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Use the Poincaré-Lindstedt method. Introduce $\theta =\Omega t$ and $x=x_0(\theta )+ϵx_1(\theta )+\cdots$. The leading equation gives ${\omega }_0=3$ and $x_0=A_0\cos \theta$, after fixing the arbitrary phase. At order $ϵ$,Here $4{\cos }^2\theta \sin \theta =\sin \theta +\sin 3\theta$. Periodicity requires the two resonant coefficients to vanish. With $A_0>0$, this givesTherefore the leading limit cycle is $x(t)=\cos (3t)+O(ϵ)$, with frequency $3+O({ϵ}^2)$. For example the nonresonant first correction contains $-ϵ\sin (9t)/24$; its homogeneous sine/cosine pieces depend on phase and amplitude conventions. The absence of a first-order frequency shift is a solvability condition, not an assumption.

The native paper prints $n\ge 1$ while specifying $x_0$ alone. This does not determine $x_1$. We use the intended convention $x_{n+1}=\sin x_n$ for $n\ge 0$; even with an independently chosen real $x_1$, the same proof gives the requested convergence.

For the sine fixed-point iteration, after one step $x_1\in [-1,1]$. On this interval, the sine has the same sign as its argument, andFor example, $t-\sin t={\int }_0^t(1-\cos s)ds>0$, since the integrand is positive for $0<s\le 1$. By oddness of the sine, $a_n=|x_n|$ consequently satisfies $a_{n+1}=\sin a_n\le a_n$ for $n\ge 1$. This monotone sequence is bounded below by zero, so it has a limit $a\in [0,1]$. The continuity of the sine gives $a=\sin a$, and the strict inequality above forces $a=0$.

Hence $x_n\to 0$, and $0$ is a fixed point of the sine. It is the only real fixed point: every such point has absolute value at most one, where the preceding strict inequality applies. ThusThe Banach fixed-point theorem does not directly apply to the sine on $[-1,1]$: the ratios $|\sin t-\sin 0|/|t-0|$ tend to one as $t\to 0$, so no uniform contraction mapping constant smaller than one exists there.

Take eastward $x$, upward $z$, positive wavenumbers $k,K$, and a real instantaneous amplitude $A$. A common growing factor does not affect the phase geometry. Put $C=\cosh (Kz)$ and $S=\sinh (Kz)$, using the hyperbolic cosine and hyperbolic sine. The prescribed streamfunction iswhereFollowing a fixed phase givesTherefore the streamfunction pattern tilts westward with height. The lower boundary leads the upper boundary eastward.

In the conventional quasi-geostrophic streamfunction normalization, hydrostatic balance gives $b^{\prime }=f_0{\psi }_z^{\prime }$. ThereforeFor the phase calculation, multiplication by the constant $f_0$ is immaterial. Define a continuous phase ${\delta }_b=atan2(\alpha C,S)$; this avoids a spurious discontinuity at $z=0$. ThenThe buoyancy perturbation pattern tilts eastward with height, opposite to the streamfunction pattern. At $z=0$, ${\psi }^{\prime }=A\cos (kx)$ while $b^{\prime }=-f_{0}AK\alpha \sin (kx)$, so the two fields are a quarter wavelength apart there.

Both perturbation amplitudes are largest at the boundaries. Indeed,which increase with $|z|$. The physical reason is particularly clear through stratified quasi-geostrophic inversion: the Eady model has no interior background potential-vorticity gradient, and a growing normal mode has zero interior potential vorticity anomaly. The independent thermal anomalies reside at the rigid upper and lower boundaries, where meridional motion advects the basic horizontal buoyancy gradient. Each Eady edge wave induces an interior balanced flow that decays away from its boundary on scale $K^{-1}$. Their interacting, phase-locked combination produces the growing baroclinic instability. Boundary confinement of the thermal sources, rather than a rigid-wall requirement ${\psi }^{\prime }=0$, explains the amplitude pattern.

