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A distribution $\pi$ is an invariant distribution whenfor every $j$, equivalently $\pi P=\pi$.

The inequality implies $u(u-1)\ge v^2\ge 0$, so $u\le 0$ or $u\ge 1$. Since $u$ is continuous and $\mathbb{C}$ is connected, its image lies wholly in one of these intervals. If $u\ge 1$, then $g$ has no zero and $|1/g|\le 1$; Liouville theorem makes $1/g$, hence $g$, constant. If $u\le 0$, then $|g-1|\ge 1$, so the same argument applied to $1/(g-1)$ proves that $g$ is constant.

The Cauchy integral formula and its differentiated form areIf $f$ is entire and $|f|\le M$, the Cauchy estimate gives $|f^{\prime }(a)|\le M/\rho$. Letting $\rho \to \infty$ yields $f^{\prime }(a)=0$ for every $a$. Thus $f$ is constant, which is Liouville theorem.

Let $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$, the symmetry group of a square. The mappreserves the relations and is an automorphism. Conjugation by a rotation sends $s$ to $r^{2j}s$, while conjugation by a reflection also inverts $r$. Therefore no conjugation fixes $r$ and sends $s$ to $rs$. This is an outer automorphism of the nonabelian group $D_8$.

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.

The equations areHencewhich is not generally zero. Conservation requires $\beta /\alpha =1$. Then the potential is $\alpha (x^2+y^2{)}^2/4$, so the action has rotational symmetry. By Noether's theorem, $J$ is the conserved angular momentum, up to the sign convention used in its definition.

The Euler-Lagrange equations are $\ddot{x}+{\omega }^{2}x=0$ and $\ddot{y}+{\omega }^{2}y=0$. They implyWritinggives the conserved value

Taking the Laplace transform in $t$ givesThe bounded resolvent solution supplied in the hint can be writtenPart (a) inverts the kernel, yielding the heat kernel

The Laplace transform isUsing the supplied Gaussian integral after $t=x^2$ gives

The cover $X={\bigcup }_{n\in X}{U}_n$ has no finite subcover, because every $U_n$ is finite. Hence $X$ is not compact.

For the stated pasting lemma for closed subspaces, if $C\subset Z$ is closed, thenis a union of two closed subsets of $Y$; therefore $f$ is continuous.

For $m,n\in X$, let $q=mn$. On $[0,\frac{1}{2}]$ define a path equal to $m$ before the endpoint and equal to $q$ at the endpoint; it is continuous because every open set containing $q$ also contains its divisor $m$. On $[\frac{1}{2},1]$ define it to equal $q$ at the first endpoint and $n$ thereafter. This is likewise continuous. The two closed-interval paths agree at $1/2$, so the pasting lemma joins them into a path from $m$ to $n$. Thus $X$ is path-connected.

Since $U_m\cap U_n=U_{\gcd (m,n)}$ when the greatest common divisor is at least $2$, and is empty otherwise, finite intersections of the $U_n$ are again among the allowed unions. Arbitrary unions are built into the definition; the empty union is empty, and $X={\bigcup }_{n\in X}{U}_n$. Thus these sets form a topology.

The space is not Hausdorff: every neighborhood of $4$ contains its divisor $2$, and every neighborhood of $2$ contains $2$, so $2$ and $4$ cannot have disjoint neighborhoods.

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.

The characteristic polynomial is $(t-\frac{1}{2}{)}^3$. For $N=A-\frac{1}{2}I$, one finds $N\ne 0$ but $N^2=0$, so the minimal polynomial is $(t-\frac{1}{2}{)}^2$. Hence the invariant factors are $t-\frac{1}{2}$ and $(t-\frac{1}{2}{)}^2$, and the rational canonical form is

Make $F^n$ an $F[X]$-module by $Xv=Av$. It is finitely generated and torsion, so the structure theorem gives cyclic summands $F[X]/(d_i)$ with $d_1\mid \cdots \mid d_k$. In the power basis of each summand, multiplication by $X$ has the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form, unique because the invariant factors are unique.

The structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated module over a Euclidean domain $R$ iswith the rank and nonunit invariant factors of a linear operator unique up to associates.

During any year, repeats may be ignored until the first non-repeat outcome. Conditional on that event, the chance of progressing isThe student must progress successfully through all three years. By the Markov property, the required probability is therefore

The three year states are transient states: from each there is positive probability of entering an absorbing state, and repeated trials make eventual departure certain because $q<1$. The singleton classes $\{L\}$ and $\{G\}$ are closed absorbing states, hence recurrent. Thus the chain has two closed recurrent classes and three transient singleton classes.

In the order $(1,2,3,L,G)$, where $L$ means left without a degree and $G$ means graduated, the transition matrix isHere $p+q+r=1$.