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A Fourier series represents a periodic function as a constant plus sine and cosine waves whose frequencies are integer multiples of its fundamental frequency. For period $2L$, use the conventionwhere the Fourier coefficients areThe coefficient formulas follow by multiplying by a basis sine or cosine and using their orthogonality over a full period. A Fourier series need not equal an arbitrary function at every point. The pointwise convergence of a piecewise smooth Fourier series theorem says that for a piecewise continuously differentiable function, the Fourier series converges at each point to the average of the two one-sided limits, and hence to $f(x)$ wherever $f$ is continuous.

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

At $x=\pi /2$, all odd-indexed cosine terms vanish, while $\cos (2k\pi /2)=(-1{)}^k$. Thus the Fourier series from part (c) givesDivide by $2(e^{\pi }-1)/\pi$ and rearrange. The hyperbolic sine identity $e^{\pi }-1=2e^{\pi /2}\sinh (\pi /2)$ yieldsThe evaluation is legitimate by the convergence established in part (c).

The product-to-sum formula immediately gives the Fourier series:In the convention $f(x)=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the only nonzero Fourier coefficients are $a_2=a_8=1/2$. This is a finite trigonometric polynomial, and the identity holds at both endpoints as well.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Mixed endpoint spectrum. For $y^{\prime \prime }+\lambda y=0$, integration against $y$ and integration by parts, using $y(0)=0,y^{\prime }(\pi /2)=0$, gives $\lambda \int y^2=\int (y^{\prime }{)}^2>0$ for every nonzero real eigenfunction. Thus $\lambda =k^2>0$ and the first condition makes $y=A\sin (kx)$. The second requires $\cos (k\pi /2)=0$, so $k=2n+1$, $n\ge 0$. Sincethe normalized eigenfunctions and eigenvalues areThe orthogonality also follows by applying integration by parts to two different eigenfunctions of the same self-adjoint differential operator.

Forced expansion. Let $k_n=2n+1$ and write $x=\sum f_n{u}_n$, $y=\sum c_n{u}_n$. The Fourier coefficients of the forcing arebecause the boundary cosine vanishes and $\sin (k_n\pi /2)=(-1{)}^n$. Applying $d^2/d{x}^2+\kappa$ to the eigenfunction expansion gives $(\kappa -k_n^2)c_n=f_n$. Since $\kappa$ is not an eigenvalue,The coefficients are $O(k_n^{-4})$, so the series and its first two differentiated series converge absolutely and uniformly on the interval. Each term satisfies the two endpoint conditions, and $y^{\prime \prime }+\kappa y$ equals the mixed-endpoint Fourier series of $x$. The forcing coefficients are $O(k_n^{-2})$, giving uniform convergence of that series as well. Uniqueness follows because the homogeneous problem has a nonzero solution only at one of the excluded eigenvalues.

Odd fourth-power sum. Set $\kappa =0$. Direct integration of $y^{\prime \prime }=x$, with the same endpoint conditions, givesAt $x=\pi /2$, the eigenfunction expansion instead gives $y(\pi /2)=-(4/\pi ){\sum }_{n\ge 0}{k}_n^{-4}$. Equating the two values yieldsAbsolute and uniform convergence justify evaluation at the endpoint.

An even function satisfying the spatial equation has $X=C\cos (kx)$. The Dirichlet boundary conditions require $\cos (kL)=0$, so the allowed wavenumbers areThere is no nonzero zero-wavenumber mode. With $T=e^{-\alpha t/2}S$, the time equation becomes $S^{\prime \prime }+{\omega }_n^{2}S=0$, where ${\omega }_n^2=c^2{k}_n^2-{\alpha }^2/4$. Thus superposition of the even normal modes givesThis uses completeness of the even Fourier series modes. Imaginary ${\omega }_n$ describe overdamping. At critical damping, ${\omega }_n=0$, the two displayed exponentials coincide: the complete time factor must instead be $e^{-\alpha t/2}(C_n+D_{n}t)$. This repeated-root exception is necessary for a genuinely general solution.