For a sketch with a quarter-wavelength difference between the boundary streamfunction patterns, chooseAn upper positive streamfunction crest is then ${\lambda }_x/4$ west of its lower counterpart, where ${\lambda }_x=2\pi /k$. The buoyancy perturbation crests slope the other way. The quarter-wavelength boundary assumption is used for the streamfunction; differentiating in $z$ fixes the buoyancy perturbation and does not generally give it the same boundary phase separation.

The connection to poleward heat transport can be checked without relying on the sketch. The meridional geostrophic flow and buoyancy perturbation areAverage over one zonal wavelength, using equal averages $1/2$ for squared sine and cosine and zero for their product:In the Northern Hemisphere this is positive: warm anomalies preferentially move northward and cold anomalies southward. For the same eastward-increasing basic thermal wind in the Southern Hemisphere, $f_0<0$ and the basic buoyancy gradient reverses; the negative northward eddy buoyancy flux is again poleward. For thermal expansion coefficient ${\alpha }_T>0$ and specific heat capacity at constant pressure $c_p$, the corresponding sensible heat flux in a thermal Boussinesq approximation is ${\rho }_0{c}_p\overline{v^{\prime }{T}^{\prime }}={\rho }_0{c}_p\overline{v^{\prime }{b}^{\prime }}/(g{\alpha }_T)$. The growing amplitude makes this flux grow as $A^2$. The tilt supplies the correlation absent in an untilted disturbance and reduces the background available potential energy.

DefineThis is a Lebesgue integrable function: on $(0,1]$ it is bounded by $x^{-1/2}$, whose integral is two, while on $[1,\infty )$ it is bounded by $e^{-x}$, whose integral is $e^{-1}$. Since $|\sin (xy)|\le x|y|$,Thus $h_y$ is Lebesgue integrable for every $y\ge 0$, including $h_0=0$; the same argument actually works for all real $y$.

The mean value theorem for the sine gives $|\sin u-\sin v|\le |u-v|$. Henceand the triangle inequality for an integral givesThus $F$ is a globally Lipschitz function, in particular continuous on $[0,\infty )$, including the endpoint zero.

For the derivative, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the integrandFor every $x>0$, $D_t(x)\to H(x)\cos (xy)$ as $t\to 0$. The same mean value theorem gives the uniform-in-$t$ bound $|D_t(x)|\le H(x)$.

The dominated convergence theorem states that if measurable functions $u_n$ converge almost everywhere to $u$ and $|u_n|\le G$ for a single Lebesgue integrable function $G$, then $u$ is Lebesgue integrable and $\int |u_n-u|\to 0$, hence $\int u_n\to \int u$. Apply it to the difference quotients along every sequence $t_n\to 0$, using the common dominator $H$. The resulting limit is independent of the sequence, soThe right-hand integral is absolutely convergent because its integrand has absolute value at most $H$. This is dominated differentiation under the integral sign, justified by an explicit integrable bound rather than a formal interchange.

The function is Lebesgue integrable. The inequalities $|\sin u|\le \min \{u,1\}$ for $u>0$ give bounds that handle the two ends of the $x$-axis differently. For $0<x<1$,For $x\ge 1$,Tonelli's theorem therefore provesThe factors of sine suppress the near-axis contributions that caused divergence in part i.

Every finite sum of the sine modes is odd, and so is a limit in the L2 norm of such sums. In contrast, $x^2-1$ is a nonzero even function. Its inner product with every odd sine mode vanishes, so all its proposed Fourier sine series coefficients would be zero. It cannot be represented by this basis on the symmetric interval. An even Fourier cosine series is the appropriate alternative there.

The nonnegative modulus $r=|z|$ must be a zero of the sine function. The restriction $r<10$ permits $r=0,\pi ,2\pi ,3\pi$, since $3\pi <10<4\pi$. Thus the locus is the origin and three concentric circles, centred at the origin with radii $\pi$, $2\pi$ and $3\pi$. A circle of radius zero is just one point.

