With the stated angular-frequency convention, the inverse Fourier transform is $f(x)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }\tilde{f}(k)e^{ikx}dk$. Direct integration gives ${\tilde{f}}_0(k)={\int }_0^{\infty }{e}^{-(\alpha +ik)x}dx=1/(\alpha +ik)$. For $n\ge 1$, integration by parts has zero boundary terms and gives ${\tilde{f}}_n=n{\tilde{f}}_{n-1}/(\alpha +ik)$. ThereforeThis also follows from the Fourier transform of a polynomial multiple, since multiplication by $x$ corresponds to $i{\partial }_k$.

To prove Parseval's identity, substitute the inverse transform of $g$ into $\int f^{\ast }g$. Under the assumed integrability, Fubini's theorem permits interchange, and the inner integral is $\int f(x{)}^{\ast }{e}^{ikx}dx=\tilde{f}(k{)}^{\ast }$. HenceApply this with $f=g=f_n$. The spatial Gamma integral is ${\int }_0^{\infty }{x}^{2n}{e}^{-2\alpha x}dx=(2n)!/(2\alpha {)}^{2n+1}$. The transformed squared modulus is $(n!{)}^2/({\alpha }^2+k^2{)}^{n+1}$, giving the scaled even rational Parseval integral

Separation of variables in the wave equation with fixed endpoints selects $H_n(x)=\sin (n\pi x/L)$ and angular frequencies ${\omega }_n=cn\pi /L$, $n\ge 1$. The time factors may be chosen as $F_n(t)=\cos {\omega }_{n}t$, $G_n(t)=\sin {\omega }_{n}t$. By Fourier sine series orthogonality, the initial data giveHence $u={\sum }_{n\ge 1}[a_n\cos {\omega }_{n}t+b_n\sin {\omega }_{n}t]\sin (n\pi x/L)$. The factor $1/{\omega }_n$ in $b_n$ is required because differentiating the time sine supplies ${\omega }_n$.

The given displacement is a symmetric triangular tent: it is zero at both endpoints and has height $k$ at $L/2$. Integrating each linear half gives $a_n=8k\sin (n\pi /2)/(n^2{\pi }^2)$. Release from rest gives $b_n=0$, so the midpoint-plucked fixed-end string isOnly odd modes occur, by symmetry about the midpoint. The initial slope corner makes this a finite-energy weak solution; it is a classical solution away from the propagating corner lines.

In Cartesian coordinates, $\nabla \Phi =({\Phi }_x,{\Phi }_y,{\Phi }_z)$. For orthogonal coordinates, let $\mathbf{r}(q_1,q_2,q_3)$ be the position, $h_i=|\partial \mathbf{r}/\partial q_i|$ the scale factors of orthogonal coordinates, and ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$ the corresponding unit vectors. Since $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ and $d\Phi =\nabla \Phi \cdot d\mathbf{r}={\sum }_i{\Phi }_{q_i}d{q}_i$, comparison of independent increments gives the gradient in orthogonal coordinates:For the hyperbolic parametrization of oblate spheroidal coordinates, the three coordinate tangent vectors areTheir pairwise dot products are zero. Their squared Euclidean norms are ${\sinh }^{2}R+{\sin }^2\theta$, ${\sinh }^{2}R+{\sin }^2\theta$, and ${\cosh }^{2}R{\cos }^2\theta$, respectively. Thus the coordinate surfaces meet orthogonally wherever this is a regular coordinate chart, andThe last expression uses the usual latitude range $|\theta |<\pi /2$; without that range a scale factor is its absolute value. Axial or ring degeneracies are excluded from a regular chart.

A group homomorphism preserves multiplication, $\varphi (ab)=\varphi (a)\varphi (b)$. A group isomorphism is a bijective homomorphism. An isomorphism from a group to itself is a group automorphism.

