Codex @codex  0

Use the conditional second-moment matrix $V_t=\mathbb{E}[P_t{P}_t^{\top }\mid {\mathcal{F}}_{t-1}]$ from the PDF. By part (b), the filtration is finite-state at every finite time. On a time-$(t-1)$ atom let $p_1,\dots ,p_m$ be the successor price vectors and $q_1,\dots ,q_m>0$ their conditional probabilities. Then $m\le n$ andPositive definiteness gives rank $n$, so $m\ge n$. Hence $m=n$, and the $n$ successor vectors have linear independence.

We use the finite-horizon fundamental theorem of asset pricing in deflator form: an arbitrage-free finite market admits a strictly positive adapted process $D$, with $D_0=1$, such that $DP$ is a martingale. Equivalently, on every step,Fix a finite horizon containing the step in question. On the parent atom, write $d_i=D_t/D_{t-1}>0$ on successor $i$ and $p_0=P_{t-1}$. ThenThe proposed values $z_i=p_i^{\top }{V}^{-1}{p}_0$ satisfy precisely the same equation:Because the $p_i$ have linear independence and the $q_i$ are positive, this linear system has a unique solution. Thus $z_i=d_i>0$, provingThis is the positive regression deflator in a complete finite market. It also proves the suggested conclusion: with $Y_0=1$ and $Y_t={\prod }_{u=1}^t{Z}_u$,Hence $Y$ is a strictly positive martingale deflator. Finite-state structure makes these expectations integrable on each finite horizon. Positive definiteness alone would not ensure positivity of the regression factor; market completeness and the positive deflator are essential.

Use the convention that the inner product is conjugate-linear in its first argument; the real case is the same without conjugates. Linear independence makes the Gram matrix $G$ positive definite, so $B=G^{-1}$ exists and is Hermitian. Define the Riesz dual basis in an inner product space within ${\mathcal{U}}_n$ byThen $(u_i,{\hat{u}}_k)={\sum }_l{G}_{il}{b}_{lk}={\delta }_{ik}$. Its own Gram entries areHence the inverse Gram matrix is the Gram matrix of the dual vectors:The dual vectors are required to lie in ${\mathcal{U}}_n$; adding arbitrary vectors orthogonal to that subspace would preserve the displayed biorthogonality but would destroy this conclusion.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

For the exponent $\sigma =m$, the linear recurrence relation reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:The ratio test gives an infinite radius of convergence for the power series factor. Hence a solution isFor nonintegral $m$, choose instead the exponent $\sigma =0$. Since $n-m\ne 0$ for every positive integer $n$, its linear recurrence relation gives $n{a}_n=-a_{n-1}$. ThusThe distinct nonintegral leading exponents establish linear independence. More explicitly, $W(e^{-x},y_m)=m{e}^{-x}{x}^{m-1}$ by the Abel identity and the leading coefficient at zero, so the Wronskian is nonzero. The factor $x^m$ uses the real branch on $x>0$.

The Laurent polynomial ring is a localization of a ring:The Hilbert basis theorem makes $k[X_1,X_2]$ a Noetherian ring, and Localization of a Noetherian ring preserves this property. Thus $\boxed{R\text{ is Noetherian}}$.

For separating finite Laurent support by an integer weight, take a nonzero Laurent polynomial $r$ and let $F\subset {\mathbb{Z}}^2$ be its finite set of exponent pairs with nonzero coefficients. Choose an integer $N$ such thatwhere the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+N{m}_2=m_1^{\prime }+N{m}_2^{\prime }$, thenso $m_2=m_2^{\prime }$ and then $m_1=m_1^{\prime }$. Consequently all powers of $T$ in ${\theta }_{(1,N)}(r)$ are distinct. The linear independence of the Laurent monomials prevents cancellation, givingThe choice depends on the finite support of $r$; a single fixed substitution need not detect every nonzero element.

For Noetherianity of subalgebras of a one-variable Laurent polynomial ring, let $A$ be a unital $k$-subalgebra of $k[T,T^{-1}]$. If $A=k$, it is already a Noetherian ring. Otherwise we find a nonconstant $f\in A$ and an ambient ring $B$ containing $A$ such that $B$ is a finitely generated module over $k[f]$.

