By the two Fourier transform rules in part (b),Consequently transforming the differential equation givesorThe inverse Fourier transform gives the reverse implication under the same regularity hypotheses. This expresses the Fourier symmetry of the quantum harmonic oscillator operator.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Set $L=d^2/d{t}^2-4$ and choose $L_{t}G(t,t^{\prime })=\delta (t-t^{\prime })$, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are $e^{\pm 2t}$. Decay at each end and continuity at $t=t^{\prime }$ therefore give $G=C{e}^{-2|t-t^{\prime }|}$. Integrating its equation across the source gives a unit jump in its first derivative:ThusThe solution of the boundary value problem is the convolutionThere is no extra homogeneous term: no nonzero combination of $e^{2t},e^{-2t}$ decays at both infinities. The forcing must of course permit the imposed decay; integrable $f$, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

The two homogeneous Dirichlet boundary conditions on $x$ force negative separation constants. The initial profile is an even function, so only the even sine-family eigenfunctions are needed:They obey $X_m(\pm 1)=0$ and ${\int }_{-1}^1{X}_m{X}_{n}dx={\delta }_{mn}$. Odd eigenfunctions have zero projection onto the even initial data. Hence the initial coefficients areThe desired eigenfunction expansion is thereforeAt $y=0$ this is the Fourier series of $1-x^2$ on the interval; its coefficients are absolutely summable. Each term satisfies the partial differential equation and side conditions. For $y>0$ the additional Gaussian decay in $m$ justifies termwise differentiation. Since $k_m^2+1>0$ and $e^{\cos y-1}\le 1$, the whole series decays uniformly as $y\to \infty$.

Use separation of variables, $u=X(x)Y(y)$, and divide by $XY$ where nonzero. A real separation constant $\kappa$ givesso $X^{\prime \prime }=\kappa X$ and $Y^{\prime }=(\kappa -1-\sin y)Y$. Normalizing the second factor to one at zero givesThe spatial ordinary differential equation has three types of real solution:Products $X{Y}_{\kappa }$ are the separated eigenfunctions. The negative, zero and positive separation constants give trigonometric, linear and hyperbolic function forms respectively. There is no discrete quantization until boundary conditions on $x$ are supplied. Since $\cos y-1$ is bounded, the envelope of $Y_{\kappa }$ decays for $\kappa <1$, grows for $\kappa >1$, and is bounded and periodic for $\kappa =1$. The scalar prefactor can also be derived by removing a variable scalar reaction rate by an integrating factor.

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

Use the normalized eigenfunctions with the stated weighted inner product. Orthogonality gives $G[\hat{y}]={\sum }_n|c_n{|}^2$. The corrected Sturm-Liouville problem and the vanishing boundary term giveFor a complex trial the numerator means $\int (p|{\hat{y}}^{\prime }{|}^2+q|\hat{y}{|}^2)dx$, agreeing with the given formula for real trials. ThereforeIt is a convex combination of the eigenvalues, with equality only for a trial proportional to the lowest eigenfunction. The Rayleigh-Ritz variational principle consequently estimates ${\lambda }_0$ by choosing a nonzero admissible trial family and minimizing this quotient over its parameters:The result is an upper bound. A more flexible trial space can lower it, while equality requires that the exact lowest eigenfunction belong to the trial space. One need not know the expansion coefficients to calculate the quotient directly from the trial function.

The original PDF has a typographical omission: its displayed equation has $\lambda w(x)$ rather than $\lambda w(x)y(x)$ on the right. Taken literally, it is an inhomogeneous boundary value problem, not the stated Sturm-Liouville problem. The quotient and the later eigenfunction expansion identify the intended equation, which the following derivation produces explicitly.

Choose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$ and a nonzero real admissible $y$. Then $G[y]>0$ because $w>0$. For an endpoint-zero variation $\eta$, integration by parts givesThe first variation of the Rayleigh quotient is thusBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiply this corrected eigenvalue equation by $y$ and integrate. The same boundary cancellation gives $F[y]=\lambda G[y]$, so the first variation vanishes. This proves the requested equivalence for the intended equation. Homogeneous Neumann boundary conditions, or a Dirichlet condition at one end and a Neumann condition at the other, also eliminate the boundary term. General Robin boundary conditions would require the corresponding boundary contribution in the numerator.

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

The azimuthal vector ${\mathbf{h}}_{\varphi }$ is automatically orthogonal to the other two. Their remaining dot product isThus orthogonal coordinates require $g{g}^{\prime }=\sinh \xi \cosh \xi$, which integrates to $g^2={\sinh }^2\xi +C$. The initial value gives $C=1$, and the smooth branch with $g(0)=1$ isThe corresponding coordinate-vector norms are $|{\mathbf{h}}_{\xi }|=|{\mathbf{h}}_{\theta }|=\sqrt{{\sinh }^2\xi +{\sin }^2\theta }$ and $|{\mathbf{h}}_{\varphi }|=\sinh \xi \sin \theta$, as in the hyperbolic parametrization of prolate spheroidal coordinates. Coordinate degeneracies on the axis and focal segment are understood.

Eliminating the angular coordinates on the requested surface givesIt is a prolate spheroid with long axis $z$, equatorial radius $\sinh 1$, polar semiaxis $\cosh 1$, and foci $(0,0,\pm 1)$.

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

A cyclic group is generated by one element: $G=⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. The hypothesis needs the qualification no proper nontrivial subgroup: every nontrivial group has the proper trivial subgroup $\{e\}$, so the literal wording admits no examples.

Under that intended hypothesis, choose $g\ne e$. The subgroup $⟨g⟩$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $⟨g^2⟩$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.

If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $⟨g^d⟩$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. ConsequentlyConversely, Lagrange's theorem shows that a group of prime order has only itself and its trivial subgroup. This proves the characterization of groups with no proper nontrivial subgroups, even without assuming finiteness at the outset.

Write $G=\{e,a,b,c\}$. By Lagrange's theorem, the order of a group element divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,so the group is Abelian. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, givingThe map $e\mapsto (0,0)$, $a\mapsto (1,0)$, $b\mapsto (0,1)$, $c\mapsto (1,1)$ preserves this table under componentwise addition modulo two. Thereforethe Klein four-group, also called the Vierergruppe.

For each $g$, the mapis surjective by definition of a coset and injective by left cancellation. Its inverse is multiplication by $g^{-1}$. ThereforeThe same argument with right multiplication proves the analogous result for right cosets. This equality of sizes is the counting step in Lagrange's theorem.

For $g\in G$, the left coset of $H$ represented by $g$ isIt is a subset of $G$ obtained by multiplying every element of $H$ on the left by the same group element. The identity representative gives $eH=H$.

Map each symmetry to the permutation it induces on the three vertices. This is a group homomorphism, because composing geometric motions composes their vertex permutations. In the labels chosen in (b), the map isAll six images are distinct and exhaust the symmetric group $S_3$. Equivalently, the kernel is trivial because a triangle symmetry fixing all three vertices is the identity. Thus the map is bijective and