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Here the autonomous planar system is $\dot{x}=v$, $\dot{v}=-\cos x$. Its equilibrium points areThe Jacobian matrix isFor even $n$, the eigenvalues are $1,-1$, so these points are unstable saddle equilibria. Writing $\xi =x-x_{\ast }$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the eigenvalues are $i,-i$.

To classify the latter points nonlinearly, differentiate the first integralAt $x_{\ast }=3\pi /2+2m\pi$ this energy has a strict local minimum. Its nearby level sets are closed curves, proving that the odd-$n$ points are stable center equilibria, with no attraction to the center. The phase portrait consists of closed oscillatory orbits for $-1<H<1$, heteroclinic orbits at $H=1$ joining consecutive saddle equilibria, and open rotating orbits for $H>1$. The separating curves haveAbove the horizontal axis $\dot{x}=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.

The covariance is symmetric, because multiplication of real random variables is commutative. Expanding with the bilinearity just proved givesIdentical probability distributions imply equal second moments and equal means, hence equal variances. Therefore the assertion is true, withThe cancellation shows exactly why no assumption of independence is needed.

Linearity of expectation gives $\mathbb{E}(X+Y)=\mathbb{E}X+\mathbb{E}Y$. Expand the centered product and use linearity of expectation again:Thus the assertion is true whenever the covariances are defined. Neither independence nor identical distributions are required.

A direct construction, valid for every $0\le t<1$, isThe first coordinates give $a_1\cdot a_2=a_1\cdot a_3=t$, andThe first two norms are one. For the third,Thus the required Gram matrix is $(1-t)I+t11^T$. The column matrix is upper triangular and has determinantso these vectors are linearly independent. This is an explicit instance of unit vectors with a common inner product; at $t=0$ it reduces to the standard orthonormal basis.

No. For every $n\ge 2$, takein ${\mathbb{R}}^n$. These form a minimally dependent vector family: the full family is dependent because $a_n-{\sum }_{j=1}^{n-1}{a}_j=0$.

If the omitted vector is $a_n$, the remaining coordinate vectors are independent. If the omitted vector is $e_k$, a relation among the retained vectors has an $e_k$ coordinate coming only from $a_n$, so its coefficient must be zero; the coefficients of all retained coordinate vectors must then be zero. Thus every $(n-1)$-element subfamily is linearly independent, as required for a counterexample. When $n=1$, the empty subfamily is independent while the one-vector family consisting of the zero vector is not, so the answer remains negative.

Yes. Suppose ${\alpha }_1{a}_1+{\alpha }_2{a}_2+{\alpha }_3{a}_3=0$. Taking its dot product with $a_2\times a_3$ eliminates the last two terms and givesThe nonzero scalar triple product forces ${\alpha }_1=0$. Cyclically using $a_3\times a_1$ and $a_1\times a_2$ forces ${\alpha }_2={\alpha }_3=0$ as well. This is the definition of linear independence. Equivalently the three-column matrix has nonzero determinant.

At $x=1/4$ the terms are $c_n$. Their decay to zero alone is insufficient for convergence. In factbecause the difference has numerator $n\ge 0$. Since $c_1=1/2$, induction gives $c_n\ge 1/(n+1)$. The comparison harmonic series diverges, so

At $x=-1/4$ the terms are $(-1{)}^n{c}_n$. Part (ii) gives positive decreasing magnitudes tending to zero, so the alternating series test applies. To see the convergence directly, let $S_N={\sum }_{n=1}^N(-1{)}^n{c}_n$. Thenwhile $S_{2k}\ge -c_1$, obtained by grouping $S_{2k}=-c_1+(c_2-c_3)+\cdots +(c_{2k-2}-c_{2k-1})+c_{2k}$. Thus the even partial sums have a finite limit. The odd partial sums differ from them by $-c_{2k+1}\to 0$, so the whole sequence of partial sums has the same limit. Absolute convergence would require convergence of $\sum c_n$, already disproved. Hence

Extend the sequence by $c_0=1$. Cancelling factorials in consecutive normalized central binomial coefficients givesThe given logarithmic inequality therefore impliesOn $[j,j+1]$, $1/t\le 1/j$, so summing its integrals gives ${\sum }_{j=1}^{n}1/j\ge \log (n+1)$. Exponentiating yields the explicit boundAlso every consecutive ratio is less than one, so $(c_n)$ is strictly decreasing, a fact needed for the alternating endpoint.