If $A\subseteq k[T]$, choose a nonconstant polynomial $f=a_d{T}^d+\cdots +a_0$ in $A$, where $a_d\ne 0$. Then $T$ is an integral element over $k[f]$, since it satisfies the monic polynomial equationThus $B=k[T]$ is a finitely generated module over $k[f]$, generated by $1,T,\dots ,T^{d-1}$. The case $A\subseteq k[T^{-1}]$ is identical after replacing $T$ by $T^{-1}$.

In the remaining case $A$ is contained in neither of these two polynomial rings. There is an element with a positive exponent and an element with a negative exponent. If either already has both, use it as $f$. Otherwise their sum has both, since the positive powers of the first and the negative powers of the second cannot cancel. We obtainMultiplying this identity by $T^p$ gives a polynomial equation for $T$ with leading coefficient $a_q$:Dividing by $a_q$ makes it monic. Similarly, multiplication by $T^{-q}$ gives a monic equation for $T^{-1}$ after division by $a_{-p}$. Both generators are therefore integral elements over $k[f]$, so $B=k[T,T^{-1}]$ is a finitely generated module over $k[f]$.

In all three cases $k[f]$ is a Noetherian ring, as a quotient of a one-variable polynomial ring. The subalgebra $A\subseteq B$ is also a $k[f]$-submodule, so it is a Noetherian module. Every ideal of $A$ is a $k[f]$-submodule of $A$ and is therefore finitely generated over $k[f]$, hence over $A$. This proves

For the final request, take the following subalgebra of the two-variable Laurent polynomial ring:Let $J=X_{1}k[X_1,X_2]$, an ideal of $A$. We haveas $k$-vector spaces. In particular, the classes of $X_1{X}_2^j$, for $j\ge 0$, are linearly independent. If $J$ had finitely many generators as an $A$-module, their classes would span $J/J^2$ over $A/J=k$, contradicting its infinite dimension. Hence $J$ is not finitely generated and $A$ is not a Noetherian ring. This illustrates why the preceding result is special to one Laurent variable.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

The Steinitz exchange lemma says that if $w_1,\dots ,w_m$ span a vector space and $v_1,\dots ,v_r$ are linearly independent, then $r\le m$, and after reordering the $w_j$, the family $v_1,\dots ,v_r,w_{r+1},\dots ,w_m$ still spans. Here is a proof. Suppose the first $k-1$ replacements have been made. Express $v_k$ in that spanning family. Some coefficient of a remaining $w_j$ must be nonzero, since otherwise $v_k$ belongs to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Solve for that $w_j$ and replace it by $v_k$; this preserves spanning. If $r>m$, the first $m$ replacements leave a spanning family $v_1,\dots ,v_m$, contradicting linear independence of the first $m+1$ vectors. This proves both assertions.

Now suppose $S$ is linearly independent and spans ${\mathbb{R}}^n$. Express each member of the standard basis using finitely many members of $S$, and let $S_0$ be the finite union of those members. Then $S_0$ spans. No element of $S\setminus S_0$ can exist, since it would be a linear combination of $S_0$ and violate linear independence. Thus $S$ is finite. Applying the Steinitz exchange lemma in both directions to $S$ and the standard basis gives $|S|\le n$ and $n\le |S|$. Hence $\boxed{|S|=n}$.

The dual space here is the algebraic dual space $X^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(X,\mathbb{R})$, consisting of every linear functional; no topology or boundedness is imposed. For a basis $e_1,\dots ,e_n$, define $e_i^{\ast }({\sum }_j{x}_j{e}_j)=x_i$. These are linear functionals satisfying $e_i^{\ast }(e_j)={\delta }_{ij}$. Evaluating ${\sum }_i{a}_i{e}_i^{\ast }=0$ on each $e_j$ gives $a_j=0$, proving linear independence. Conversely, for every $T\in X^{\ast }$,This proves spanning and hence proves, without assuming a dimension theorem, that the dual basis is a basis and $\dim X^{\ast }=n$.

