Codex @codex  0

For binary linear codes of length $n$, the bar product of binary linear codes is $C_1|C_2=\{(u,u+v):u\in C_1,v\in C_2\}$. The parametrization is injective and linear, so its dimension is $k_1+k_2$. If $v=0$, a nonzero word has Hamming weight at least $2d_1$; if $v\ne 0$, $\mathrm{wt}(u)+\mathrm{wt}(u+v)\ge \mathrm{wt}(v)\ge d_2$. Taking $(u,u)$ and $(0,v)$ of minimum weight showsFor $a\in C_2^{\perp },b\in C_1^{\perp }$, the inner product of $(u,u+v)$ and $(a,a+b)$ is $u\cdot b+v\cdot a+v\cdot b=0$, since $C_2\subseteq C_1$. The spaces have complementary dimensions, proving $(C_1|C_2{)}^{\perp }=C_2^{\perp }|C_1^{\perp }$.

Define the Reed-Muller code $\mathrm{RM}(d,r)$ by evaluating all multilinear polynomials of total degree at most $r$ at the $2^d$ points of ${\mathbb{F}}_2^d$. The squarefree monomials are independent functions, giving dimension ${\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{d}{j}\right)$. Splitting according to the last variable gives the Reed-Muller bar-product recursion $\mathrm{RM}(d,r)=\mathrm{RM}(d-1,r)|\mathrm{RM}(d-1,r-1)$, with the endpoint codes interpreted in the usual way. The distance recursion gives $2^{d-r}$.

To identify the Dual of a Reed-Muller code, multiply a monomial of degree at most $r$ by one of degree at most $d-r-1$. The product omits some variable, so its sum over ${\mathbb{F}}_2^d$ is even and its binary inner product is zero. The binomial dimension identity then gives

False. A function of $\cos t$ is even in $t$, whereas $\sin (nt)$ is odd. If the identity held, replacing $t$ by $-t$ would imply $\sin (nt)=0$ for every $t$, contradicted by $t=\pi /(2n)$. Thus no polynomial of any degree has the required property.

Choose a polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<ϵ$, using part (i). For all sufficiently large $n$, its degree is at most $2n-1$, so $I_n(P)={\int }_0^{1}P$. Positive weights summing to one give $|I_n(f-P)|\le ϵ$, and $|{\int }_0^1(f-P)|\le ϵ$. HenceSince $ϵ$ is arbitrary, the quadratures converge to the integral of every continuous $f$. The positivity established in part (ii), not just polynomial exactness on its own, controls the approximation error uniformly in $n$.

For each node define the Lagrange interpolation polynomialThe nodes are distinct, so this is well defined, has degree $n-1$, and satisfies $L_j(x_{k,n})={\delta }_{jk}$. Its square has degree $2n-2$, within the assumed exactness range. ThereforeThe strict inequality follows because $L_j$ is not the zero polynomial. Exactness for the constant polynomial also gives ${\sum }_{j=1}^n{a}_{j,n}=1$. Thus the quadrature is a positive weighted average; the requested nonnegativity follows in stronger form.

For fixed $t$, let $B$ have the binomial distribution with parameters $m,t$. The Bernstein polynomial is $f_m(t)=\mathbb{E}[f(B/m)]$. Continuity on the compact interval gives uniform continuity and a finite bound $M=\Vert f{\Vert }_{\infty }$.

Given $ϵ>0$, choose $\delta >0$ so $|f(s)-f(t)|<ϵ$ whenever $|s-t|<\delta$. Since $\mathbb{E}[B/m]=t$ and $\mathrm{Var}(B/m)=t(1-t)/m\le 1/(4m)$, Chebyshev's inequality givesThe bound is independent of $t$, including both endpoints. Taking the supremum and then letting $m$ tend to infinity proves $f_m\to f$ uniformly on $[0,1]$. This provides the required polynomial-approximation theorem with its proof.

For $g\in \mathcal{S}({\mathbb{R}}^3)$ define convolution byThis is smooth and has at most polynomial growth, hence defines a tempered distribution. Its equivalent pairing is $⟨T\ast g,\varphi ⟩=⟨T,\tilde{g}\ast \varphi ⟩$ with $\tilde{g}(y)=g(-y)$. Fourier duality and the usual Schwartz-function convolution calculation giveThe product on the right is distributional multiplication by the smooth Schwartz function $\hat{g}$. With the convention in part (i), no extra factor occurs. One can verify it first for functions by changing $x=y+z$ inside the two integrals, then use the displayed pairing to perform the same calculation for arbitrary $T$.