The complex sine identity isIt follows either from the exponential definition of sine or its addition formula, using $\sin (iy)=i\sinh y$ and $\cos (iy)=\cosh y$. ConsequentlyThe angles $2$ and $3$ in these functions are dimensionless; the real trigonometric argument is measured in radians.

Expand the sine of the complex number using hyperbolic functions:Its squared complex modulus isTaking nonnegative square roots proves $\boxed{|\sin z|\ge |\sin x|}$. Equality holds exactly when $y=0$.

Take $m,k,\gamma ,F_0>0$ and $\omega >0$, and write $D_{\omega }=k-m{\omega }^2$ and $B_{\omega }=m\gamma \omega$. Substitution of $z=A\cos \omega t$ into the forced harmonic oscillator equation givesEquating the independent sine and cosine coefficients yieldsThus the required phase matching for a displacement referenced driven oscillator isFor a positive amplitude $A$, one must also choose the correct quadrant: $\cos {\varphi }_0=D_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$ and $\sin {\varphi }_0=B_{\omega }/\sqrt{D_{\omega }^2+B_{\omega }^2}$. Squaring and adding the coefficient equations givesFor arbitrary forcing phase, the steady response is instead $z=A\cos (\omega t+{\varphi }_0-\delta )$, where $\delta$ has the sine and cosine just specified. A change of time origin can put this displacement into $A\cos \omega t$, but it also changes the stated forcing phase.

The source's requested parameter list for $A$ omits $\gamma$; in general it cannot be omitted. As a concrete counterexample, take $m=k=F_0=\omega =1$ and ${\varphi }_0=\pi /2$. The prescribed cosine response exists for every $\gamma >0$, but $A=1/\gamma$, so fixing all four listed parameters does not fix the amplitude. Likewise, with fixed ${\varphi }_0=0$, positive damping and nonzero $\omega$, the second coefficient equation prevents a nonzero pure-cosine response. The forcing phase is an essential part of the condition.

Write $\tan x=a_{1}x+a_3{x}^3+a_5{x}^5+O(x^7)$. Multiplying by the cosine Maclaurin series and comparing with the sine series givesCoefficient comparison yields $a_1=1$, $a_3-a_1/2=-1/6$ and $a_5-a_3/2+a_1/24=1/120$. HenceSubstitute $u=\tan x$ into $e^u=1+u+u^2/2+u^3/6+O(u^4)$. Keeping one extra order to make the term-counting explicit,Counting the constant term, the first three nonzero terms are $\boxed{1+x+x^2/2}$. If the phrase is taken to mean the first three nonconstant terms, they are $x+x^2/2+x^3/2$; the displayed expansion supplies both conventions. The cubic coefficient is $1/3+1/6=1/2$, combining the cubic term in $\tan x$ with the cubic exponential term.

Use $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$ with $\theta =\pi x/l$. The initial shape becomesThe sine functions are orthogonal on $[0,l]$, so this decomposition directly identifies the two nonzero spatial normal mode coefficients. Since the string is released from rest, every time-dependent sine coefficient vanishes and the subsequent motion isThus the signed mode amplitudes are $A_1=A$, $A_2=A/2$ and $A_n=0$ for $n\ge 3$; their magnitude amplitudes are $|A|$ and $|A|/2$. The factor of two in the second angular frequency is fixed by the same boundary conditions as its spatial wavelength.

The Maclaurin series of the exponential function and sine giveBoth the numerator and denominator first become nonzero at order $x^4$. Dividing by that common power gives the limit

Use the Galerkin method, rather than pointwise equality: products generate harmonics outside the retained five-dimensional space. Put $X=\alpha x$, $Z=\pi z$, and denote the five prefactors byThe retained streamfunction is $P\sin X\sin Z$, the retained temperature perturbation is $T\cos X\sin Z+C\sin 2Z$, and the retained magnetic-flux perturbation is $D\sin X\cos Z+E\sin 2X$. These satisfy all the stated boundary conditions. Let $J(f,g)=f_x{g}_z-f_z{g}_x$, the streamfunction advection bracket. Orthogonality of the sine and cosine modes gives the following coefficients of the retained harmonics:For example the $C$ contribution to the temperature bracket is $\alpha \pi PC[\cos X\sin 3Z-\cos X\sin Z]$. The magnetic bracket follows by writing $A=A_D+A_E$: $J(A,\Delta A)=({\beta }^2-4{\alpha }^2)J(A_D,A_E)$ and $J(A_D,A_E)=\alpha \pi DE(\sin 3X-\sin X)\sin Z$. These identities exhibit both the retained terms and discarded harmonics.

