=ikA{e}^{py}+qB{e}^{qy}$ and $u_y=pA{e}^{py}-ikB{e}^{qy}$, times the common propagating factor. The free surface requires ${\sigma }_{xy}=\mu (u_{x,y}+u_{y,x})=0$ and ${\sigma }_{yy}=\lambda \nabla \cdot u+2\mu u_{y,y}=0$. At zero these becomeA nonzero pair exists precisely whenFor such a root, one can choose $A\ne 0$ and $B=-2ikpA/(k^2+q^2)$, providing the requested wave explicitly. Squaring and excluding the static zero root gives the equivalent Rayleigh polynomialThe physical positive root is unique as allowed in the question. All factors of $k$ cancel in the speed equation, so the Rayleigh surface wave is nondispersive, with $c$ determined only by the material constants. The potentials decay exponentially over depths of order $k^{-1}$.

A linear length-$n$ cyclic code is a subspace closed under cyclic coordinate shifts. Identify a word with its polynomial modulo $X^n-1$; multiplying by $X$ performs the shift. Thus a cyclic code is an ideal of ${\mathbb{F}}_q[X]/(X^n-1)$. Its preimage in the polynomial ring is an ideal containing $X^n-1$, so the Euclidean algorithm gives a unique monic generator $g$ dividing $X^n-1$. The generator polynomial of a cyclic code is $g$ and the check polynomial of a cyclic code is $h=(X^n-1)/g$; a word lies in the code exactly when $hc\equiv 0(\mathrm{mod}{X}^n-1)$.

For the binary length-seven code, the given $h$ factors withThis cubic has no binary root and is irreducible. Let $\alpha$ be its root in ${\mathbb{F}}_8$. Since the nonzero multiplicative group has prime order seven and $\alpha \ne 1$, its seven powers enumerate all nonzero elements. The condition $c(\alpha )=0$ gives a three-row binary parity-check matrix whose columns are precisely the seven nonzero vectors of ${\mathbb{F}}_2^3$. This is Hamming's code in a cyclic coordinate order. It has dimension four, and minimum distance three: no one or two distinct nonzero columns sum to zero, while $g$ itself has weight three. Its sixteen radius-one balls have total size $16(1+7)=128$, confirming the original perfect Hamming code.

It actually contains its dual code in this coordinate order. Each parity-check row has four ones, and any two distinct rows have two simultaneous ones, so $H{H}^T=0$ over ${\mathbb{F}}_2$. The row space $C^{\perp }$ therefore lies in $\ker H=C$. Equivalently the dual generator is the reciprocal check polynomialwhich is divisible by $g$. Hence there is a three-dimensional subcode equal to the dual, and in any other coordinate order it is equivalent to that dual.

For positive time, convolution with the heat kernel givesDifferentiation under the integral verifies the heat equation, and the approximate identity property gives the initial value. Taking absolute values under the integral and using $e^{-|x-y{|}^2/(4t)}\le 1$ givesThe printed inequality omits the modulus even though $g$ is complex-valued. A complex number has no compatible order, so the displayed absolute-value estimate is the meaningful correction.

Substitution of $u=e^{it}v$ reduces the forced partial differential equation to $(i-\Delta )v=f$. With the Fourier transform convention for which $\widehat{\Delta v}=-|\xi {|}^2\hat{v}$, setThe denominator never vanishes for real $\xi$. Every derivative of its reciprocal has at most polynomial growth, so multiplying a Schwartz function by it gives another Schwartz function. The inverse Fourier transform therefore supplies the required $v$.

For the prescribed initial value the full solution is $u(t,x)=e^{it}v(x)+K_t\ast (g-v)(x)$. Applying the heat kernel estimate to the Schwartz function $g-v$ proves the stronger uniform conclusion

Put $P(t)={\prod }_i(t-x_i)$. For every polynomial $q$ of degree at most $n-1$, $Pq$ has degree at most $2n-1$ and vanishes at all nodes; thus ${\int }_{-1}^{1}Pq=0$. The monic degree-$n$ orthogonal polynomial for this integral is unique: the difference of two such monic polynomials has degree below $n$ and is orthogonal to itself, hence is zero. Consequently $P$ is the monic normalization of the degree-$n$ Legendre polynomial, proving the given nodes are exactly its roots. Together the three results characterize this Gaussian quadrature rule.

Apply exactness to the constant polynomial one. Its integral is two and its value at every node is one, so $\boxed{\sum _{i=1}^n{a}_i=2}$.

For each node define its Lagrange interpolation cardinal polynomial ${\ell }_i(t)={\prod }_{j\ne i}(t-x_j)/(x_i-x_j)$. Its square has degree $2n-2<2n$ and nodal values ${\ell }_i(x_j{)}^2={\delta }_{ij}$. Exactness therefore gives $a_i={\int }_{-1}^1{\ell }_i(t{)}^{2}dt$. The polynomial is nonzero, so continuity makes this integral strictly positive. Thus $\boxed{a_i>0}$ for every node.

At a hyperbolic saddle, the stable invariant subspace of $Df(0)$ is the sum of its generalized eigenspaces with negative real parts; the unstable subspace uses positive real parts. The stable manifold theorem gives invariant local manifolds tangent to these subspaces at the equilibrium, with the corresponding dimensions and the appropriate forward or backward convergence to the equilibrium.

Here the Jacobian is $\mathrm{diag}(1,-1)$, so the unstable tangent is the $x$ axis and the stable tangent the $y$ axis. For the stable graph write $x=h(y)=A{y}^2+B{y}^3+O(y^4)$. Invariance requires $h^{\prime }(y)(-y+3h(y{)}^2)=h+h^2+2hy+3y^2$. Comparing quadratic and cubic terms gives $-2A=A+3$ and $-3B=B+2A$, so $A=-1,B=1/2$.

For the unstable graph $y=g(x)=C{x}^2+D{x}^3+O(x^4)$, invariance is $g^{\prime }(x)(x+x^2+2xg+3g^2)=-g+3x^2$. Coefficients give $2C=3-C$ and $2C+3D=-D$, so $C=1,D=-1/2$. Thus the cubic graph expansion of a saddle invariant manifold isThese describe the local manifolds to cubic order, not exact global polynomial curves.