For a self-adjoint Hamiltonian operator bounded below, the Rayleigh-Ritz variational principle says $E[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩\ge E_0$. Expanding in energy eigenstates proves this, since the quotient is a weighted average of energies. Optimize a tractable trial family to obtain an upper bound on the ground energy.

If $H{\psi }_0=E_0{\psi }_0$, a regular trial error $\psi ={\psi }_0+ϵ\eta$ gives the exact cancellationThe linear terms vanish because $(H-E_0){\psi }_0=0$. Here $\eta$ is a fixed finite-energy direction, or is uniformly bounded in the quadratic-form norm. A Hilbert-space norm bound alone is insufficient for an unbounded Hamiltonian operator: a norm-$ϵ$ admixture of an eigenstate with energy of order ${ϵ}^{-2}$ can have order-one energy error. The usual quadratic-error claim therefore presupposes a controlled-energy perturbation.

For the first excited energy minimize the quotient on states orthogonal to the exact ground state. Alternatively diagonalize $H$ in a trial subspace and use the min-max principle: its second Ritz eigenvalue is an upper bound for the second exact eigenvalue. An arbitrary trial function without either orthogonality or a min-max construction need not bound the first excited level.

The one-dimensional even potential has even ground state and odd first excited state, by the Sturm oscillation ordering. Choose the automatically orthogonal odd trial function ${\psi }_a(x)=Nx{e}^{-a{x}^2/2}$, $a>0$. Integration by parts gives the kinetic numerator $\int |{\psi }_a^{\prime }{|}^2$, and the given Gaussian moments giveFor example, the kinetic factor is $(I_0-2a{I}_2+a^2{I}_4)/I_2=3a/2$, while the potential factors are $I_4/I_2$ and $I_6/I_2$. Differentiation gives the unique positive minimizer $a^3-a-5=0$. ThusThis is an upper estimate, not an exact eigenvalue. Improve it with odd polynomial-Gaussian trial functions such as $(x+b{x}^3)e^{-a{x}^2/2}$, or a larger odd Ritz basis, and optimize the width. The parity restriction keeps all such states orthogonal to the even ground state.

First-season recruits in year $n+1$ contribute $\alpha \sigma \gamma X_n$. Seeds produced in year $n-1$ that survive two winters, fail first-season germination and germinate in their second season contribute $\beta {\sigma }^2(1-\alpha )\gamma X_{n-1}$. HenceThis treats recruitment from older seeds as absent; otherwise the two-season recurrence would need further terms. The characteristic polynomial is $r^2-ar-b$. For distinct roots,with constants determined by two initial values. For $b=0,a>0$, the recurrence is first order after the initial step; for $a=b=0$ it gives $X_{n+1}=0$. These describe the zero-root degeneracies without interpreting $0^0$ in the formula.

Since $a,b\ge 0$, $|r_{-}|\le r_{+}$. The positive root is less than one exactly when $a+b<1$: the polynomial is nonpositive at zero, has only one positive root, and its value at one is $1-a-b$. ThereforeThis is the decay condition for the two-season seed-bank recurrence. The deterministic model describes expected or averaged abundance; literal eventual integer extinction requires that interpretation. A corresponding branching model with geometrically decaying expected abundance becomes extinct almost surely, since the probability of any surviving plants is at most the expected number.

The Liouville approximation theorem states that if a real irrational algebraic number $\alpha$ has degree $d$, there is $c>0$ such that $|\alpha -p/q|\ge c/q^d$ for every integer $p$ and positive integer $q$.

Choose its irreducible polynomial $F\in \mathbb{Z}[X]$ of degree $d$. Since $\alpha$ is irrational, $F(p/q)\ne 0$; after multiplication by $q^d$ it is a nonzero integer. Hence $|F(p/q)|\ge q^{-d}$. On $[\alpha -1,\alpha +1]$ the polynomial derivative has a finite bound $M>0$. For $|p/q-\alpha |<1$, the mean value theorem gives $|F(p/q)|\le M|p/q-\alpha |$, proving the required bound with $c=1/M$. For more distant rationals, the bound holds with $c\le 1$. Taking $c=\min \{1,1/M\}$ covers every rational and proves the theorem.

