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Treat the white noise as innovations with conditional mean zero and conditional second moment $v$, independent of the chosen past control. This is the convention needed for the dynamic programming calculation. The printed stage cost is $\frac{1}{2}{x}^2+u^2$, not one-half of both terms.

With terminal value $V_0(x)=x^2$, induction in the Bellman equation givesIf $V_{h-1}(z)=z^2+(h-1)v$, completing the square yieldsThus $\boxed{V_h(x)=x^2+hv,u^{\ast }(x)=-x/2}$ for every horizon, in particular $V_6(x_0)=x_0^2+6v$. The same completion gives the average-cost Bellman equation solution $\boxed{\varphi (x)=x^2,\lambda =v}$, up to an additive constant in $\varphi$.

For every policy in $P$, conditional second moments satisfy $\mathbb{E}[x_{t+1}^2\mid {\mathcal{F}}_t]=(x_t+u_t{)}^2+v\le 0.9x_t^2+v$. Iterating givesThe Bellman inequality implies ${\mathbb{E}}_{\pi }{\sum }_{t<h}(\frac{1}{2}{x}_t^2+u_t^2)\ge hv+x_0^2-{\mathbb{E}}_{\pi }{x}_h^2$. Dividing by $h$ and using the bound proves that every admissible policy has lower limiting average cost at least $v$, even if its ordinary limit does not exist. The feedback $u=-x/2$ belongs to $P$ because $(x+u{)}^2=x^2/4\le 0.9x^2$. It makes the Bellman inequality an equality, and its bounded terminal second moment proves its average cost tends to $\boxed{v}$. Therefore it is optimal over $P$.

Merely uncorrelated noise with the displayed unconditional moments would not ensure the conditional identities for arbitrary adapted controls. The innovation interpretation specifies the hypothesis used in the calculation.

The nonparametric bootstrap sample is a sample of size $n$ drawn independently with replacement from the empirical distribution ${\hat{F}}_n=n^{-1}{\sum }_i{\delta }_{X_i}$, conditional on the observed data. To simulate it, take independent uniform $U_j$ and set $X_j^b=X_{1+\lfloor n{U}_j\rfloor }$, assigning $U_j=1$ to the last index if it occurs in a finite-precision generator.

Let $G_n^b$ be the conditional cumulative distribution function of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$, and let $q_n(p)=\inf \{t:G_n^b(t)\ge p\}$. A basic bootstrap confidence interval isIn practice the conditional quantiles are estimated from many simulated bootstrap means; the asymptotic argument below describes exact bootstrap quantiles or a simulation count tending to infinity.

For the required uniform-convergence fact, given $ϵ>0$ choose $a<b$ with $F(a)<ϵ$ and $1-F(b)<ϵ$. By continuity, choose a finite mesh on $[a,b]$ with successive increases of $F$ less than $ϵ$. Pointwise convergence at the mesh points makes $|F_n-F|<ϵ$ there for large $n$. Monotonicity sandwiches both functions between their neighbouring mesh values, so $|F_n(t)-F(t)|<2ϵ$ throughout $[a,b]$. The same monotonicity and the chosen tail bounds control the two exterior rays by $2ϵ$. Hence ${\sup }_t|F_n(t)-F(t)|\to 0$.

The bootstrap consistency theorem for the sample mean states that, for iid observations with $0<{\sigma }^2<\infty$, the conditional law of $\sqrt{n}({\bar{X}}_n^b-{\bar{X}}_n)$ converges weakly, in probability, to $N(0,{\sigma }^2)$. By continuity of that limiting distribution and the preceding monotonicity argument, the convergence of conditional distribution functions is uniform in probability. Since the limit is strictly increasing, $q_n(p)\to \sigma {\Phi }^{-1}(p)$ in probability for $0<p<1$. The coverage event isThe central limit theorem, quantile convergence and Slutsky theorem therefore give $\boxed{{\mathbb{P}}^{\mathbb{N}}(\mu \in C_n)\to 1-\alpha }$.

Positive variance is needed for this exact coverage conclusion. If the finite variance allowed in the wording is zero, all observations equal $\mu$ almost surely and the interval collapses to $\{\mu \}$, with coverage one. The bootstrap distribution is still consistent with the degenerate limit, but a nonrandomized usual interval cannot then have limiting coverage $1-\alpha$ for $0<\alpha <1$.