Before rescaling time, the projected coefficient equations areSubstitute the prefactors and divide time derivatives by ${\beta }^2$. With $\tau ={\beta }^{2}t$, $\varpi =4{\pi }^2/{\beta }^2$ and $q={\pi }^{2}Q/{\beta }^4$, the resulting five-mode vertical-field magnetoconvection equations areHere primes denote differentiation in $\tau$. The factor $3-\varpi$ is $(3{\alpha }^2-{\pi }^2)/{\beta }^2$, which checks the magnetic nonlinear sign. There is a terminology qualification in the source: on this thermal diffusion time scale, its $\zeta$ is the magnetic-to-thermal diffusivity ratio $\eta /\kappa$. The conventional magnetic Prandtl number is $\nu /\eta =\sigma /\zeta$, where $\sigma$ is the Prandtl number. The displayed equations and subsequent calculation use the source's $\zeta$ unchanged.

Put $h=1/(m+1)$ and impose homogeneous Dirichlet boundary conditions, as required by the stated continuum eigenvalues. Let $T=\mathrm{tridiag}(1,-2,1)$ be the unscaled one-dimensional Dirichlet discrete Laplacian. For $1\le k\le m$, the vectors $v_j^{(k)}=\sin (jk\pi h)$ satisfyThe identity follows by adding the two neighboring sines; the values at $j=0,m+1$ vanish. These $m$ mutually orthogonal vectors form the discrete sine transform basis. Consequently their tensor products give all $m^2$ two-dimensional modes, not merely a few candidate eigenvectors.

In a compatible grid ordering the axial operator is $A_5=h^{-2}(T\otimes I+I\otimes T)$. The nine-point operator isIndeed, the added Kronecker product supplies the diagonal neighbors with weight $1/6$, subtracts $1/3$ from each axial neighbor and adds $2/3$ to the central coefficient. Thus it has exactly the coefficients of the second discretization. On $v^{(k)}\otimes v^{(l)}$ the eigenvalues areandEquivalently, $h^2{\lambda }_{9,kl}=-10/3+4[\cos (k\pi h)+\cos (l\pi h)]/3+2\cos (k\pi h)\cos (l\pi h)/3$.

For fixed mode numbers $k,l$, expansion of the sines giveswhereas the mixed Kronecker product correction givesThus both eigenvalue approximations are second order for fixed modes. The leading nine-point error includes an additional positive term ${\pi }^4{h}^2{k}^2{l}^2/6$.

There is also an exact comparison, not just an asymptotic one. Since $\sin x<x$ for $0<x<\pi /2$, and both sine factors in the correction are positive,Both estimates lie on the less-negative side of the continuum Laplacian eigenvalue, so the five-point formula is the better eigenvalue estimate for every corresponding interior mode. The five-point versus nine-point Laplacian eigenvalue accuracy comparison does not contradict the useful isotropy or special Poisson accuracy properties of a nine-point stencil. Those properties concern a different error criterion. The fixed-mode expansion is not uniform for modes with $k$ or $l$ comparable to $h^{-1}$; wavelengths close to the mesh scale need not have small relative spectral error.

Use the conformal map of a vertical half-strip by sine followed by a Cayley transform:We verify that this maps the entire half-strip bijectively onto the unit disc. First,so the sine image lies in the upper half-plane. Also $\cos z\ne 0$ in the domain, making the map locally a conformal map. If $\sin z=\sin w$, the identity for a difference of sines gives either $z-w=2\pi n$ or $z+w=(2n+1)\pi$. The real-part bounds force $n=0$ in the first case, while positive imaginary parts exclude the second. Thus sine is injective.