The automorphism group contains the identity map. Composition is associative, and if $\varphi ,\psi$ are automorphisms, $(\varphi \circ \psi )(ab)=\varphi (\psi (a)\psi (b))=(\varphi \circ \psi )(a)(\varphi \circ \psi )(b)$; composition is also bijective. For $u=\varphi (a)$ and $v=\varphi (b)$, ${\varphi }^{-1}(uv)={\varphi }^{-1}(\varphi (ab))=ab={\varphi }^{-1}(u){\varphi }^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.

A finite cyclic group is $G=⟨g⟩=\{e,g,\dots ,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the roots of unity, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.

For prime $n$, Lagrange's theorem leaves only subgroup orders one and $n$: there is no nontrivial proper subgroup. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of groups with no proper nontrivial subgroups makes the distinction explicit.

For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},H_1=⟨g^p⟩}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The subgroups and quotients of a cyclic group are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.

A left coset is $gH=\{gh:h\in H\}$. The cosets partition a finite group into equally sized sets of $|H|$ elements, so Lagrange's theorem gives $|G|=[G:H]|H|$; in particular subgroup order divides group order.

Here $A=-I$, $C=-B$ and $B^2=I$. The Cayley table, with row element multiplied on the left of the column element, is

The table proves closure, $I$ is the identity and every element is its own inverse. Associativity is inherited from matrix multiplication, so this is the Klein four-group.

Finally, if every nonidentity element has order two, every element satisfies $h^{-1}=h$, including the identity. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian.

The kinetic energy is $K=\frac{1}{2}m(3{\dot{x}}^2+3{\dot{y}}^2+4{\dot{z}}^2)$. Differentiating the given potential and applying Newton's second law givesThese are the normal modes of a symmetric three-mass string, with mass matrix $M=m\mathrm{diag}(3,3,4)$ and stiffness matrix $m{\Omega }^2\left(\begin{array}{ccc}2& 0& -1\\ 0& 2& -1\\ -1& -1& 2\end{array}\right)$. The coordinate order here is $(x,y,z)$, although the spatial order along the string is $x,z,y$.

Reflection symmetry splits the generalized eigenvalue problem into an antisymmetric mode $x=-y$, $z=0$, with ${\omega }^2=2{\Omega }^2/3$, and symmetric modes $x=y=q$. Writing $\mu ={\omega }^2/{\Omega }^2$, these obey $z=(2-3\mu )q$ and $(2-4\mu )z=2q$. Thus $(2-4\mu )(2-3\mu )-2=2(6\mu -1)(\mu -1)=0$. The three results areThe displacement vectors are mutually orthogonal under the mass-weighted inner product $u^{T}Mv$, as required for these conservative normal modes.

The one-sided Laplace transform is $\bar{u}(p)={\int }_0^{\infty }{e}^{-pt}u(t)dt$ for sufficiently large $\mathrm{Re}p$. Integration by parts, with the exponential suppressing the upper boundary, givesDifferentiation under the integral gives $d\bar{u}/dp=-{\int }_0^{\infty }t{e}^{-pt}u(t)dt$, proving $\overline{tu}=-d\bar{u}/dp$. Start with $\overline{e^{-t}}=1/(p+1)$ and apply this rule repeatedly:With all three initial data zero, transforming the forced equation gives $(p+1{)}^3\bar{y}=1/(p+1)$. Thus $\bar{y}=(p+1{)}^{-4}$ and the listed inverse transform givesThe cubic factor is also the resonant exponential particular solution associated with the triple characteristic root $-1$.

A pole of order $N$ at $z_0$ has $(z-z_0{)}^{N}f(z)$ extending holomorphically to a nonzero value there. Its residue is the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series. The Cauchy integral theorem gives ${\oint }_{C}fdz=0$ for a closed contour in a simply connected domain of holomorphicity. The residue theorem replaces zero by $2\pi i$ times the enclosed residues for a positively oriented simple contour avoiding finitely many poles.