Regard complex matrices as a real vector space of dimension $2n^2$ and Hermitian matrices as a real vector space of dimension $n^2$. Set $F(A)=A{A}^{\ast }$. Its derivative is $D{F}_A(H)=H{A}^{\ast }+A{H}^{\ast }$. At a unitary $A$, any Hermitian $S$ is reached by $H=SA/2$. Hence $I$ is a regular value, and the regular level set theorem makes $\mathrm{U}(n)=F^{-1}(I)$ a smooth real manifold of dimension $\boxed{n^2}$.

The tangent space is the kernel of this derivative. Equivalently,At $I$, therefore, $H$ is skew-Hermitian. The explicit geodesic with initial point $I$ and velocity $H$ is $\gamma (t)=e^{tH}$. The matrix exponential is unitary because $e^{t{H}^{\ast }}{e}^{tH}=I$, and $\dot{\gamma }(0)=H$. Its acceleration is $\gamma H^2$. Since $H^2$ is Hermitian and every tangent vector at $\gamma$ is $\gamma K$ with $K$ skew-Hermitian, the induced real inner product satisfiesIndeed the trace of a Hermitian matrix times a skew-Hermitian matrix is purely imaginary. Its acceleration is therefore always normal to the manifold, which is the geodesic equation for an induced Euclidean metric. Diagonalizing $H=W\mathrm{diag}(i{\lambda }_j)W^{\ast }$ gives the equivalent explicit construction $W\mathrm{diag}(e^{it{\lambda }_j})W^{\ast }$.

Write $n=2^{r}m$ with $m$ odd. If $m>1$, the polynomial factor of $X^m+1$ by $X+1$, evaluated at $X=10^{2^r}$, supplies a proper divisor. Therefore $n$ must be a power of two. Conversely this is only a necessary condition for primality testing.

For a prime factor $p$ of the number, $p$ is odd and does not divide $10$. Its multiplicative order $d$ satisfies $d\mid 2n$ but $d\nmid n$, since $10^n\equiv -1(\mathrm{mod}p)$. When $n$ is a power of two, every proper divisor of $2n$ divides $n$, so $d=2n$. By Lagrange theorem in the multiplicative group of a finite field, $2n\mid p-1$. Thus $\boxed{p\equiv 1(\mathrm{mod}2n)}$.

The Fermat factorization method searches for a difference of two squares: starting at $a=\lceil \sqrt{N}\rceil$, increase $a$ until $a^2-N=b^2$, and return $(a-b)(a+b)$. Every factorization of an odd integer has this form, with $a=(d+e)/2$ and $b=(e-d)/2$. For $N=10001$, the successive differences at $a=101,102,103,104,105$ are $200,403,608,815,1024$. The last is $32^2$, giving $\boxed{10001=73\cdot 137}$. Trial division by the primes up to their respective square roots confirms that both factors are prime.

Use an inner product linear in the first variable. For each $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ is represented uniquely by a vector $T^{\ast }y$ under the Riesz representation theorem, so $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Uniqueness and the conjugate-linearity in the second argument imply $T^{\ast }(ay+bz)=a{T}^{\ast }y+b{T}^{\ast }z$. FurthermoreThus the adjoint operator exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.

A normal operator satisfies $T^{\ast }T=T{T}^{\ast }$. The unilateral unilateral shift operator $S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )$ on ${\ell }^2$ is bounded but not normal: $S^{\ast }S=I$, whereas $S{S}^{\ast }=I-P_{e_1}$. For general $T$,Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the polarization identity then gives $T^{\ast }T-T{T}^{\ast }=0$.

An approximate eigenvalue $\lambda$ has unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. It is necessarily in the spectrum, since a bounded inverse would prevent this. Conversely suppose $\lambda$ is not an approximate eigenvalue. Then $A=T-\lambda I$ is bounded below, $\Vert Ax\Vert \ge c\Vert x\Vert$. It is injective and has closed range: a convergent sequence $A{x}_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\Vert A^{\ast }x\Vert =\Vert Ax\Vert$ and $\ker A^{\ast }=0$. The identity $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$ makes its range dense. Closed and dense means surjective, and its inverse is bounded by $1/c$. Hence $\lambda$ is in the resolvent. This proves $\boxed{\sigma (T)={\sigma }_{\mathrm{ap}}(T)}$.