For surjectivity, given any $w$ in the upper half-plane, solveNo root is on the unit circle, since $w=(t-t^{-1})/(2i)$ would then be real. The product of the two roots is $-1$, so exactly one root has $|t|<1$. For this root,which forces $\mathrm{Re}t>0$. Choose its logarithm with argument in $(-\pi /2,\pi /2)$ and put $z=-i\log t$. Then $\mathrm{Im}z=-\log |t|>0$, its real part lies in the required interval, and $\sin z=w$. This proves that sine maps onto the upper half-plane.

Finally, for $\mathrm{Im}w>0$, $|w-i|<|w+i|$, and the Möbius transformation $(w-i)/(w+i)$ is a bijection to the unit disc. Its inverse is $w=i(1+\zeta )/(1-\zeta )$. Both constituent derivatives are nonzero, completing the proof for the displayed conformal map.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

Let the photon momenta be ${\mathbf{p}}_1,{\mathbf{p}}_2$, with $|{\mathbf{p}}_i|=E_i/c$. Four-momentum conservation gives total energy $E_1+E_2$ and total momentum ${\mathbf{p}}_1+{\mathbf{p}}_2$ for the pion. Its mass shell relation therefore givesSince $0\le \theta \le \pi$, the sine is nonnegative, and taking the positive square root yields

On any compact set avoiding the integers, writeFor all sufficiently large $|n|$, uniformly on that compact set this is $O(|n{|}^{-2})$. The summands in the higher series are $O(|n{|}^{-k})$ for $k\ge 2$. The Weierstrass M-test therefore gives locally uniform convergence away from the integer poles and permits termwise differentiation. Near an integer $m$, isolate its singular summand; all the others converge uniformly on a small neighborhood. Thus the first function has a simple pole of residue one at each integer, and the $k$th function has a pole of order $k$ there with principal part $(z-m{)}^{-k}$. There are no other singularities, so all are meromorphic functions.

Pairing positive and negative indices is legitimate because the regularized first series converges absolutely. It becomesHere is a direct derivation of its cotangent partial-fraction expansion. Fix $z\notin \mathbb{Z}$ and integrate $F(\zeta )=\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with real and imaginary coordinates $\pm R$, $R=N+1/2$, taking $N$ large enough to enclose $\pm z$. On the vertical sides $|\cot \pi \zeta |\le 1$, and on the horizontal sides it is at most $\coth (\pi R)$. Thus the cotangent is uniformly bounded and $|{\zeta }^2-z^2|\ge R^2-|z{|}^2$, so the contour integral is $O(R^{-1})\to 0$.

The residues at integers are $1/(n^2-z^2)$; the two residues at $z$ and $-z$ sum to $\pi \cot (\pi z)/z$. The residue theorem and absolute convergence of the integer residue sum consequently giveUsing the exponential definitions of sine and cosine now yields, for $w=e^{2\pi iz}$,The same locally uniform convergence gives ${\epsilon }_k^{\prime }=-k{\epsilon }_{k+1}$, including $k=1$. Hence if ${\epsilon }_k(z)={\phi }_k(e^{2\pi iz})$, the chain rule givesThis is a rational function whenever ${\phi }_k$ is, proving the requested descent for every $k$ by induction, withIn particular periodicity is a consequence of the explicit descent, rather than an assumption about the summation order.

Finally, for $|z|<1$ expand each paired summand as a geometric series:On $|z|\le r<1$ the double series converges absolutely, since its absolute sum is at most $2r{\sum }_{n\ge 1}{n}^{-2}/(1-r^2)$. Interchanging the sums and using the Riemann zeta function gives the Laurent seriesThus the $z^{-1}$ coefficient is one; the positive odd coefficients are $-2\zeta (2j+2)$, and all other coefficients vanish.