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The vector triple product identity gives $\mathbf{n}\times (\mathbf{n}\times \mathbf{x})=\mathbf{n}(\mathbf{n}\cdot \mathbf{x})-\mathbf{x}$, and henceThus $\Phi$ is the identity on the orthogonal complement ${\mathbf{n}}^{\perp }$ and multiplies the component parallel to $\mathbf{n}$ by $\alpha$. It is invertible exactly when $\alpha \ne 0$, and then

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

The fundamental theorem of calculus says first that if $h$ is continuous on $[a,b]$ and $H(x)={\int }_a^{x}h(t)dt$, then $H^{\prime }(x)=h(x)$. Indeed,whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F^{\prime }$ shows that $F(x)-{\int }_a^x{F}^{\prime }(t)dt$ has zero derivative and is constant. Hence

A derivative need not be Riemann integrable. Define $f(0)=0$ and $f(x)=x^2\sin (x^{-2})$ for $x\ne 0$. Then $f$ is differentiable at zero with $f^{\prime }(0)=0$, while for $x\ne 0$,which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, $g=f^{\prime }{|}_{[0,1]}$ need not be Riemann integrable.

Assume $a_n\to 0$. The function is bounded, and every interval has lower Darboux sum zero because it contains an irrational. Given $\epsilon >0$, choose $N$ so that $a_n<\epsilon /2$ for $n>N$. Put small intervals around the finitely many points $q_1,\dots ,q_N$ whose total length makes their contribution to the upper sum less than $\epsilon /2$, and refine their endpoints to a partition. On every remaining subinterval the function is at most $\epsilon /2$, so the upper sum is less than $\epsilon$. The Riemann integrability criterion proves that $f$ is Riemann integrable with integral zero.

The condition $a_n\to 0$ is sufficient but not necessary. Assign $a_n=1$ when $q_n=1/k$ for some positive integer $k$, and $a_n=1/n$ otherwise. Then $(a_n)$ does not tend to zero. Outside a short interval near zero only finitely many unit spikes occur and can be enclosed in intervals of arbitrarily small total length; all remaining values can be made uniformly small except for finitely many further points. The same Darboux-sum argument proves that this $f$ is Riemann integrable with integral zero.

At a nonzero rational $p/q$ in lowest terms, irrational points approach $p/q$ with function value zero, whereas $f(p/q)=1/q>0$, so $f$ is discontinuous. At zero, $|p/q|\ge 1/q$ for nonzero $p$, hence $0\le f(x)\le |x|$ and continuity follows.

At an irrational $x$, given $\epsilon >0$, only finitely many rationals in a bounded neighbourhood have reduced denominator $q\le 1/\epsilon$. Choose a neighbourhood avoiding those points; throughout it, $f<\epsilon$. Consequently the continuity set is $(\mathbb{R}\setminus \mathbb{Q})\cup \{0\}$. This is a variant of the Thomae function.

For $x\ne 0$, $f(x)=|x{|}^x=e^{x\log |x|}$, so $f^{\prime }(x)=|x{|}^x(\log |x|+1)$. Although $x\log |x|\to 0$ makes $f$ continuous at zero, $(f(x)-1)/x\sim \log |x|\to -\infty$, so it is not differentiable there.

Since the cosine function is even, $g(x)=\cos |x|=\cos x$, so it is differentiable everywhere with $g^{\prime }(x)=-\sin x$. Finally, $h(x)=x|x|$ is differentiable everywhere, including at zero, and $h^{\prime }(x)=2|x|$. Therefore

The mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f^{\prime }(c)(b-a)$ for some $c\in (a,b)$.

For $F(t)=\cos (e^{-t})$ and $t\ge 0$,The mean value theorem gives $|\cos (e^{-x})-\cos (e^{-y})|\le |x-y|$.

For the second inequality put $t=\sqrt{1+x}\ge 1$. It becomes $2\log t\le t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ andThus $\log (1+x)\le x/\sqrt{1+x}$.

Differentiability at $x_0$ meansHence $f(x)-f(x_0)=(x-x_0)(f^{\prime }(x_0)+o(1))\to 0$ as $x\to x_0$. Therefore every differentiable function is continuous at each point where it is differentiable.

