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The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes $3k\psi (0)$. The jump condition for an impulse is thereforeContinuity at zero gives $C_{+}=C_{-}=C$. The two one-sided derivatives are $-3kC/2$ and $+3kC/2$, whose difference is exactly $-3kC$. Hence the matching condition introduces no further restriction on $C$, and the globally bounded solution isThis continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.

Away from zero the Dirac delta distribution is absent. On either half-line, use $u=1+k|z|$ and the power-law ansatz $\psi =u^{\alpha }$. The Euler-Cauchy equation becomesThe growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are

For real nonintegral $\alpha$, the Parseval identity and the coefficient formula giveSolving for the sum yieldsAt $\alpha =5/2$, the sine square is one and the cotangent is zero, giving $2{\pi }^2/25$.

Evaluate the absolutely convergent complex Fourier series at $x=\pi$. The factors $(-1{)}^n{e}^{in\pi }$ equal one, and the $n=0$ term is $1/{\alpha }^2$. Pairing positive and negative integers givesSince $\sin (\pi \alpha )\ne 0$ for nonintegral complex $\alpha$, rearranging provesThe last term means $\pi \cot (\pi \alpha )/(2\alpha )$, so half-integer arguments are included by the finite cotangent value zero.

Integrate the two exponentials in $\cos (\alpha x)$ against $e^{-inx}$. For $\alpha \notin \mathbb{Z}$ there are no zero denominators, andThus the complex Fourier series isThis is the periodic extension, not a globally nonperiodic cosine formula. For each fixed complex $\alpha$, the coefficients are $O(n^{-2})$, giving absolute and uniform convergence. The extension is continuous at the endpoints because the two endpoint values agree; piecewise smooth pointwise convergence of a piecewise smooth Fourier series identifies the sum with it, including those endpoints.

The Parseval identity for a complex Fourier series isIt holds for every square-integrable function, with complex conjugation expressed by the absolute squares. It is the Pythagorean theorem extended to the complete orthonormal system of complex exponentials.

The complex Fourier series of a $2L$-periodic function isThe coefficient formula follows from orthogonality of the exponential functions. For an arbitrary square-integrable function, this representation means convergence in the $L^2$ norm. With appropriate additional regularity it also gives pointwise convergence, to the mean of the one-sided limits at a jump. No pointwise representation is asserted for an unrestricted function merely from periodicity.

In $\mathbb{R}$, takeThe sets are nonempty, closed and strictly descending, but their intersection is empty. Their diameters are infinite, so the shrinking-diameter assumption is essential.

In $\mathbb{R}$, takeThese are nonempty descending sets of diameter $1/n$, with empty intersection. They are not closed sets, so there is no contradiction with the Cantor intersection theorem for the complete metric space $\mathbb{R}$.

Take the incomplete metric space $X=(0,1)$ with its usual distance, andEach is nonempty and closed relative to $X$: it is $X\cap (-\infty ,1/(n+1)]$. The sequence is strictly descending and its diameters are $1/(n+1)\to 0$. Its intersection is empty, since every positive point exceeds $1/(n+1)$ for sufficiently large $n$. The excluded limiting point is zero.

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.