In characteristic two, repeated squaring gives $x^8-1=(x-1{)}^8$. There is only the root one, already in the base field. Thus $\boxed{K={\mathbb{F}}_2,[K:{\mathbb{F}}_2]=1}$. Repeated roots do not require a larger splitting field.

The cubic $x^3+x+1$ has no zero in ${\mathbb{F}}_2$, so is irreducible. Its quotient field has eight elements. Every nonzero element of that field satisfies $x^7=1$ by the order of its multiplicative group, so the polynomial splits there. It does not already split in the base field; a root of that irreducible cubic, which divides $x^7-1$, is needed. Hence

A primitive eighth root satisfies ${\zeta }_8=(1+i)/\sqrt{2}$, while $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$. Thus $K=\mathbb{Q}(i,\sqrt{2})$. The real elements of $\mathbb{Q}(i)$ are rational, so it does not contain $\sqrt{2}$. The tower law gives

A primitive seventh root generates all seventh roots, so the splitting field is $\mathbb{Q}({\zeta }_7)$. Its minimal polynomial is the given irreducible cyclotomic polynomial $1+x+\cdots +x^6$. Therefore $\boxed{[K:\mathbb{Q}]=6}$.

All sixth roots of unity lie in $\mathbb{Q}({\zeta }_6)$, with ${\zeta }_6=(1+\sqrt{-3})/2$. Its quadratic polynomial $x^2-x+1$ has no rational root, so

Independent increments give independent Poisson variables $N[0,s]$ and $N(s,t]$ with means $\lambda s$ and $\lambda (t-s)$. Dividing their joint probability by $\mathbb{P}(N[0,t]=n)$ yieldsThus the conditional distribution is binomial distribution with parameters $n,s/t$.

The event $T_i>s$ means that at most $i-1$ events have occurred by $s$. ConsequentlyFor $1\le i\le n$, the conditional event time lies in $[0,t]$. Integrate its survival function and substitute $u=s/t$:The index range $i\le n$ is implicit in the source's last assertion. For $i>n$, the $i$th event is after $t$, and independent future increments instead give expectation $t+(i-n)/\lambda$; the displayed $it/(n+1)$ formula does not apply.

The state space is $\{0,1,2,\dots \}$. Write $q=1-p$. The nonzero transition probabilities areThe stationary equations give ${\pi }_1={\pi }_0$ and ${\pi }_j=q^{j-1}{\pi }_0$ for $j\ge 1$. Normalization gives ${\pi }_0[1+{\sum }_{j\ge 1}{q}^{j-1}]={\pi }_0(1+1/p)=1$, soThe remaining equation ${\pi }_0=p{\sum }_{j\ge 1}{\pi }_j$ is satisfied, confirming stationarity.

The chain is irreducible: every positive state can jump to zero, and zero can reach every state through consecutive immigration steps. A stationary probability distribution makes it positive recurrent. Returns to zero can have lengths two and three with positive probability, so it is also aperiodic. The Markov-chain ergodic theorem now gives the almost-sure long-run time proportion for $X_n>K$ as the stationary tail:This time-average conclusion applies to the specified start at zero.

First simplify the composite: $(S\circ \mathcal{I}\circ S)(A)=((A^2{)}^{-1}{)}^2=A^{-4}$. Applying the product rule to four factors of $A^{-1}$, and the preceding inverse derivative to one factor at a time, givesThe four terms have left and right exponents $(1,4),(2,3),(3,2),(4,1)$. The same expression follows by composing $D{S}_{A^{-2}}$, $D{\mathcal{I}}_{A^2}$ and $D{S}_A$ in that order. Only when $\Delta$ commutes with $A$ can it be simplified to $-4A^{-5}\Delta$.

For small $\Vert \Delta \Vert$, the matrix $A+\Delta =A(I+A^{-1}\Delta )$ remains invertible and its inverse is locally bounded, for example by the convergent Neumann series when $\Vert A^{-1}\Delta \Vert <1$. Applying the resolvent identity twice givesThe last term is $O(\Vert \Delta {\Vert }^2)$ by that local bound. Hence the inverse map is differentiable, withThis verifies the differentiability and its remainder, rather than only formally differentiating the identity $A{A}^{-1}=I$.

The Fréchet derivative at $a$ is the linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ satisfyingIt is unique. When both derivatives exist, the chain rule is $D(g\circ f{)}_a=D{g}_{f(a)}\circ D{f}_a$.

For the square map,The remainder has norm at most $\Vert \Delta {\Vert }^2$ in a submultiplicative matrix norm, hence is $o(\Vert \Delta \Vert )$. ThereforeMatrix multiplication is not commutative, so the two terms cannot generally be combined as $2A\Delta$.

The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.

A quotient map $\pi :X\to Y$ is a surjection such that $U\subseteq Y$ is open if and only if ${\pi }^{-1}(U)$ is open in $X$. Thus the topology on $Y$ is exactly the final topology induced by $\pi$.