Let $d_n$ be the even Fourier coefficient of the initial displacement, with the normalization ${\int }_{-L}^L{\cos }^2(k_{n}x)dx=L$. Writing $s_n=\sin (n\pi /2)=(-1{)}^{(n-1)/2}$ givesZero initial velocity requires the real modal time factorOne consistent choice in the question's complex representation is thereforeFor real ${\omega }_n$, $B_n=\overline{A_n}$; the real-part notation otherwise makes the complex coefficients nonunique. For imaginary ${\omega }_n$, the same formula is interpreted using hyperbolic functions. Its limit at ${\omega }_n=0$ is $d_n{e}^{-\alpha t/2}(1+\alpha t/2)$.

For $l/L\to 0$ at fixed mode number,which can be substituted into $A_n,B_n$ above. The profile specified at $t=0$ is actually nonzero at $x=\pm L$. Thus for finite $l/L$ it cannot be the initial data of a solution continuous at the initial endpoint corners with fixed endpoints. The Fourier series reproduces it in the open interval, with zero endpoint values; it is an L2 space initial-data interpretation. The assumption $l\ll L$ makes this compatibility error exponentially small, rather than exactly zero.

Put $\Phi (r,\theta )=R(r)\Theta (\theta )$ for separation of variables. Multiplication by $r^2/(R\Theta )$ givesSingle-valuedness requires a $2\pi$-periodic angular function. The nonzero possibilities are $\lambda =n^2$ for integers $n\ge 0$. For $n\ge 1$, $\Theta$ is a combination of $\cos n\theta$ and $\sin n\theta$, while the radial Euler-Cauchy equation has solutions $r^n,r^{-n}$. At $n=0$, periodicity excludes an angular term proportional to $\theta$, and $(r{R}^{\prime }{)}^{\prime }=0$ gives $R=A_0+B_0\ln r$.

Superposing the separated harmonic functions yieldsTo obtain the general solution, expand any sufficiently regular harmonic function in a Fourier series on circles; each Fourier coefficient satisfies the corresponding radial equation. The coefficients must give the required convergence on compact annuli. Because no condition at zero or infinity is imposed in this part, neither radial branch is discarded.

Boundedness at the origin allows only $r^n$ for nonconstant interior modes; decay at infinity allows only $r^{-n}$ outside. Write the common surface potential in its Fourier series as ${\sum }_{n\ge 1}{v}_n(\theta )$, where $v_n=A_n\cos n\theta +B_n\sin n\theta$. Continuity then makes the corresponding interior and exterior factors $(r/R{)}^n$ and $(R/r{)}^n$.

Their outward radial derivatives at the surface are $n{v}_n/R$ inside and $-n{v}_n/R$ outside. Thus the jump is $-2n{v}_n/R$. Equating it to minus the prescribed surface-charge mode givesTherefore the potential of a charged cylindrical shell with angular Fourier data isEither branch gives the same value at $R$. The mean angular mode is zero: outside, neither a constant nor a logarithm can survive the condition at infinity; continuity then forces the interior constant to zero. Consequently $\Phi (0)=0$.

The absence of a mean term in the surface charge density is essential. A nonzero mean would require a logarithmic exterior potential and would be incompatible with decay at infinity in this two-dimensional reduction. The normalization follows the stated derivative jump, with no additional permittivity factor introduced.

For separation of variables, let $\Phi =S(\sigma )T(\tau )$. The Laplace equation givesA nonzero $2\pi$-periodic angular solution requires $\lambda =n^2$, $n=0,1,\dots$. For $n\ge 1$, $S$ is a linear combination of $\cos (n\sigma )$ and $\sin (n\sigma )$, while $T^{\prime \prime }-n^{2}T=0$ gives hyperbolic sine and cosine. For $n=0$, periodicity makes $S$ constant, and $T=A_0+B_0\tau$. Negative separation constants would give nonperiodic real exponentials in $\sigma$.

