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The set is the ideal quotient $K=(I:J)$. It contains zero. If $r,s\in K$, then $(r-s)J\subseteq I$, since $I$ is an subgroup. If $a\in R$ and $r\in K$, then $(ar)J=a(rJ)\subseteq I$ because $I$ is an ideal. Thus $K$ is an ideal.

Write $J=(a)$ and $K=(b)$. Because $I\subseteq J$, each $x\in I$ has the form $ar$. Then $rJ=r(aR)=xR\subseteq I$, so $r\in K$, and $x\in aK$. Conversely $aK\subseteq I$ by the definition of $K$. Thereforeso $I$ is a principal ideal. This argument uses only the commutative ring convention, not cancellation or an integral domain hypothesis.

With the usual commutative, unital conventions, the chain isA field is an integral domain, and every nonzero ideal contains an invertible element and is consequently the whole ring. Its only ideals are $(0)$ and $(1)$, so it is a principal ideal domain.

To prove the middle implication, first show that a principal ideal domain has the ascending chain condition on ideals. The union of an ascending chain is an ideal, hence is $(a)$; its generator belongs to one member of the chain, and all later members equal $(a)$. Now suppose a nonzero nonunit admitted no factorization into irreducible elements. It would factor into two nonunits, at least one of which again has no such factorization. Repeating would produce a strictly increasing chain of principal ideals: if $a=bc$ with $c$ a nonunit, $(a)⊊(b)$, since equality and cancellation would make $c$ invertible. This contradicts the ascending chain condition. Existence of factorization follows. By the permitted fact that every irreducible element is a prime element in a principal ideal domain, any irreducible factor in one factorization divides a factor in the other; cancellation then proves uniqueness up to units and order. This is a unique factorization domain. The final implication is part of the definition of a unique factorization domain.

None of the adjacent implications reverses. The principal ideal domain $\mathbb{Z}$ is not a field, since $2$ is not invertible. The polynomial ring $\mathbb{Q}[x,y]$ is a unique factorization domain, by applying the polynomial extension theorem for unique factorization domains twice, but its ideal $(x,y)$ is not principal: a generator would divide both $x$ and $y$, so would be a unit, whereas every member of this ideal vanishes at the origin. Finally, $\mathbb{Z}[\sqrt{-5}]$ is an integral domain as a subring of $\mathbb{C}$, but is not a unique factorization domain. Its multiplicative norm is $N(a+b\sqrt{-5})=a^2+5b^2$. There is no element of norm $2$ or $3$. Hence $2,3,1+\sqrt{-5},1-\sqrt{-5}$, of norms $4,9,6,6$, are all irreducible: a nontrivial factorization would require one of those missing norms. The only units have norm one and are $\pm 1$. The two factorizationsare therefore not related by associates and reordering.

Here $u=v=x-y$. The streamlines are $x-y=\text{constant}$, straight lines parallel to $(1,1)$. The entire line $x=y$ is stationary. On the $x-y>0$ side the motion is northeast, and on the $x-y<0$ side it is southwest. In orthonormal coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, the velocities are $\dot{p}=2q$, $\dot{q}=0$: this is simple shear flow. The right panel shows both moving families and the stationary line.

Here $u=-y$, $v=x$. The streamlines are $x^2+y^2=\text{constant}$. This is rigid rotation counterclockwise with angular velocity one; the origin is a stagnation point and all other trajectories are circles. The middle panel shows their orientation.

Here $u=x$, $v=-y$. The streamlines are $xy=\text{constant}$, rectangular hyperbolas, together with the coordinate axes as separatrices. Particles satisfy $x(t)=x(0)e^t$, $y(t)=y(0)e^{-t}$: the flow stretches along the $x$ axis and compresses along the $y$ axis, with a saddle stagnation point at the origin. The left panel shows the flow directions.

Every sine term vanishes at zero, so $G(0)=0$. The one-sided values of the periodic extension of $f^{\prime }$ are $f_{\mathrm{per}}^{\prime }(0-)=-1$ and $f_{\mathrm{per}}^{\prime }(0+)=1$. The Fourier series converges at this jump to their midpoint,This midpoint need not equal the arbitrarily assigned value of the periodic derivative at the endpoint.

The endpoint agreement $f(1)=f(0)=0$ makes the boundary terms vanish when integrating the Fourier coefficients of $f^{\prime }$ by parts. If $a_n^{\prime },b_n^{\prime }$ denote these coefficients, thenThese are exactly the coefficients obtained by differentiating the preceding Fourier series:On every compact subinterval of $(0,1)$ the differentiated series converges uniformly by the Dirichlet test, since partial sums of the sine factors are uniformly bounded there. The Termwise differentiation of a Fourier series is therefore valid away from integers and gives $1-2x$ on $(0,1)$. At an integer the periodic original has a corner, so the differentiated series is not asserting the existence of an ordinary derivative there.

Use the real Fourier series conventionThe constant coefficient is $a_0=2{\int }_0^{1}x(1-x)dx=1/3$. Reflection about $x=1/2$ changes the sine factor's sign and preserves the quadratic, giving $b_n=0$. Two integrations by parts, with $q=2\pi n$, giveConsequently the Fourier series of a periodically extended quadratic arch isThe periodic function is continuous, including at integers where its value is zero. Its piecewise smoothness gives pointwise Fourier series convergence everywhere, and the summable coefficient bound gives uniform and absolute convergence.

