The $j$th column of the matrix of $Df(a)$ in the standard bases is $Df(a)e_j$. Its $i$th component is $\partial f_i/\partial x_j(a)$, found by taking increments $t{e}_j$. Thus the matrix of the total derivative is the Jacobian matrix, with entry $(i,j)$ equal to $\partial f_i/\partial x_j(a)$.

The directional derivative in the specified vector direction isThe equality follows from the total-derivative remainder estimate; for the zero direction the result is zero. The convention keeps the supplied vector and does not replace it by a unit vector.

The total derivative is a linear map $Df(a):{\mathbb{R}}^n\to {\mathbb{R}}^n$ such thatIt is unique: the difference of two such linear maps, evaluated on $h=tv$, must vanish on every fixed vector $v$ after division by $t$ and passage to the limit.

A useful differentiation theorem says: if $f(t,x)$ is integrable in $x$, differentiable in $t$ for almost every $x$, and on a neighborhood of each parameter value the derivative is bounded in absolute value by one integrable function of $x$, then the integral is differentiable and its derivative is the integral of ${\partial }_{t}f$. The proof applies dominated convergence to difference quotients, bounded by that same majorant by the mean value theorem.

Fix $t_0>0$ and choose $0<a<t_0<b$. For $t\in [a,b]$,The first part is integrable by part (i). The second is integrable by the permitted hypothesis for $h_b$; alternatively exponential decay dominates its polynomial and logarithmic factors. The theorem therefore applies locally at every $t_0$, giving

Since $\log x\le 0$ on $(0,1]$, substitute $x=e^{-s}$ to obtainThe functions are measurable, so this proves Lebesgue integrability. Also $0<e^{-x}\le 1$ gives $|h_t(x)|\le |g_t(x)|$. Hence $h_t$ is integrable on $(0,1]$ as well.

Take the middle-thirds Cantor set $C$. Its $n$th stage is the union of $2^n$ closed intervals of length $3^{-n}$, and contains $C$. Their total length is $(2/3{)}^n\to 0$; slightly enlarging them to open intervals gives covers of arbitrarily small total length. Therefore $C$ is null.

For uncountability map every binary sequence $({\epsilon }_j{)}_{j\ge 1}$ to ${\sum }_{j\ge 1}2{\epsilon }_j{3}^{-j}\in C$. If two sequences first differ at index $k$, their image difference in absolute value is at least $2\cdot 3^{-k}-{\sum }_{j>k}2\cdot 3^{-j}=3^{-k}>0$. Thus the map is injective. The set of binary sequences is uncountable by Cantor's diagonal argument, so $C$ is uncountable.

Let $E_n$ be null. Given $\epsilon >0$, cover $E_n$ by intervals of total length less than $\epsilon 2^{-n}$, for $n\ge 1$. The union of all these interval families is countable, covers ${\bigcup }_n{E}_n$, and has total length less than ${\sum }_n\epsilon 2^{-n}=\epsilon$. Thus the countable union is null.

A set $E\subseteq [0,1]$ is null when for every $\epsilon >0$ it can be covered by countably many open intervals whose lengths sum to less than $\epsilon$. Equivalently its Lebesgue outer measure is zero. This definition does not presuppose measurability of $E$.

Euler's criterion implies multiplicativity of the symbol on nonzero residues and gives $(-1/p)=(-1{)}^{(p-1)/2}$. Apply quadratic reciprocity with the odd prime 3:The only nonzero square modulo three is one. Since $p>3$,

If $m=uv$ with both factors of positive degree, evaluation gives $u(a)v(a)=0$. Since $K$ is a field, one factor vanishes at $a$, contrary to the minimal degree of $m$. Thus $m$ is an irreducible polynomial over $F$.

For the final unnumbered request, $\zeta =e^{\pi i/3}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ satisfies ${\zeta }^2-\zeta +1=0$. The monic quadratic $x^2-x+1$ has no rational root, since its discriminant is $-3$. It is therefore an irreducible polynomial over $\mathbb{Q}$, and

Divide $f$ by $m$ in the polynomial ring over the field: $f=qm+r$, where $r=0$ or $\mathrm{deg}r<\mathrm{deg}m$. If $f(a)=0$, then $r(a)=0$, and minimality forces $r=0$. Conversely a multiple of $m$ vanishes at $a$. ThereforeThis includes the zero polynomial.

An element is algebraic over $F$ when some nonzero polynomial in $F[x]$ vanishes at it. Its minimal polynomial is the monic polynomial of least positive degree with this property; scaling a least-degree nonzero vanishing polynomial makes it monic.

If $m_1,m_2$ are two such polynomials, they have the same least degree. Their difference has smaller degree because both are monic, and it also vanishes at $a$. Minimality forces the difference to be zero. Hence the monic minimal polynomial is unique.

Independence of the slope estimate and residual variance, together with their normal and chi-square laws, gives the Student pivotal quantityThus an exact Gaussian-model two-sided 95% confidence interval isUsing a normal quantile without estimating the variance uncertainty would not give this exact finite-sample coverage.

For $n>2$, the usual unbiased estimator of variance isThe fitted mean lies in the two-dimensional span of the constant design vector and $x$. Orthogonal projection of the Gaussian error onto its complement gives $n-2$ independent standardized normal coordinates. HenceThis residual projection is independent of the fitted-coefficient projection. If maximum likelihood for variance is intended instead, its estimator is ${\hat{\sigma }}_{\mathrm{ML}}^2=\mathrm{RSS}/n$, with $n{\hat{\sigma }}_{\mathrm{ML}}^2/{\sigma }^2\sim {\chi }_{n-2}^2$; the denominator differs, not the residual degrees of freedom.

For nondegenerate design let $S_{xx}={\sum }_i{x}_i^2>0$. Maximizing the Gaussian likelihood over the mean parameters is equivalent to minimizing ${\sum }_i(Y_i-\alpha -\beta x_i{)}^2$. Since ${\sum }_i{x}_i=0$, the normal equations separate:Its error is the linear combination $\hat{\beta }-\beta ={\sum }_i{x}_i{ϵ}_i/S_{xx}$. Independence and normality therefore giveIf every design point is zero, the slope is not identifiable and no such slope estimator can be inferred; the standard regression answer assumes the stated nondegeneracy.

There are six pooled classes, one normalization constraint, and one fitted Poisson distribution parameter. Thus the reference degrees of freedom are $\boxed{6-1-1=4}$. Under the null model the Pearson statistic has approximately the Chi-squared distribution with four degrees of freedom. Its supplied 5% upper critical value is $9.49$.

Because $\boxed{2.07<9.49}$, do not reject the Poisson count distribution at the 5% level. This is consistency with that distribution, not proof that it is exactly true. A test of the marginal count frequencies also does not by itself verify independence of successive time windows.

Separate high-count classes have very small expected frequencies, making the chi-square approximation unreliable. Pooling the upper tail gives a class with expected count about $20.9$, comfortably above the customary small-cell threshold. The pooling protects the asymptotic test approximation, while retaining the tail count as part of the goodness-of-fit comparison.