Superposition yieldsCompleteness follows by expanding a harmonic function in a Fourier series in the periodic coordinate: its Fourier coefficients must satisfy exactly these radial equations. As usual, the coefficients are restricted by the convergence required of the series on compact subregions. The bipolar coordinates domain imposes no boundary condition at $\tau =0$ or $\tau =\infty$, so both hyperbolic branches remain available.

Use separation of variables with $\Phi =M(\mu )N(\nu )$. A nonzero product givesThe angular factor must be a periodic function of period $2\pi$. If $\lambda <0$, its exponential solutions cannot be periodic unless they vanish. If $\lambda =0$, $N=c+d\nu$ forces $d=0$, while $M=A_0+B_0\mu$. If $\lambda >0$, periodicity requires $\sqrt{\lambda }=n\in \mathbb{N}$, giving angular factors $\cos (n\nu ),\sin (n\nu )$ and radial factors $\cosh (n\mu ),\sinh (n\mu )$.

Superposing these separated harmonic functions yieldsTo see completeness, expand an arbitrary smooth harmonic function in a Fourier series in $\nu$ on each ellipse. Substitution into the equation forces every Fourier coefficient to obey the corresponding radial equation above. The series is interpreted with the convergence needed for a harmonic function, for example locally on compact subannuli. No boundedness condition at infinity or extension across $\mu =0$ has been imposed, so there is no reason to discard either radial branch in these elliptic coordinates.

Write the boundary Fourier series as $f=A_0+{\sum }_{n\ge 1}(a_n\cos n\varphi +b_n\sin n\varphi )$. Since $R=2$, its coefficients obey $a_n=2^n{A}_n$ and $b_n=2^n{B}_n$. The mean isFor $n\ge 1$, the integral of $\cos n\varphi$ on the first semicircle is zero. On the second semicircle, use ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ and cosine orthogonality. This gives $a_2=1/4$ and every other $a_n=0$.

For the sine coefficients, every even $n$ gives zero on both semicircles. For odd $n$, product-to-sum formulas giveThus the requested radial-expansion coefficients are

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 $y\to \infty$.

Equal pitch need not mean equal frequency spectrum. For an ideal periodic sound wave of fundamental frequency $f$, a Fourier series has the formChanging relative amplitudes $A_n$ changes the waveform and generally its timbre without changing its fundamental period. Attack, decay and the time evolution of the frequency spectrum also matter for real musical notes.

The middle ear transfers vibration into the cochlea. The basilar membrane supports a frequency-selective travelling wave: higher-frequency responses occur nearer the base and lower-frequency responses nearer the apex. Hair cells in the organ of Corti transduce mechanical stimulation into signals transmitted through the auditory nerve. Different spectral and temporal patterns therefore produce different sensory-activity patterns.^[ref]

This supplies a physical basis for timbre discrimination. It does not make the cochlea a set of independent strings, nor does excitation position alone explain every perceptual distinction: the cochlear amplifier and temporal neural coding also contribute.

For a Bravais lattice with primitive translation vectors ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$, define the reciprocal lattice by vectors $\mathbf{G}$ satisfying $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct-lattice translation $\mathbf{R}$. A convenient basis iswith cyclic formulas for ${\mathbf{b}}_2,{\mathbf{b}}_3$. Then ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ and $\mathbf{G}=h{\mathbf{b}}_1+k{\mathbf{b}}_2+l{\mathbf{b}}_3$ for integers $h,k,l$. The reciprocal primitive volume is $(2\pi {)}^3/V_{\mathrm{cell}}$.

These vectors are the spatial Fourier series frequencies of a periodic lattice potential or density. Elastic coherent crystal diffraction occurs when the scattering wavevector transfer equals a reciprocal lattice vector, with intensities additionally controlled by the basis crystal lattice structure factor. This is the reciprocal-space version of Bragg's law. A lattice of points and its basis should not be confused: a basis can make allowed reciprocal-lattice diffraction intensities vanish.