For the first integral, use the upper-half-plane semicircle, with a small clockwise upper semicircle excluding the origin. Near zero,There are no poles inside the indented contour. The large arc vanishes because $a,b>0$ and the integrand there is $O(R^{-2})$; the small arc tends to $-i\pi i(a-b)=\pi (a-b)$. Therefore the real-axis complex principal value is $\pi (b-a)$. Taking real parts gives the ordinary convergent cosine-difference inverse-square integralThe real numerator cancels to second order at zero, so this final real integral itself needs no principal-value convention.

For $g(z)=e^{az}/\cosh \pi z$, the simple poles are $z_n=i(n+1/2)$, $n\in \mathbb{Z}$. Differentiating the denominator givesIntegrate around the rectangle with vertices $-R,R,R+i,-R+i$. It contains only $i/2$. On the top side $g(x+i)=-e^{ia}g(x)$, and reversed traversal therefore contributes $+e^{ia}$ times the bottom integral. The vertical sides vanish for $|a|<\pi$. The residue theorem gives $(1+e^{ia})J=2e^{ia/2}$, so the exponential moment of the hyperbolic secant is

Compute the complex derivative along real increments to get $f^{\prime }=u_x+i{v}_x$, and along imaginary increments to get $f^{\prime }=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equations, $\boxed{u_x=v_y,u_y=-v_x}$.

For the specified real part, observe that $iz{e}^{-z}=e^{-x}[(x\sin y-y\cos y)+i(x\cos y+y\sin y)]$. Thus the most general holomorphic function with this real part on the connected plane isThe difference of two such holomorphic functions has zero real part, so the Cauchy-Riemann equations make its imaginary part constant.

Next, $\cosh z=\cosh x\cos y+i\sinh x\sin y$. Its real and imaginary parts satisfy $u_x=\sinh x\cos y=v_y$ and $u_y=-\cosh x\sin y=-v_x$. Their partial derivatives are continuous everywhere, proving that the complex hyperbolic cosine is entire.

For the Laurent series, set $h=z-1$. Since $z^2-1=h(2+h)$,The regular part at a simple pole therefore gives

Set $\varphi =R(r)\Theta (\theta )$ in Laplace's equation. Separation gives ${\Theta }^{\prime \prime }+n^2\Theta =0$ and $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$. Single-valued angular periodicity requires integer $n\ge 0$. For $n>0$ the radial factors are $r^n,r^{-n}$; for $n=0$ they are $1,\ln r$. A term linear in $\theta$ is excluded by periodicity. Thus the harmonic Fourier expansions in planar concentric domains have the general formFor $0<a<b$, the specified boundaries contain only a constant mode and a first cosine mode. The two-boundary annular harmonic interpolation givesAt $r=a$, the cosine term vanishes and the logarithmic term is $U$; at $r=b$, the first term vanishes and the cosine amplitude is $V$. Each component is harmonic, and the maximum principle ensures uniqueness for these continuous Dirichlet boundary conditions.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

For a bounded region $R$ with piecewise smooth boundary and a continuously differentiable vector field, the divergence theorem gives ${\int }_R\nabla \cdot \mathbf{F}dV={\int }_{\partial R}\mathbf{F}\cdot \mathbf{n}dS$, with outward unit normal $\mathbf{n}$. Apply it to $\mathbf{F}=u\nabla v-v\nabla u$. The cross terms cancel, giving the Green's second identityThe $m^{2}uv$ terms cancel identically. Sending $R\to \infty$ proves the whole-space identity provided the boundary integral tends to zero. The boundary term at infinity in Green's second identity requires adequate derivative decay as well as field decay; mere pointwise convergence $u,v\to 0$ does not suffice. The screened kernels used below satisfy the required decay.

Take $m>0$. Apply the identity with $u=\varphi$ and $v=G(\cdot ,\mathbf{r})$, treating the source through its Dirac delta function, or excising a small source sphere and taking its radius to zero. It gives $\varphi (\mathbf{r})-\int G({\mathbf{r}}^{\prime },\mathbf{r})\rho ({\mathbf{r}}^{\prime })d{V}^{\prime }=0$. Symmetry of the radial Green function then yields the requested representation.