Regularity at the centre requires finite ${\rho }_c,P_c$ and $M(0)=0$. Then $M(r)=4\pi {\rho }_c{r}^3/3+o(r^3)$, so $P^{\prime }(0)=0$. At a free stellar surface, $P(R)=0$; more generally it matches the external pressure. For $P=K{\rho }^2$ with $K>0$ and vacuum outside, this also gives $\rho (R)=0$.

Inside the star, cancel positive $\rho$ to obtain $2K{\rho }^{\prime }=-GM/r^2$. Differentiate and use mass conservation:Thus the Lane-Emden equation of polytropic index one isFor example, the regular solution is $\rho ={\rho }_c\sin x/x$, with its central value interpreted by continuity, and the first vacuum surface is at $x=\pi$. This also verifies the stated central and surface boundary conditions.

For a thin spherical shell, the radial pressure force per unit volume is $-dP/dr$. By the shell theorem, the inward gravitational acceleration is $GM(r)/r^2$, because exterior spherical shells contribute no force. Balancing force densities in hydrostatic equilibrium givesThe second equation is mass conservation; it closes the pressure-support equation once an equation of state is supplied.

The count weights correct the loss of frequencies for a deliberately coarsened binary response, but do not supply a third outcome category. Thus this remains inappropriate for the full contingency table. With the base R factor convention, it could fit the chosen first damage level versus the other two combined, using a logistic regression and the frequencies as weights. That is a different scientific question. The supplied log-linear model retains all three categories and their counts without this information loss.

This does not model the full three-category response. A binomial distribution has two outcome classes, whereas the observed damage factor has three. In base R, a factor response to a binomial generalized linear model is coded as first level versus all other levels; this would silently collapse the categories. It would also treat the nine table rows as nine individual observations, ignoring the frequencies. A deliberate binary analysis would require specifying the collapse and retaining the counts, but cannot recover the full damage distribution.

This is not a valid ordinary R glm family specification. A multinomial distribution is natural for the three damage counts conditional on the row total, but base R's generalized linear model interface does not supply a family named multinomial. Furthermore, the written response is one cell count, not a three-category response or grouped multinomial vector. A properly specified multinomial logistic regression could model category probabilities by group, but it is not the model fit by this command. The independent-count log-linear model in part (a), conditional on the margins, is the appropriate supplied alternative.

This is the appropriate command for the full contingency table. It fits a log-linear model to the nine cell counts, with logarithmic link function and no interaction:With baseline constraints on $g_i,d_j$, there are five parameters. The factorized means express independence of treatment group and damage category. The maximum-likelihood estimates areAlthough the experiment fixes each row total at 30, conditioning these independent Poisson random variables on those totals produces the appropriate row-wise multinomial distribution. The Poisson trick therefore gives the correct likelihood inference for this independence log-linear model for a two-way contingency table; the margins are nuisance parameters. It tests whether treatment changes the distribution of damage, rather than modelling a binary outcome.

Let $K$ be the diagonal halting set, which is undecidable. The pumping argument below actually holds for every subset $K\subseteq \mathbb{N}$; nonregularity needs the stated convention, or at least a set of exponents whose unary language is nonregular.

The pumping lemma for regular languages holds with pumping length one. For an accepted nonempty word, pump its first symbol. A word containing only $1$s remains in $1^{\ast }$ even if that symbol is deleted. If there is a zero and a nonempty prefix before its last zero, pumping the first symbol leaves the last zero and its following $n\in K$ ones unchanged. If the word is $01^n$, pumping its initial zero produces $0^i{1}^n$; this is accepted for $i\ge 1$, and for $i=0$ it belongs to $1^{\ast }$. In every case the pumped substring is nonempty and lies in the first position.

If $L$ were a regular language, closure under intersection would makeregular. A finite-state automaton for this intersection would decide membership of $n$ in the diagonal halting set by reading $01^n$, a contradiction. Thus $L$ satisfies the pumping lemma but is not regular, under the usual meaning of $\mathbb{K}$. If $K$ were arbitrary, the printed conclusion would be false, for example for $K=\mathbb{N}$.