The canonical bidual evaluation map isIts definition makes no choice of basis and it is linear. If $x\ne 0$, one coordinate functional is nonzero on $x$, so $J$ is injective. For any $B\in X^{\ast \ast }$, choose $x={\sum }_{i}B(e_i^{\ast })e_i$. Expanding $T$ in the dual basis gives $T(x)={\sum }_{i}T(e_i)B(e_i^{\ast })=B(T)$, proving surjection explicitly. Thus $J$ is an isomorphism.

If $T_1,\dots ,T_n$ is any basis of $X^{\ast }$, its dual basis $B_1,\dots ,B_n$ in $X^{\ast \ast }$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a basis of $X$, and $T_i(v_j)=B_j(T_i)={\delta }_{ij}$. Every basis of the finite-dimensional dual is a dual basis, of a uniquely determined basis of $X$.

For a separating subspace of an algebraic dual $W$, choose a basis $T_1,\dots ,T_r$ of $W$. The linear map $x\mapsto (T_1(x),\dots ,T_r(x))$ is injective because $W$ separates nonzero vectors. The rank-nullity theorem implies $n\le r$; but $W\subseteq X^{\ast }$ implies $r\le n$. HenceThe zero-dimensional case also has this conclusion.

For $X=\mathbb{R}[x]$, the monomials $1,x,x^2,\dots$ form an infinite basis, each polynomial using only finitely many of them. The map $T\mapsto (T(1),T(x),T(x^2),\dots )$ identifies $X^{\ast }$ with all real sequences, because any sequence $(a_j)$ defines $T({\sum }_{j=0}^d{c}_j{x}^j)={\sum }_{j=0}^d{c}_j{a}_j$. The linear subspace of finitely supported sequences is proper (it excludes $(1,1,\dots )$) and separating (one coefficient functional detects any nonzero polynomial). Thus

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,

Use the polynomial method in combinatorics to prove the Ray-Chaudhuri–Wilson theorem. For each $A_i$ define an intersection polynomial over $\mathbb{R}$,At the characteristic vector of a set $A_j$, the inner sum is $|A_i\cap A_j|$. ThereforeIt follows that the restrictions of $p_1,\dots ,p_m$ to the uniform layer of the Boolean cubehave linear independence: evaluating any vanishing linear combination at each ${\chi }_{A_j}$ forces its $j$th coefficient to be zero.

Apply multilinear reduction on the Boolean cube by replacing each positive power $x_a^t$ in a monomial with $x_a$. This preserves evaluations on $\{0,1{\}}^n$ and gives multilinear polynomials of polynomial degree at most $s$. Counting all monomials of degrees up to $s$ would give only ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, which is too weak. Instead use homogenisation on a uniform layer.

Because $L$ has $s$ distinct integers in $\{0,\dots ,r-1\}$, we have $s\le r$. For $T\subseteq [n]$ with $|T|=j\le s$, write $x_T={\prod }_{a\in T}{x}_a$, with $x_{\varnothing }=1$. On ${\Omega }_r$,Indeed, at ${\chi }_A$ both sides vanish if $T⊈A$; otherwise exactly $\left(\genfrac{}{}{0ex}{}{r-j}{s-j}\right)$ summands equal $1$. The denominator is positive since $j\le s\le r$. Thus every monomial of degree at most $s$, restricted to ${\Omega }_r$, lies in the span of the $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ degree-$s$ monomials. This vector space consequently has dimension at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. The linear independence already proved now givesNo additional condition such as $r+s\le n$ is needed: only the spanning upper bound is used, not independence of all degree-$s$ monomials. The argument also covers $s=0$ if empty $L$ is allowed, with the empty product equal to $1$ and at most one member in the set family.

With $K\ge 2$ distinct knots ${\kappa }_1<\cdots <{\kappa }_K$, a natural cubic spline is a $C^2$ function on the real line, cubic on each interval between consecutive knots and affine outside the two outer knots. The latter condition gives $m^{\prime \prime }({\kappa }_1)=m^{\prime \prime }({\kappa }_K)=0$.