The first Brillouin zone is the Wigner-Seitz cell around the reciprocal origin: it consists of points closer to $\mathbf{G}=0$ than to any other reciprocal point. Its faces lie on the perpendicular bisectorsTo see the condition, equate $|\mathbf{k}{|}^2$ and $|\mathbf{k}-\mathbf{G}{|}^2$. These are also free-electron Bragg-degeneracy planes, where a periodic potential can open a band gap. Higher zones classify regions beyond successive Bragg planes; the reduced-zone description folds them into the first zone while retaining a band label.

In one dimension with spacing $a$, reciprocal points occur at $2\pi j/a$, and the first zone is $[-\pi /a,\pi /a]$. A square lattice of spacing $a$ has reciprocal basis $2\pi \hat{\mathbf{x}}/a,2\pi \hat{\mathbf{y}}/a$, and a square first zone with edges at $k_x,k_y=\pm \pi /a$.

The reciprocal of a face-centered cubic lattice is a body-centered cubic lattice and vice versa, with the scale fixed by the primitive-vector definition. In a Bloch state, wavevectors differing by $\mathbf{G}$ describe equivalent crystal momenta modulo the lattice periodicity. A crystal with $N$ primitive cells has $N$ allowed wavevectors in one Brillouin zone, hence $N$ states in each band per spin. This state count links reciprocal geometry to band filling, metallic conduction and band insulating behaviour. Reciprocal space also labels phonons and makes conservation of crystal momentum modulo $\hslash \mathbf{G}$ transparent.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

Choose the Fourier series conventionFor a real potential energy, $V_{-1}=V_1^{\ast }$. The symbol $V(G_1)$ means the Fourier coefficient $V_1$, with units of energy, rather than the pointwise value of $V(x)$ at a position $G_1$.

Retain the two plane waves $|1⟩=L^{-1/2}{e}^{ikx}$ and $|2⟩=L^{-1/2}{e}^{i(k-G_1)x}$ in the nearly-free electron model. Their kinetic energies are ${ϵ}_1={\hslash }^2{k}^2/(2m)$ and ${ϵ}_2={\hslash }^2(k-G_1{)}^2/(2m)$. The matrix representation of a linear map for the Quantum Hamiltonian projected onto this two-dimensional basis isThus $\det (H-EI)=0$ givesSolving this quadratic equation yieldsChoosing the zero of potential energy so that its spatial average is $V_0=0$ reproduces the requested expression. If that average has not been removed, both energy bands are shifted by $V_0$; no band gap, group velocity or effective mass is changed by this constant shift.

The two kinetic energies coincide at $k=G_1/2$. Their coupling gives an avoided crossing with band gap $2|V_1|$. The result is exact within the projected two-wave Quantum Hamiltonian. It is an approximation to the full crystal problem, appropriate near this crossing for a weak periodic potential whose other Fourier coefficients do not cause strong mixing with remote plane waves. Small $|V_1|$ alone would not justify ignoring arbitrarily large other Fourier coefficients.

A Fourier series separates a periodic physical signal into harmonics. For example, a periodic voltage applied to a linear circuit can be treated one frequency at a time; the harmonic responses then superpose to give the output waveform.

Write $w=\Delta T$ for the pulse width and $\delta =w/T$ for its duty cycle, with $0<w<T$. The waveform is a train of positive rectangular pulses centred on $t=jT$, each with area one. It is even about $t=0$, so every sine coefficient vanishes. Cosine orthogonality givesand, for $n\ge 1$,This is the Fourier series of a periodic rectangular pulse:The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a jump discontinuity, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.

Counting the constant as the first term, the first three terms areso their sum isIf “first three” instead means the first three nonconstant harmonics, add $a_3\cos (6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.

The original sketches use the illustrative duty cycle $\delta =0.15$ and cover $-T\le t\le T$. The coefficient panel also displays the first spectral zero.

For $n\delta \ll 1$, expand the sinc function:Thus the low-index coefficients, including $a_0$, have nearly equal values, although their constant-series contribution still has the factor $1/2$. The first sinc zero occurs at $n\delta =1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta =1+\epsilon$ gives $a_n\simeq -(2/T)\epsilon$. This pulse-width scale of Fourier coefficients means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.