For an open continuous surjection, continuity gives one implication. For the other, if ${\pi }^{-1}(U)$ is open then $U=\pi ({\pi }^{-1}(U))$ is open by openness and surjectivity. Hence it is a quotient map.

If $\pi :X\to Y$ and $\sigma :Y\to Z$ are quotient maps, they are surjective andThis set is open exactly when ${\sigma }^{-1}(U)$ is open, exactly when $U$ is open. Thus the composition is a quotient map.

For the factorization, continuity of $g$ implies continuity of $f=g\circ \pi$. Conversely, if $f$ is continuous and $U$ is open in $Z$, then ${\pi }^{-1}(g^{-1}(U))=f^{-1}(U)$ is open. The quotient property makes $g^{-1}(U)$ open. This holds for every open $U$, so $g$ is continuous.

The theorem givesIntegration by parts has zero boundary term: the sine vanishes at zero and the Gaussian tends to zero at infinity. Hence $F^{\prime }(t)=-2t{\int }_0^{\infty }{e}^{-x^2}\cos (2xt)dx=-2tF(t)$.

The same Gaussian majorant and dominated convergence give continuity at zero, where the allowed Gaussian integral gives $F(0)=\sqrt{\pi }/2$. The differential equation implies $(e^{t^2}F(t){)}^{\prime }=0$; continuity at zero fixes that constant. Thus

We first prove the differentiation theorem. Fix $t\in (0,b)$ and take nonzero $h\to 0$ with $t+h\in (0,b)$. The measurable difference quotientsconverge pointwise to $f_t(x,t)$. The mean value theorem and the assumed derivative bound give $|q_h(x)|\le g(x)$. In particular $f_t(\cdot ,t)$ is measurable as a limit of quotients along a sequence, and its bound by $g$ proves integrability. The dominated convergence theorem, in its parameter-limit form, says that pointwise convergence with a common integrable bound permits convergence of the integrals. Applying it yieldsThe parameter-limit statement follows from the ordinary sequential theorem by applying it to every sequence of nonzero $h$ tending to zero.

For the specific function, $|f(x,t)|\le e^{-x^2}$, which is integrable by the allowed fact. It is differentiable in $t$, withThe last majorant has integral one on $[0,\infty )$, using the substitution $u=x^2$. All three hypotheses therefore hold. The indicator of the parameter domain is one at every point under consideration and does not affect this derivative.

Take $a=[x]$. Every quotient class is a polynomial expression in $a$ with coefficients from $F$, so $K=F[a]$. Substitution gives $f(a)=[f(x)]=0$. ThusThis constructs a field containing a root even when no root existed in the original field.

Polynomial division gives a unique remainder of degree less than $n$ in every coset. Thus the classesspan the quotient as an $F$-vector space. A linear combination of them that vanishes would be a polynomial of degree less than $n$ divisible by $f$, so it is the zero polynomial. They are therefore linearly independent, proving

An irreducible polynomial has positive degree, so its ideal is proper: otherwise $1=fh$ would contradict degrees. Every nonzero class $[g]$ has a representative not divisible by $f$. Irreducibility gives $\gcd (g,f)=1$, and the Euclidean algorithm supplies polynomials $u,v$ with $ug+vf=1$. Passing to the quotient gives $[u][g]=1$. Thus the quotient is a nonzero commutative ring in which every nonzero element has an inverse, hence a field. Constants embed injectively because no nonzero constant lies in the positive-degree ideal $(f)$.

FactorThe roots are $\pm \sqrt{2},\pm i$, so the splitting field is $\mathbb{Q}(\sqrt{2},i)$. The real quadratic field $\mathbb{Q}(\sqrt{2})$ does not contain $i$, which adds a further degree two. Therefore

Let $\zeta =e^{\pi i/4}$. Its odd powers are exactly the four roots, so $\mathbb{Q}(\zeta )$ is the splitting field. The polynomial $x^4+1$ is irreducible: after the invertible substitution $x\mapsto x+1$, it becomes $x^4+4x^3+6x^2+4x+2$, which satisfies Eisenstein criterion at two. HenceEquivalently the field is $\mathbb{Q}(\sqrt{2},i)$, since $\zeta =(1+i)/\sqrt{2}$, ${\zeta }^2=i$ and $\zeta +{\zeta }^{-1}=\sqrt{2}$.

First address the general splitting-field request. A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. For a degree-$m$ polynomial, adjoin a root of an irreducible factor. The quotient construction in the preceding parts gives an extension of degree at most $m$. Remove the resulting linear factor and repeat, now with a polynomial of degree $m-1$ over that extension. Induction and the tower law giveThe resulting field is generated by the roots and is a splitting field. Repeated roots or inseparability can only reduce the required degrees; separability was not used.

For the specified cubic, put $a=\sqrt[3]{2}$ and $\omega =e^{2\pi i/3}$. Its splitting field is $\mathbb{Q}(a,\omega )$. Eisenstein criterion at two gives $[\mathbb{Q}(a):\mathbb{Q}]=3$. This cubic field is real and therefore does not contain the nonreal $\omega$, whose quadratic polynomial is $x^2+x+1$. Consequently