True. Let $g=f^{-1}$. For every open $U\subseteq X$,which is open in $Y$ because $f$ is an open map. Thus $g$ is continuous. Together with the given continuity and bijectivity of $f$, this establishes a homeomorphism. This is inverse continuity from an open bijection.

False. Give $X=\mathbb{R}$ the discrete topology and $Y=\mathbb{R}$ its usual topology. The identity bijection $f:X\to Y$ is continuous because every inverse image is a subset of the discrete space and hence is open. But $f^{-1}:Y\to X$ is not continuous: the open set $\{0\}$ of $X$ has inverse image $\{0\}$, not open in $Y$. Thus a continuous bijection need not be a homeomorphism.

False. The constant continuous function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=0$, sends the open set $\mathbb{R}$ to $\{0\}$, which is not open. Therefore continuity alone does not imply an open map.

For the near-origin probability in a maximal-angular-momentum hydrogen state, integrate the normalized radial probability density:Throughout this small interval the exponential is $1+O(r_p/a)$ for fixed $l$. HenceUsing the normalization above,The high power of the small radius reflects the near-origin behavior $R\propto r^l$ of the hydrogen atom state.

The normalized spherical harmonics separate the angular variables. The Radial Schrodinger equation for the hydrogen atom isWrite $R=A{r}^l{e}^{-\kappa r}$. ThenSubstitution separates a constant term and a $1/r$ term. The latter vanishes precisely when ${\hslash }^2\kappa (l+1)/m=e^2/(4\pi {ϵ}_0)$. With $\kappa =1/[a(l+1)]$ this determines the Bohr radiusThe constant term determinesThese are the maximal-angular-momentum hydrogen states, with principal quantum number $n=l+1$.

Because the angular integral is one, normalization requiresThereforeThe phase of $A$ remains arbitrary.

The Duhamel principle superposes the response to source impulses:This expression uses the source at the origin, is causal, and has zero initial data in the distributional sense. Its integral over the plane is ${\int }_0^{t}p(s)ds$, consistent with the injected total amount.

For the finite-duration source and $t>t_0$, set $b=(x^2+y^2)/(4D)$. The causal diffusion from a finite-duration planar point source becomesWith $\eta =b/\tau$, or by differentiating the supplied exponential integral, this isIf $b/(t-t_0)\ll 1$, both arguments are small. Their logarithmic constants and $\ln b$ cancel, leavingAt the origin this logarithmic expression is exact for $t>t_0$, obtained as the limit $b\to 0$. The leading spatial correction is $-p_{0}b{t}_0/[4\pi Dt(t-t_0)]$. If also $t\gg t_0$, the leading value is $p_0{t}_0/(4\pi Dt)$.

Take the Fourier transform in $x$. The transform of the source is $\delta (t-t^{\prime })e^{-ik{x}^{\prime }}$, so the Green function equation becomesChoose the causal Green function, zero for $t<t^{\prime }$. Away from $t^{\prime }$ the solution is a decaying exponential, and integrating across $t^{\prime }$ gives the jump $\tilde{G}(t^{\prime }+)-\tilde{G}(t^{\prime }-)=e^{-ik{x}^{\prime }}$. ThereforeThe phase shifts the kernel's centre to $x^{\prime }$. Applying the inverse Fourier transform and the Fourier transform of a Gaussian with $a=D(t-t^{\prime })$ givesFor positive elapsed time this heat kernel solves the homogeneous diffusion equation and has unit integral; as elapsed time tends to zero it approaches the spatial delta distribution, supplying the required impulse.

True, and the closedness assumption on $Y$ is unnecessary. The Minkowski sum has the representationEach translate of the open set $X$ is open, and an arbitrary union of open sets is open. If $Y$ is empty the union is empty, which is also open.

False. In $\mathbb{R}$, letEach is closed: its distinct points escape to infinity, so it has no finite accumulation point. Their Minkowski sum contains $1/n$, tending to zero. However zero is not in the sum. If $m-n+1/n=0$ with positive integers $m,n$, then $n-m=1/n$. For $n>1$ the right side is strictly between zero and one, while the left side is an integer; for $n=1$ the equation would require $m=0$. Thus the sum of two closed sets need not be closed. The same counterexample embeds along a coordinate axis in ${\mathbb{R}}^n$.

True. The same subsequence argument applies to the sum of a compact set and a closed set. For a convergent sequence $z_j=x_j+y_j$, boundedness of $X$ provides a convergent subsequence of $x_j$, with limit $x\in X$. Then $y_j\to z-x\in Y$ along that subsequence. Consequently $z\in X+Y$. No boundedness of $Y$ is needed.

True. Boundedness follows from $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$. To prove closedness of the Minkowski sum, let $z_j=x_j+y_j\in X+Y$ converge to $z$. By the Bolzano-Weierstrass theorem, the bounded sequence $x_j$ has a subsequence converging to $x$. Closedness gives $x\in X$. Along that subsequence, $y_j=z_j-x_j\to z-x$, which belongs to $Y$ by closedness. Thus $z=x+(z-x)\in X+Y$. This proves that the sum is bounded and closed.

For the determinant of a complex block representation, putThe change-of-basis matrix $T$ is invertible, and direct block multiplication givesThus $T^{-1}MT$ is block diagonal. Using multiplicativity of the determinant and the preceding block formula,The identity follows from this constant change of basis for arbitrary complex $A,B$.