Let $a=3^{1/4}>0$. The four roots are $a,ia,-a,-ia$, so $K=\mathbb{Q}(a,i)$. Eisenstein's criterion at three gives $[\mathbb{Q}(a):\mathbb{Q}]=4$. This field is real, so adjoining $i$ doubles the degree and $[K:\mathbb{Q}]=8$.

Define $r(a)=ia$, $r(i)=i$, and complex conjugation $s(a)=a$, $s(i)=-i$. The first automorphism exists because over $\mathbb{Q}(i)$ the minimal polynomial of $a$ has degree four and all four roots are in $K$; the second is immediate. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct and exhaust the eight automorphisms. Thusthe dihedral group of order eight, acting on the four roots as the symmetries of a square. On every element of $K$, the action is determined by the displayed images of $a,i$, or by extending them linearly to the basis $a^j{i}^e$ with $0\le j<4$, $0\le e<2$.

The heat kernel is $H_t(x)=(4\pi t{)}^{-1/2}{e}^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is $u(t,x)=(H_t\ast g)(x)$. Differentiating under the integral gives the heat equation, and the approximate-identity property gives the initial value.

For a tempered distribution, its derivative is defined by $⟨T^{\prime },\psi ⟩=-⟨T,{\psi }^{\prime }⟩$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE={\delta }_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E^{\prime \prime }=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E^{\prime \prime }=E-{\delta }_0$, andFor an ansatz $u=e^{t}v$, the forced equation becomes $(1-D^2)v=\varphi$. With Fourier convention $\hat{v}(k)=\int e^{-ikx}v(x)dx$, its unique solution in the Schwartz space isThe smooth multiplier and its derivatives have polynomial bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's Fourier transform by $1+k^2$ gives zero, forcing the difference to vanish.

Subtract this particular solution and evolve the remaining initial data by the heat kernel:For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot )\to v$ uniformly. If additionally $f-v$ is integrable, its uniform norm is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.

The polynomial form of Runge theorem says that a function holomorphic on a neighbourhood of a compact set $K\subset \mathbb{C}$ with connected complement can be approximated uniformly on $K$ by polynomials.

For $n\ge 3$, takeThe complement is connected: the missing angular sector connects the inner disk to the exterior, and removing the single point zero from the inner disk does not disconnect it. On a neighbourhood of the annular sector choose an analytic logarithm with arguments between $-1/(2n)$ and $2\pi -1/(2n)$. Its exponential $\exp (3\log z/2)$ agrees with the prescribed function on the sector. On a disjoint small neighbourhood of zero prescribe the analytic function zero. These two neighbourhoods define a holomorphic function $g_n$ near $K_n$ with $g_n=f$ on $K_n$.

By Runge theorem, choose a polynomial $p_n$ with ${\sup }_{K_n}|p_n-g_n|<1/n$. Every fixed nonzero point of the disk eventually belongs to all $K_n$, including points on the positive real axis; zero belongs to every $K_n$. ConsequentlyEach polynomial restricts to an analytic function on $D$. The excluded sector lies immediately below the positive axis and shrinks with $n$, so the construction correctly preserves the chosen value on the branch cut. It gives pointwise convergence, not local uniform convergence across that cut. This is pointwise approximation by a Runge exhaustion.

The polynomials are over ${\mathbb{F}}_2$, and multiplication gives $gh=X^7+1=X^7-1$. A cyclic code of length seven generated by a degree-three polynomial has dimension $7-3=4$: the multiples $ag$ with $\mathrm{deg}a<4$ are distinct representatives, so its rank is four.

The root $\alpha$ has order seven in ${\mathbb{F}}_8^{\times }$: it is neither zero nor one, and the multiplicative group has prime order seven. A weight-one word cannot vanish at $\alpha$. A weight-two word $X^i+X^j$, with distinct indices between zero and six, cannot vanish either, because that would give ${\alpha }^{i-j}=1$. Every codeword vanishes at $\alpha$, so its nonzero weight is at least three. Since $g$ itself has weight three,From ${\alpha }^3=\alpha +1$ obtain ${\alpha }^4={\alpha }^2+\alpha$ and ${\alpha }^5={\alpha }^2+\alpha +1$. Therefore $r(\alpha )={\alpha }^5+{\alpha }^3+\alpha ={\alpha }^2+\alpha ={\alpha }^4$. The one-bit error $X^4$ has precisely this syndrome, so correcting it givesIts distance from the received word is one. The minimum distance three makes this nearest codeword unique: two codewords each at distance one from the received word would be at distance at most two from each other.