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The characteristic polynomial of the homogeneous equation isso $y_h=C_0+(C_1+C_{2}x)e^{2x}$. For a particular integral put $y=e^{2x}v$. ThenTaking $v^{\prime \prime }=x/2-1/4$ gives $v=x^3/12-x^2/8$ after discarding homogeneous terms. Therefore

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the vector identity and contraction with $n$ leavesThe boundary integrand satisfies $(n\times v)\cdot dx=v\cdot (dx\times n)$, proving the formula. The vector $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin \theta e_{\theta }$. The projected divergence is $2\cos \theta$, so the left side isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

The given integral matrices form a group because products and inverses remain integral. The inverse formula shows that an integral inverse exists exactly when the determinant divides every cofactor; taking determinants shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.

Because the parameter enters through $\lambda wy$,Use this as $f$ in part (a). The necessary orthogonality condition isWith $y=\epsilon y_0+{\epsilon }^2{y}_1$ and $\mu =O(1)$,ConsequentlyThe normalization makes the second integral one. Divide by ${\epsilon }^{m+1}$ to obtainThis is the leading nonlinear eigenvalue shift in a Sturm-Liouville problem.

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrödinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

Forthe parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a ring.

It is not an integral domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial rings. Indeed, an idempotent in ${\mathbb{Z}}^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.

To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.

Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

For a partitionputThe upper and lower Darboux sums areThe upper and lower Darboux integrals areThe bounded function is Riemann integrable when these two numbers agree; their common value is its integral.

No. Enumerate the rationals in $[0,1]$ asand defineThe function is nonnegative and unbounded, while it is identically zero on $[0,1]\setminus A$. For any $r>0$, however, $f_r$ is positive on the dense set of rationals and zero on the dense set of irrationals. Every interval therefore has positive oscillation, and the upper and lower Darboux integrals differ. Thus even the truncations need not be Riemann integrable.

No. DefineIt is continuous at every point of $(0,1)$ and is unbounded. Its truncation is Riemann integrable, butThus it is not improperly integrable with a finite integral.

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

The first field isso its integral around $C$ is zero and it is conservative with potential $e^{x^2}\cos (yz)$.

Along $C$, $dr=(-\sin s,\cos s,0)ds$ andsoThus $B_2$ is not conservative and cannot be a gradient on the stated punctured region.

Finally,Its integral around $C$ is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectivelyand exactly $B_1,B_3$ are conservative gradient fields.

Comparing the prepared energy with the formula in part (ii),so $l+1=3$ and $l=2$. The radial probability density, apart from normalization, isUsing the given factorial integral, its mean is the ratioFor $l=2$ this gives the mean radius of a circular Coulomb bound state

Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are

For $0<a<2$, the beta-integral evaluation givesDifferentiation under the integral sign near $a=1$ is justified by domination at zero and infinity. ThereforeWriting $a=1+t$ givesSincethe logarithmic moments of the Cauchy kernel yieldThe same expansion gives $I^{\prime \prime }(1)={\pi }^3/8$, agreeing with the supplied identity.

In a Boolean ring, $(r+1{)}^2=r+1$. Using $r^2=r$ and $1^2=1$ givesso

If $R$ is also a nonzero integral domain, thenforces every $r$ to equal $0$ or $1$. Hence $R\cong {\mathbb{F}}_2$.

If $I$ is prime in a Boolean ring, then $R/I$ is a nonzero Boolean integral domain by part (i), and hence is ${\mathbb{F}}_2$. Since the quotient is a field,This is the prime ideals of a Boolean ring are maximal property.