Use state elimination for finite automata on a generalized finite automaton, whose directed edges are labelled by regular expressions. Add a fresh initial state $s$ with an $\epsilon$-edge to the old initial state and a fresh final state $f$ with $\epsilon$-edges from all old accepting states. Every missing edge has label $\varnothing$, the empty language; parallel edges are merged by union. The label $\epsilon$ denotes the language containing just the empty word.

When eliminating a state $k$, replace the label on each surviving edge $i\to j$ byHere juxtaposition means concatenation of formal languages, and the Kleene star allows zero or more traversals of the loop at $k$. Then delete $k$ and its incident edges. Eliminate every original state; the remaining $s\to f$ label is a regular expression for exactly the accepted formal language. The update accounts for paths avoiding $k$ and paths entering $k$, looping there any number of times, and leaving it.

Use a Fourier mode $u_m^n={\zeta }^n{e}^{im\theta }$. Substitution givesIts roots have product one. They lie on the unit circle for all spatial frequencies precisely when $|1-2\mu {\sin }^2(\theta /2)|\le 1$, givingFor $\mu >1$, the Nyquist frequency has a real reciprocal root pair with one root of modulus greater than one, giving exponential growth and instability.

The endpoint convention needs care in this two-step wave scheme: repeated unit roots occur at zero frequency, and also at the Nyquist frequency when $\mu =1$. They are not power-bounded for arbitrary two independent displacement sequences with a mesh-independent difference. With the usual initialization from displacement and velocity,each Fourier component satisfies ${\hat{u}}^n=\cos (n\eta ){\hat{u}}^0+k[\sin (n\eta )/\sin \eta ]{\hat{v}}^0$, where $\cos \eta =1-2\mu {\sin }^2(\theta /2)$. Taking the limiting values at repeated roots and using $|\sin (n\eta )|\le n|\sin \eta |$ gives $|{\hat{u}}^n|\le |{\hat{u}}^0|+nk|{\hat{v}}^0|$. This is a uniform finite-time stability bound in the physical displacement/velocity data and includes $\mu =1$. A strict root-condition test for arbitrary two-level data must record the double-root qualification rather than calling that endpoint unconditionally power-bounded.

Let $L=\mathrm{tridiag}(1,-2,1)$. The zero Dirichlet boundary values introduce no forcing terms, so the scheme isThe common orthogonal sine eigenvectors have $L$-eigenvalues $-4{\sin }^2[\pi j/(2(M+1))]$. Thus $B$ is invertible for every $\mu >0$, and each amplification eigenvalue isThe matrix $B^{-1}C$ is orthogonally diagonalizable, so its powers have Euclidean norm at most one, uniformly in the mesh and time step. Therefore Crank-Nicolson diffusion scheme is stable for every $\mu >0$. Negative amplification factors at large $\mu$ can create temporal sign oscillations, which do not contradict this norm stability.

For $-\beta /{\ell }^2<V\le 0$, there is one positive stationary wavenumber $k_0\in (0,\ell ]$ satisfying $\beta (k_0^2-{\ell }^2)/(k_0^2+{\ell }^2{)}^2=V$, and its negative. Since ${\omega }^{\prime \prime }(k_0)>0$, their paired stationary-phase contribution isAt $V=0$ the finite stationary points are $\pm \ell$ and are nondegenerate, so the same formula applies, assuming sufficient decay to control the large-wavenumber tail.

For $V>\beta /(8{\ell }^2)$ or $V<-\beta /{\ell }^2$, there are no real stationary points: no group-velocity ray reaches those slopes. Integration by parts gives much smaller contributions; for a Schwartz amplitude they decay faster than every inverse power of $t$ at a fixed such $V$. For less regular data the rate depends on the amplitude, and the printed even function assumption alone does not specify it. At the excluded endpoint values of the group velocity stationary points coalesce and an ordinary $t^{-1/2}$ formula fails, typically requiring an Airy-type transition rather than the expressions above.

For $0<V<\beta /(8{\ell }^2)$ there are two positive solutions $k_1\in (\ell ,\sqrt{3}\ell )$ and $k_2>\sqrt{3}\ell$, together with their negatives. The second derivative isEach negative stationary point has the conjugate contribution of its positive partner, because $A$ is real even and ${\Phi }_V$ is odd. ThereforeIf a leading amplitude vanishes, this gives a zero coefficient and a higher-order term supplies that branch's first nonzero contribution.