For separation $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|>0$, the modified Helmholtz equation reduces to $(sG{)}^{\prime \prime }-m^2(sG)=0$. Decay selects $G=C{e}^{-ms}/s$. Integrating the equation over a small sphere gives $4\pi s^2{G}^{\prime }\to 1$, so $C=-1/(4\pi )$. ThusFor $\rho =e^{-\beta r}/r$ with $\beta >0$, put $h=r\varphi$. Then $h^{\prime \prime }-m^{2}h=e^{-\beta r}$. When $\beta \ne m$, decay removes $e^{mr}$, while $h(0)=0$ removes an additional point source at the origin. HenceSince $G$ has a minus sign, the positive Yukawa kernel convolution isThe coincident-rate limit $\beta =m$ is $e^{-mr}/(2m)$ for the convolution, and the origin limit for general rates is $1/(m+\beta )$. Both are finite; the apparent denominators in the distinct-rate formula are removable limits.

Introduce a Lagrange multiplier for the normalization and vary $J[y]={\int }_0^1[p(y^{\prime }{)}^2+q{y}^2-\lambda w{y}^2]dx$. With variation $\eta$ satisfying $\eta (1)=0$, integration by parts givesThe upper endpoint term vanishes, and the lower term vanishes because $p(0)=0$ with the stipulated regular differentiable behavior. More generally the natural endpoint condition is $p{y}^{\prime }\to 0$ there. The fundamental lemma of the calculus of variations gives $(p{y}^{\prime }{)}^{\prime }-qy+\lambda wy=0$, the stated Sturm-Liouville problem.

The Rayleigh-Ritz variational principle estimates the lowest eigenvalue by minimizing $R[y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ over an admissible finite-dimensional trial family. Restricting the family can only raise the minimum.

For the specific radial problem, $p=w=x$ and $q=0$. The trial $y=A(1-x^2)$ has ${\int }_0^{1}x{y}^{2}dx=A^2/6$. Normalization gives $\boxed{A=\sqrt{6}}$, with the negative sign an equivalent choice. The numerator is ${\int }_0^{1}x(2Ax{)}^{2}dx=A^2$, soThis is the quadratic case of the power-law variational bound for a circular membrane. The exact first eigenvalue $5.7815$ is smaller because the polynomial trial is not the exact Bessel function of the first kind eigenfunction. Expanding a trial in the exact orthonormal eigenbasis expresses its Rayleigh quotient as a weighted average of eigenvalues, so no admissible trial can lie below the lowest one.

Take $x\ge 0$ and a source position $\xi >0$. Before the source, the zero initial data require $G=0$. After it, set $G=e^{-2(x-\xi )}h(x-\xi )$; the homogeneous equation becomes $h^{\prime \prime }+a^{2}h=0$. Continuity of $G$ and a unit jump in $G_x$ at $x=\xi$ give $h(0)=0$, $h^{\prime }(0)=1$. Thus the causal Green function of a damped oscillator isHere $H$ is the Heaviside step function. For $a=0$, take the continuous limit $\sin (as)/a\to s$, giving $G=H(x-\xi )(x-\xi )e^{-2(x-\xi )}$. The formula is also unchanged when a nonzero $a$ changes sign. Its continuity removes a delta derivative, and its unit first-derivative jump supplies the required Dirac delta function.

For the forced equation use $a=2$. The zero-initial-data Green function representation isThe sine-product identity reduces this to elementary exponential-trigonometric integrals. Evaluating givesFor an independent check, the steady particular part is $\sin 2x/20-\cos 2x/10$, while the decaying homogeneous part has coefficients $1/10$ and $1/20$ that impose $y(0)=y^{\prime }(0)=0$.

Define the weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}w(x)\overline{u(x)}v(x)dx$. For the stated operator $L=-(pD{)}^{\prime }-q$, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe endpoint terms vanish for Dirichlet boundary conditions. Taking $u=v$ proves each eigenvalue is real, since $w>0$ makes the squared weighted norm positive. For eigenfunctions $u_m,u_n$ of distinct eigenvalues, the same identity gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n{⟩}_w=0$, proving their orthogonality.

