The probabilists' Hermite polynomials satisfywhich follow by differentiating the Rodrigues formula. Thusso ${\alpha }_n=0$ and ${\beta }_n=n$.

For the squared norm, substitute the Rodrigues formula and integrate by parts $n$ times. All boundary terms vanish under the Gaussian weight, and ${\mathrm{He}}_n^{(n)}=n!$, giving

Inductively, $p_n$ is monic of degree $n$, because the leading term of $p_{n+1}$ comes only from $x{p}_n$. Assume $p_0,\dots ,p_n$ are mutually orthogonal. The definitions giveUsing the preceding recurrence in the formgivesso $⟨p_{n+1},p_{n-1}⟩=0$ by the definition of ${\beta }_n$. If $j\le n-2$, then $x{p}_j$ lies in the span of $p_{j-1},p_j,p_{j+1}$, all orthogonal to $p_n$, and the remaining recurrence terms are also orthogonal to $p_j$. Hence $⟨p_{n+1},p_j⟩=0$. Induction proves that the recurrence defines monic orthogonal polynomials.

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.

If $K<0$ everywhere, compactness makes the integral negative, so $\chi (S)<0$. For every triangulation, $3F=2E$ again yields the general identitySince $V\ge 1$ and $\chi (S)<0$,This supplies bounds depending only on the surface.

A geodesic triangulation of a closed smooth surface decomposes it into finitely many triangular discs whose edges are geodesic arcs and whose intersections are common faces. Its Euler number isThe Gauss-Bonnet theorem for a closed surface statesFor any triangulation of a closed surface, each triangular face has three edges and every edge belongs to two faces, so $3F=2E$. The sum of all vertex valencies is $2E$. Hence the average valency isThe torus has Euler characteristic $\chi =V-E+F=0$. Using $F=2E/3$ gives $V-E/3=0$, so $E=3V$ and

A local form of the Picard-Lindelöf theorem is as follows. Let $F(t,y)$ be continuous on a rectangle about $(t_0,y_0)$ and uniformly Lipschitz in $y$ there. Then some interval about $t_0$ admits a unique continuously differentiable solution ofThe solution is the unique fixed point of the associated integral operator. If $F$ is continuous on $[a,b]\times {\mathbb{R}}^n$ and globally Lipschitz in $y$ with one constant, the solution exists uniquely throughout $[a,b]$.

Factor the equation in the Eisenstein integers:Let $\pi =1-\omega$. Its norm is $3$, andso $3$ is a unit times ${\pi }^2$. Suppose $x\equiv 2(\mathrm{mod}3)$, say $x=3k+2$. Since $\omega =1-\pi$,The first term is divisible by ${\pi }^2$ and the second has discrete valuation exactly one, so $v_{\pi }(x+\omega )=1$. Conjugation gives $v_{\pi }(x+{\omega }^2)=1$ as well. ConsequentlyBut $v_{\pi }(y^3)=3v_{\pi }(y)$ is divisible by three, a contradiction. Hence

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

The Eisenstein integers are $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$. Their multiplicative algebraic norm isGiven $\alpha ,\beta \ne 0$, choose an Eisenstein integer $q$ nearest to the complex number $\alpha /\beta$. The hexagonal lattice has covering radius $1/\sqrt{3}$, soWith $r=\alpha -q\beta$, multiplicativity givesThus division with remainder decreases the nonnegative integer norm, proving that $\boxed{\mathbb{Z}[\omega ]\text{ is a Euclidean domain}}$.

For any action $a$, decompose the posterior expected quadratic loss around the posterior mean:The unique minimizer is therefore the Bayes estimator under squared error loss. Using the mean of the posterior Beta distribution givesUnder repeated sampling at parameter value $\theta$, $S$ has the binomial distribution with expected value $n\theta$. Consequentlyso its bias of an estimator is $(1-2\theta )/(n+2)$.

The quadratic loss, also called squared-error loss, for reporting an action $a$ when the parameter is $\theta$ is

Let $S={\sum }_{i=1}^n{X}_i$ be the number of heads. The likelihood function for the Bernoulli distribution sample is proportional to ${\theta }^S(1-\theta {)}^{n-S}$. A uniform distribution prior on $[0,1]$ is the $\mathrm{Beta}(1,1)$ distribution, so Beta-binomial conjugacy gives the posterior distribution

First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.

Suppose for contradiction that the continuous image of a connected space $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the subspace topology on $Y$, with $Y=U\cup V$. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by surjectivity both are nonempty, and their union is $X$. This is a separation of a topological space of $X$, contradicting that $X$ is a connected space. Hence $Y$ is connected.

At a point with $xy\ne 0$, the determinant of the derivative isThe inverse function theorem therefore supplies neighbourhoods on which $f$ has a continuously differentiable, hence differentiable, inverse. Concretely, if $(u,v)=f(x,y)$, the local inverse is the restriction ofwhich is continuously differentiable near the image whenever $y\ne 0$.

For $x\ne 0$, all component functions are continuously differentiable and the Jacobian matrix isAt a point $(0,y_0)$, the first component cannot be differentiable: along the first coordinate axis its difference quotient contains $h^{1/3}/h=h^{-2/3}$, which is unbounded as $h\to 0$. Therefore $f$ is continuously differentiable exactly at the points with $x\ne 0$.