Rearranging the assumed positive-term ratio inequality givesInduction therefore yields $a_n/b_n\le a_1/b_1$ for every $n\ge 1$. In particularThe positive partial sums of the first series are nondecreasing and bounded above, so the bounded monotone sequence theorem proves that they converge. This is the ratio comparison test, and establishes convergence of $\sum a_n$.

Partial fractions give $f(z)=-1/(z-1)+1/(z-2)$. In the interior disk, expand both terms as geometric series:Thus the Laurent series is a Taylor series in this region:The nearer pole at $z=1$ determines the radius of convergence.

The uniform limit theorem first gives continuity of $f$ on $D$. For any triangle $\Gamma$ whose closed interior is in $D$, the Cauchy integral theorem gives ${\int }_{\Gamma }{f}_n(z)dz=0$. Moreover,Hence ${\int }_{\Gamma }f=0$, and Morera's theorem proves that $f$ is analytic on $D$. In fact, the proof only needs locally uniform convergence on compact subsets, since each triangle is compact; the stated uniform convergence on the entire domain is stronger.

Morera's theorem states that a continuous complex-valued function on an open set is holomorphic if its integral around every triangle whose closed interior lies in the set is zero. It is enough to impose this condition locally on disks.

Fix a disk $B$ contained in the open set and a base point $z_0\in B$. Define $F(z)$ as the integral of $f$ along the straight segment from $z_0$ to $z$. The zero triangle integral gives, for sufficiently small $h$,Continuity makes the quotient by $h$ tend to $f(z)$, so $F$ is holomorphic and $F^{\prime }=f$. To justify that its derivative is holomorphic, apply the Cauchy integral formula for $F$ on a smaller circle:The right side may be differentiated again inside the circle because its denominator stays uniformly away from zero on the contour. Thus $F^{\prime }=f$ is holomorphic on $B$. Such disks cover the open set, proving the theorem.

For the single-frequency response, use separation of variables with $u=R(r)\sin \omega t$. Set $k=\omega /c$ and $V=rR$; then $V^{\prime \prime }+k^{2}V=0$, so $V=C\sin kr+D\cos kr$. A finite origin value forces $D=0$, since $\cos kr/r$ is singular. Since $\sin kr/r\to k$, the prescribed origin amplitude gives $C=A/k$. The regular time-harmonic spherical wave is thereforewith the quotient defined by its limit at zero.

This answers the literal finite-origin condition in the class of regular single-frequency responses. It is a standing combination of inward and outward waves: the numerator is proportional to $\cos (kr-\omega t)-\cos (kr+\omega t)$. A purely outgoing wave from an ideal point source would instead be proportional to $r^{-1}\sin [\omega (t-r/c)]$, which has no finite prescribed value $u(0,t)$. Such a source must be specified by a singularity strength, for example ${\lim }_{r\downarrow 0}ru$. The source's wording supplies neither an initial field nor an outgoing singular-source normalization; without restriction to the stated harmonic response, a general initial-value problem is not uniquely determined by the one origin condition.

Put $v(r,t)=ru(r,t)$. Multiplication of the radial wave equation by $r$ gives $v_{tt}=c^2{v}_{rr}$ for $r>0$. In the characteristic coordinates $\xi =r+ct$, $\eta =r-ct$, this becomes $v_{\xi \eta }=0$. Integrating first in $\xi$ and then in $\eta$ gives $v=f(\xi )+g(\eta )$. ConsequentlyThe two terms are incoming and outgoing radial disturbances. If the field is finite at the origin, their numerators must cancel there; arbitrary independent choices generally produce a $1/r$ singularity.