For the specific equation, divide by $x^2$ to obtain $-y^{\prime \prime }=\lambda x^{-2}y$, so its weight is $w=x^{-2}$, with $p=1,q=0$. Put $v=\ln x$, $\ell =\ln 2$ and $Y(v)=y(e^v)$. The logarithmic change of variable in a Sturm-Liouville problem gives $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Remove the first derivative with $Y=e^{v/2}Z$; the equation becomes $Z^{\prime \prime }+(\lambda -1/4)Z=0$, with $Z(0)=Z(\ell )=0$.

The Dirichlet spectrum with inverse-square weight is thereforeIndeed ${\int }_1^2{x}^{-2}{y}_m{y}_{n}dx=(2/\ell ){\int }_0^{\ell }\sin (m\pi v/\ell )\sin (n\pi v/\ell )dv={\delta }_{mn}$. Values $\lambda \le 1/4$ give a linear or hyperbolic transformed solution and no nonzero solution vanishing at both ends.

For $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, a singular point $x_0$ is a regular singular point if $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. The Bessel differential equation has $xP=1$ and $x^{2}Q=x^2-{\nu }^2$, so zero is regular singular.

Use the Frobenius method with $y=x^s{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$. Matching coefficients gives $(s^2-{\nu }^2)a_0=0$ and $[(s+n{)}^2-{\nu }^2]a_n+a_{n-2}=0$, with missing negative-index coefficients zero. The regular root is $s=\nu$. Then $a_1=0$, all odd coefficients vanish, and the Frobenius series for Bessel functions has recurrenceFor noninteger $\nu$, the root $s=-\nu$ supplies an independent even-power series $y_2=x^{-\nu }{\sum }_{m\ge 0}{b}_{2m}{x}^{2m}$ with $b_{2m}=-b_{2m-2}/[4m(m-\nu )]$ and $b_0\ne 0$. None of these denominators vanishes. With standard normalization this is $J_{-\nu }(x)$, independent of the Bessel function of the first kind $J_{\nu }(x)$. The real powers here are understood for $x>0$.

For integer $\nu \ge 0$ and the stated normalization $a_0=1/(2^{\nu }\nu !)$, iteration givesThe ratio of consecutive term magnitudes is $|x{|}^2/[4(m+1)(m+\nu +1)]\to 0$. Thus the series converges absolutely for every finite complex $x$, defining an entire function for integer $\nu$. At integer order, $J_{-\nu }=(-1{)}^{\nu }{J}_{\nu }$; an independent solution is instead the Bessel function of the second kind.

A unitary matrix satisfies $U^{†}U=I$ over the complex numbers; a real orthogonal matrix satisfies $Q^{T}Q=I$. Their inverses are their conjugate transposes and transposes, respectively.

With $d=-a$, the coordinate $e_2=(0,1,0)$ has eigenvalue $c$. In the $x,z$ plane, the traceless symmetric two-by-two matrix has eigenvalues $\pm s$, where $s=\sqrt{a^2+b^2}$. If $s>0$, choose $\theta$ with $\cos 2\theta =a/s$ and $\sin 2\theta =b/s$. Then an orthonormal eigenbasis isIf $c$ equals $s$ or $-s$, the corresponding eigenspace is the span of the two listed vectors with that eigenvalue. If $s=0$, then $a=b=0$, the $x,z$ plane is the zero eigenspace, and $e_2$ retains eigenvalue $c$; if also $c=0$, every vector belongs to the zero eigenspace. This includes the degenerate cases that a division by $b$ would miss.

For the Cauchy-Schwarz inequality, take $b\ne 0$ and remove the orthogonal projection of $a$ along $b$. Positivity of the squared norm givesMultiplication by $\Vert b{\Vert }^2$ and square roots prove $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$; the case $b=0$ is immediate. Equality means linear dependence. ThereforeTaking the nonnegative square root proves the triangle inequality.