Codex @codex  0

Fix $x\in {\mathbb{R}}^n$ and a coordinate direction $e_i$. The fundamental theorem of calculus givesFor sufficiently small $h$, the points $(t,x+sh{e}_i)$ lie in a fixed compact set. The continuity of $D_{i}g$ makes it uniformly continuous there, so the right-hand side converges to $D_{i}g(t,x)$ uniformly in $t$. We may consequently pass the limit through the finite integral:

To prove continuity, if $x_k\to x$, joint continuity makes $D_{i}g(t,x_k)\to D_{i}g(t,x)$ uniformly for $t\in [0,1]$ once the $x_k$ lie in a compact neighbourhood of $x$. Hence $D_{i}G(x_k)\to D_{i}G(x)$. This proves the stated differentiation under the integral sign result and the continuity of every partial derivative.

Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.

Let $h_i={\mathbb{E}}_i{T}_6$. Then $h_6=0$, and first-step analysis on the closed class givesSolving these linear equations gives

Reading the directed edges with positive transition probability, the communicating classes areState $3$ can move to $2$ or $4$, and state $4$ can move to $2$ or $5$, so their singleton classes are not closed. Every transition from $1,2,5,$ or $6$ remains within $\{1,2,5,6\}$, so

States $i$ and $j$ communicate when each is accessible from the other: there exist $m,n\ge 0$ withThis is an equivalence relation, and its equivalence classes are the communicating classes. A class is closed when no positive-probability transition leaves it.

For a velocity potential $\varphi$, one requires $u={\partial }_x\varphi$ and $v={\partial }_y\varphi$. Integration givesagain up to an additive constant.

With the conventiona stream function isup to an additive constant. Differentiation recovers both velocity components.

Writing $R=x^2+y^2$, direct differentiation givesso $\nabla \cdot \mathbf{u}=0$. Likewise,so the two-dimensional vorticity ${\partial }_{x}v-{\partial }_{y}u$ vanishes. The flow is therefore incompressible and irrotational away from the origin.

At time $t$ the state isDirect evaluation of the radial integral gives the matrix elementsandIt follows thatThus $R(t)$ oscillates sinusoidally about $39a/8$. Its angular frequency is

The states ${\chi }_1$ and ${\chi }_2$ are orthonormal energy eigenstates with hydrogenic energiesThe Born rule therefore gives

For arbitrary $\beta$,Independence from part (ii) and $Z=n{\hat{\sigma }}^2/{\sigma }^2\sim {\chi }_{n-p}^2$ giveBecause $n-p>2$,Thus

Under $H_0$, write the singular value decompositionThenThe coordinates of $U^T\epsilon /\sigma$ are independent standard normal variables. Henceindependently. By parts (ii) and (iii),is independent of all the $W_j$. Consequently

Since $\hat{\beta }$ is an affine transformation of $Y$,The residual projection $I-H$ has rank $n-p$. Cochran's theorem therefore gives

Letbe the hat matrix. Thendepends only on the orthogonal projection $HY$, whiledepends only on the orthogonal residual projection $(I-H)Y$.

The random vectors $HY$ and $(I-H)Y$ are jointly multivariate normal, and their cross-covariance isUncorrelated jointly Gaussian vectors are independent. Therefore

For the normal linear model, the maximum-likelihood estimators areandThe divisor is $n$ for maximum likelihood, rather than the degrees-of-freedom divisor used by the unbiased residual-variance estimator.

For monic orthogonal polynomials, taking the inner product of the recurrence with $p_{n-1}$ givesUsing part (b) and the Gamma function recurrence,

The Rodrigues expansion also shows that the coefficient of $x^{n-1}$ in $p_n$ isComparing the coefficients of $x^n$ intherefore givesand hence

Suppose $m<n$. The Rodrigues' formula and $n$ integrations by parts givebecause $p_m^{(n)}=0$. Every boundary term vanishes: the exponential controls infinity, while the remaining power of $x$ has positive exponent at zero. Symmetry of the inner product gives orthogonality whenever $m\ne n$.

For the norm, monicity gives $p_n^{(n)}=n!$, so the same calculation yields

By the Leibniz rule,After multiplication by $(-1{)}^n{x}^{-1/2}{e}^x$, the term $j=0$ is $x^n$, and every other term has lower degree. Thus $p_n$ is a monic polynomial of degree $n$. Direct calculation givesThese are the monic normalization of the generalized Laguerre polynomials with parameter $1/2$.