The Mayer–Vietoris sequence for $K=M\cup N$ is the long exact sequence in homologyWe use integer coefficients. A simplicial cycle $c$ in $K$ can be written $c=m+n$ with chains in $M,N$. Since $\partial m=-\partial n$, this chain lies in $M\cap N$ and is a cycle. Define ${\partial }_n[c]=[\partial m]$; changing the splitting or the cycle representative changes this by a boundary, giving a well-defined connecting homomorphism.

First consider two acyclic subcomplexes. Their intersection is either empty or acyclic. If empty, positive-degree homology of their disjoint union vanishes. If nonempty, the reduced Mayer–Vietoris sequence implies the same conclusion, including degree one since the intersection has zero reduced $H_0$. This proves the claim for two subcomplexes.

Inductively let $U=M_1\cup \cdots \cup M_{n-1}$, $V=M_n$, and $W=U\cap V={\bigcup }_{i<n}(M_i\cap M_n)$. Discard empty members as necessary. Both $U$ and $W$ satisfy the same intersection hypothesis with at most $n-1$ members, so their homology vanishes in degrees at least $n-2$. The exact segment $H_i(U)\oplus H_i(V)\to H_i(K)\to H_{i-1}(W)$ then gives $\boxed{H_i(K)=0\text{ for }i\ge n-1}$. For $n\ge 3$ these degrees are positive and the induction applies directly; empty $W$ again means a disjoint union.

For sharpness, take the boundary of an $(n-1)$-simplex, covered by its $n$ facets. Every nonempty proper intersection of facets is a simplex and thus has the homology of a point; the intersection of all facets is empty. The union is a sphere $S^{n-2}$ and $H_{n-2}$ is nonzero. For $n=2$ it is two isolated vertices and $H_0\cong {\mathbb{Z}}^2$; for $n\ge 3$, $H_{n-2}\cong \mathbb{Z}$. The degree bound cannot be lowered.

For a number field $K$, let ${\mathcal{O}}_K$ be its ring of integers of a number field, $r_1$ the number of real embeddings and $r_2$ the number of conjugate pairs of nonreal complex embeddings, so $[K:\mathbb{Q}]=r_1+2r_2$. The Dirichlet unit theorem sayswhere $\mu (K)$ is the finite group of roots of unity in $K$. Thus an imaginary quadratic number field has only roots of unity as units: usually $\{\pm 1\}$, with four for $d=-1$ and six for $d=-3$. A real quadratic number field has units $\pm {ϵ}^n$ for a fundamental unit $ϵ>1$ and integers $n$.

For $K=\mathbb{Q}(\sqrt{26})$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{26}]$ and the field norm is $x^2-26y^2$. The element $ϵ=5+\sqrt{26}$ has norm $-1$ and is a unit. It is the smallest positive unit greater than one. Indeed for $1<u<ϵ$, its conjugate is $\pm 1/u$, so the coefficient $y=(u\mp 1/u)/(2\sqrt{26})$ is a positive integer less than two. Therefore $y=1$, and $x^2=26\pm 1$ forces $x=5$, giving $u=ϵ$, a contradiction. The Dirichlet unit theorem now gives $\boxed{{\mathcal{O}}_K^{\times }=\{\pm (5+\sqrt{26}{)}^n:n\in \mathbb{Z}\}}$. Its least positive norm-one generator is ${ϵ}^2=51+10\sqrt{26}$.

For the generalized Pell equation, write $\alpha =x+y\sqrt{26}$ with norm $N=\pm 10$. Change the sign if necessary and multiply by a power of $ϵ$ to place it in $\sqrt{10}\le \alpha <ϵ\sqrt{10}$. These operations preserve integrality and the absolute norm, though the latter operation can change the norm's sign. Since ${\alpha }^{\prime }=N/\alpha$, we have $x=(\alpha +N/\alpha )/2\ge 0$ and $y=(\alpha -N/\alpha )/(2\sqrt{26})\ge 0$. The upper bound gives $y<4$. Checking $y=0,1,2,3$ in $x^2=26y^2\pm 10$ leaves only $(x,y)=(4,1),(6,1)$. Both reduced elements lie in the chosen interval. Therefore all integral solutions of either sign areNegative powers are integral because ${ϵ}^{-1}=\sqrt{26}-5$. In the first family the norm is $-10(-1{)}^n$ and in the second it is $10(-1{)}^n$, so this also separates the two equations. The reduction proves exhaustion rather than merely generating examples.