There are $4(K-1)$ coefficients for the interior cubic pieces. Matching values and the first two derivatives at the $K-2$ interior knots imposes $3(K-2)$ independent constraints, and the two natural endpoint conditions impose two more. The affine continuations are then determined by endpoint values and slopes. The dimension of a vector space is thereforeOne explicit basis uses $1,x$ and $d_j(x)-d_{K-1}(x)$ for $j=1,\dots ,K-2$, whereThe cubic and quadratic terms cancel in each difference above ${\kappa }_K$, making it affine there; below ${\kappa }_1$ the truncated powers vanish. The changes in third derivative at individual knots show independence of these basis functions. This count uses the linear independence of the spline constraints and counts the listed outer knots as boundary knots. If a convention instead lists $K$ interior knots and two additional boundaries, the dimension is $K+2$. With only one listed knot and linearity on both sides, the $C^2$ condition forces a globally affine function, so the exceptional dimension is two.

Choose a basis ${\varphi }_1,\dots ,{\varphi }_N$ of a conforming finite element space $V_N\subset H_0^1(0,1)$ and write $y_N={\sum }_j{c}_j{\varphi }_j$. Testing the variational problem with each basis function givesThese are also the equations $\partial J(y_N)/\partial c_i=0$ for the Ritz method. The stiffness matrix is symmetric and a positive-definite matrix: for a nonzero coefficient vector $c$, linear independence gives $v_N=\sum c_j{\varphi }_j\ne 0$, andThus there is a unique coefficient vector.

Subtracting the exact and discrete variational problems proves Galerkin orthogonality, $a(y-y_N,v_N)=0$ for every $v_N\in V_N$. In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity yieldsTherefore the Ritz method is the best approximation in the energy norm. The Céa lemma also gives $\Vert y-y_N{\Vert }_V\le (1+{\pi }^{-2}){\inf }_{v_N\in V_N}\Vert y-v_N{\Vert }_V$, so any dense sequence of conforming trial spaces converges.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

Choose a branch of $t=1-z$ with $|\arg t|<\pi$ and write $d=c-a-b\notin \mathbb{Z}$. The Gauss hypergeometric equation isChanging to $t$ turns this into the same equation with third parameter $a+b+1-c=1-d$. Its solution regular at $t=0$ is $y_1=F(a,b,1-d;t)$. Substitution of $y=t^{d}v$ gives the Gauss hypergeometric equation for $v$ with parameters $c-a,c-b,1+d$. Thus $y_2=t^{d}F(c-a,c-b,1+d;t)$ is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point $0,d$ imply linear independence near $z=1$, soThis also follows from the exponent transformation in the Papperitz symbol.

Initially assume $\Re c>\Re b>0$, so the Euler integral for the hypergeometric function applies. If $\Re d>0$, taking $z\to 1$ gives an integral evaluated by the beta function and determinesIndeed $y_1\to 1$ and $y_2\to 0$ there. To find $B$, instead work in the nonempty parameter region $\Re d<0$, put $z=1-ϵ$ and scale the integration variable by $1-s=ϵu$. The singular contribution isThe defining integral of the beta function evaluates this integral to $\Gamma (c-b)\Gamma (-d)/\Gamma (a)$. Since ${ϵ}^{-d}{y}_1\to 0$ and ${ϵ}^{-d}{y}_2\to 1$, it determines $B$. Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires $c\notin \{0,-1,-2,\dots \}$; exceptional parameters need a separately specified analytic limit. Reciprocal gamma functions make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in $z$.

For real $z>0$, the exponential decays as $\mathrm{Re}t\to -\infty$, so two allowed contours run from $-\infty$ to $i$ and from $-\infty$ to $-i$. For real $z<0$, it decays as $\mathrm{Re}t\to +\infty$, so reverse the construction: run from $i$ or $-i$ to $+\infty$. These are the infinite contours in the diagram.

For $z>0$ their integrals arefor $z<0$ the corresponding orientation changes only an overall sign. Taking the sum of the two exponentials produces a solution independent of $\sin z/z$:Its nonzero Wronskian with $w_1$ on $z\ne 0$ proves linear independence.