Let $\alpha =\pi (m+1)/n$, which lies strictly between zero and $\pi$. On the lower ray $z=x$, the integral is $I_R={\int }_0^R{x}^m/(1+x^n)dx$. On the upper ray $z=x{e}^{2\pi i/n}$, the positive boundary orientation runs from $R$ down to zero. Since $z^n=x^n$, that contribution is $-e^{2i\alpha }{I}_R$. The circular contribution vanishes by part (b); henceUse $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$ to obtainThe result is positive as expected. Convergence near zero follows from $m\ge 0$, so both endpoint limits used in the contour argument are justified.

On the circular arc, $|z|=R$, the reverse triangle inequality gives $|1+z^n|\ge R^n-1$. The arc length is $2\pi R/n$, so the ML inequality givesThe strict assumption $n>m+1$ is exactly what makes the exponent negative; it also ensures convergence of the positive-real-axis integral at infinity.

The sector contains exactly one simple pole, at $z_0=e^{i\pi /n}$. All roots of $1+z^n$ are simple, and its residue issince $z_0^n=-1$. For the positively oriented boundary, the residue theorem givesThe vertex at zero is harmless: $f$ is analytic there for nonnegative integer $m$, or a small indent may be removed because its integral is $O({\rho }^{m+1})$.

Let $m=\dim U$, $n=\dim V$, $r=\mathrm{rank}\alpha$ and $r_{\ast }=\mathrm{rank}{\alpha }^{\ast }$. The isomorphism of part (c) preserves dimensions, so the formulas of part (a), applied also to ${\alpha }^{\ast }:V^{\ast }\to U^{\ast }$, giveIf $m>0$, cancellation proves $\mathrm{rank}\alpha =\mathrm{rank}{\alpha }^{\ast }$. If $m=0$, both maps are zero and the equality remains true. This argument derives dual-rank equality from the requested subspace correspondence without assuming it beforehand.

Here $\Phi$ acts on $\beta :V\to U$, so it is the same dualization construction with the source and target interchanged: ${\Phi }_{V,U}:L(V,U)\to L(U^{\ast },V^{\ast })$. This is the natural family interpretation of the symbol used in part (b).

If $\beta \alpha =0$, composition reversal gives ${\alpha }^{\ast }{\beta }^{\ast }=0$, so ${\beta }^{\ast }\in M^r({\alpha }^{\ast })$. Conversely, if $\delta \in M^r({\alpha }^{\ast })$, the isomorphism in part (b) writes $\delta ={\beta }^{\ast }$ uniquely. Then ${\alpha }^{\ast }{\beta }^{\ast }=0$ implies $(\beta \alpha {)}^{\ast }=0$; injectivity of dualization gives $\beta \alpha =0$. ThusSimilarly, $\alpha \beta =0$ is equivalent to ${\beta }^{\ast }{\alpha }^{\ast }=0$, givingEquivalently, the reverse implications follow by taking double duals and using the canonical double-dual identification. The equalities are of subspaces under an isomorphism, not only inclusions.

The dual map is defined by ${\alpha }^{\ast }(\lambda )=\lambda \circ \alpha$. Suppose ${\alpha }^{\ast }=0$. Then $\lambda (\alpha u)=0$ for every $u\in U$ and every $\lambda \in V^{\ast }$. The dual space separates vectors: a nonzero vector can be included in a basis and a coordinate functional takes value one on it. Thus $\alpha u=0$ for every $u$, and $\alpha =0$. This proves injectivity of $\Phi$.

Its source and target have the same finite dimension:An injective linear map between equal finite dimensions is surjective by the rank-nullity theorem. Therefore $\Phi$ is an isomorphism. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.